11- Partial Differential Equations

11 Partial Differential Equations aaaaa 11.1 INTRODUCTION A relation between the variables (including the dependent on

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Partial Differential Equations aaaaa

11.1 INTRODUCTION A relation between the variables (including the dependent one) and the partial differential coefficients of the dependent variable with the two or more independent variables is called a partial differential equation (p.d.e.) For example:

x

∂u ∂u +y = u + xy ∂x ∂y

…(1)

∂2 u ∂2 u + =0 ∂ x2 ∂ y2

…(2)

3

2  ∂2u   ∂2u   2  +  2  = u ∂x  ∂y 

∂u = p, ∂x ∂2u = r, ∂x2 K

…(3)

∂u =q ∂y ∂2u = s, ∂x∂y K

…(4)    2  ∂u = t  ∂y2  etc K  as standard notations for partial differentiation coefficients. The order of a partial differential equation is the order of the highest order differential coefficient occuring in the equation and the degree of the partial differential equation is the degree of the highest order differential coefficient occurring in the equation. For example, equation (1) is of Ist order Ist degree, equation (2) is of 2nd order Ist degree whereas equation (3) is of 2nd order 3rd degree. If each term of the equation contains either the dependent variable or one of its derivatives, it is said to be homogeneous, otherwise, non-homogeneous. For example, equation (2) is homogeneous, whereas equation (1) is non-homogeneous. with

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The partial differential equation is said to be linear if the differential co-efficients occurring in it are of the Ist order only or in other word if in each of the term, the differential coefficients are not in square or higher powers or their product, otherwise, non-linear. e.g. x2p + y2q = z is a linear in z and of first order Further, a p.d.e. is said to be quasi-linear if degree of highest order derivative is one, no product of partial derivatives are present e.g. z – zxx + (zy)2 = 0 is a quasi-linear 2nd order. 11.2 FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS These equations are formed either by the elimination of arbitrary constants or by the elimination of the arbitrary functions from a relation with one dependent variable and the rest two or more independent variables. Observations: When p.d.e. formed by elimination of arbitrary constants 1. If the number of arbitrary constants are more than the number of independent variables in the given relations, the p.d.e. obtained by elimination will be of 2nd or higher order. 2. If the number of arbitrary constants equals the number of independent variables in the given relation, the p.d.e. obtained by elimination will be of order one. Observations: When p.d.e. formed by elimination of arbitrary functions. When n is the number of arbitrary functions, we may get several p.d.e., but out of which generally one with two least order is selected. e.g.

z = x f(y) + yg (x) involves two arbitrary functions, f and g. Here

∂4 z =0 ∂x2∂ y2

…(i)

and xys = xp + yq – z (second order) …(ii) are the two p.d.e. are obtained by elimination of the arbitrary functions. However, 2nd equation being in lower in order to 1st is the desired p.d.e. Example 1: Form a partial differential equation by eliminating a, b, c from the relation x2 y2 z2 + + =1 a2 b2 c2

[NIT Kurukshetra, 2003; KUK, 2000]

Solution: Clearly in the given equation a, b, c are three arbitrary constants and z is a dependent variable, depending on x and y. We can write the given relations as:

 x2 y2 z2  f (x, y, z) =  2 + 2 + 2 − 1 = 0 a  b c then differentiating (1) partially with respect to x and y respectively, we have ∂f ∂f ∂ z + ⋅ = 0, ∂x ∂z ∂x

∂y   = 0  Keeping  ∂x

and

∂f ∂f ∂ z + ⋅ = 0, ∂y ∂z ∂y

  Keeping

or

2x 2z ∂z + ⋅ =0 a2 c2 ∂x



…(1)

 ∂x = 0 ∂y 

c2x + a2zp = 0

…(2)

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2y 2z ∂z + =0 ⇒ c2y + b2zq = 0 b2 c2 ∂ y Again differentiating (2) with respect to x, we have

and

…(3)

2

∂z ∂2 z c2 + a2   + a2 z 2 = 0  ∂x  ∂x

On substituting −

2

z ∂z  ∂z  ∂2 z +  +z 2 =0   x ∂x ∂x ∂x

xz ⋅

or

c2 z ∂z =− from (2) in above equation, we get a2 x ∂x

2

∂2 z ∂z ∂z + x   − z =0 2   ∂x ∂x ∂x

…(4)

Similarly, differentiating (3) partially with respect to y and substituting the value of from (3) in the resultant equation, we have

c2 b2

2

 ∂z  ∂2 z ∂z yz 2 + y   − z =0 ∂y ∂y  ∂y 

…(5)

Thus equations (4) and (5) are ‘partial differential equations’ of first degree and second order. Example 2: Form partial differential equation from z = x f1(x + t) + f2(x + t). Solution: Clearly z is a function of x and t

p=

∂z = f1(x + t) + x f1′(x + t) + f2′(x + t) ∂x

q=

∂z = x f1′(x + t) + f2′(x + t) ∂t

r=

Now

∂2 z = f1′(x + t) + x f1′′(x + t) + f1′(x + t) + f2′′(x + t) ∂ x2 = 2 f1'(x + t) + x f1''(x + t) + f2''(x + t)

s=

∂ ∂z ∂2 z = = f1′ (x + t) + x f1′′ (x + t) + f2′′ (x + t) ∂ t ∂ x ∂ x∂ t

t=

∂2 z = x f1′′(x + t) + f2′′(x + t) ∂ t2

(r + t) = 2 f1'(x + t) + 2x f1''(x + t) + 2 f2''(x + t) = 2s

∂2 z ∂2 z ∂2 z + 2 −2 =0 2 ∂x ∂t ∂ x∂ t Example 3: Form the partial differential equation by eliminating the arbitrary function, or

F(x + y + z, x2 + y2 + z2) = 0

[KUK, 2004-05, 2003-04]

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Solution: Let where

F ((x + y + z), (x2 + y2 + z2 )) = 0 be F(u, v) = 0

…(1)

(x2

…(2)

u = (x + y + z) and v =

+

y2

+

z2)

Clearly F(u, v) = 0 is an implicit function.

0 = Fx =



∂F ∂u ∂F ∂v + ∂u ∂x ∂v ∂x

∂F ∂u ∂F + ∂u ∂y ∂v ∂ u ∂ u ∂ x ∂ u ∂y = + + ∂x ∂x ∂x ∂y ∂x 0 = Fy =

whereas

∂v ∂y

…(i)    …(ii)  

∂u ∂z = (1 + p) ∂z ∂x

…(3) …(4)

  ∂y ∂x  since ∂x = 0 = ∂y as x and y are two independent variables

and

Similarly,

∂u ∂u ∂x ∂u ∂y ∂u ∂z = + + = (1 + q) ∂y ∂x ∂y ∂y ∂y ∂z ∂y

…(5)

∂v = (2x + 2zp) ∂x

…(6)

∂v = (2y + 2zq) ∂y

…(7)

Thus, on substituting the values of

0=

∂u ∂u ∂v ∂v , , and in equation (3), we get ∂x ∂y ∂x ∂y

∂F ∂F (1 + p) + (2x + 2pz) ∂u ∂v

∂F ∂F 0= (1 + q) + (2y + 2qz) ∂u ∂u

…(i)    …(ii) 

…(8)

Eliminating ∂F and ∂F , we get ∂u ∂v

(1 + p) (2x + 2pz) =0 (1 + q) (2y + 2qz)

⇒ p(y – z) + q(z – x) = (x – y)

which is the desired p.d.e. Example 4: Form the partial differential equation (by eliminating the arbitrary function) from: F(xy + z2, x + y + z) = 0. [NIT Kurukshetra, 2007; KUK, 2002-03] Solution: Let F(xy + z2, x + y + z) = 0 be F(u, v) = 0 where u = xy + z2

…(1)

and

…(2)

v=x+y+z

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Clearly F(u, v) = 0 is an implicit relation, so that

∂F ∂u ∂F ∂v + …(i)  ∂u ∂x ∂v ∂x   ∂F ∂u ∂F ∂v  0 = Fy = + ∂u ∂y ∂v ∂y …(ii)  0 = Fx =

whereas

∂u ∂u ∂x ∂u ∂y ∂u ∂z  ∂u ∂u  p = (y + 2zp) = + + = + ∂x ∂x ∂x ∂y ∂x ∂z ∂x  ∂x ∂z 

∂u ∂u ∂x ∂u ∂y ∂u ∂z  ∂u ∂u  q = (x + 2zp) = + + = + ∂y ∂x ∂y ∂y ∂y ∂z ∂y  ∂y ∂z  ∂v = (1 + p) Similarly, ∂x ∂v = (1 + q) ∂y ∂u ∂u ∂v ∂v On substituting the values of ∂x , ∂y , ∂x , ∂y in equation (3), we get

and

∂F ∂F (y + 2pz) + (1 + p)   ∂u ∂v  ∂F ∂F 0= (x + 2qz) + (1 + q)   ∂u ∂v ∂F ∂F and , we get On eliminating ∂u ∂v

…(3)

…(4) …(5) …(6) …(7)

0=

…(8)

y + 2pz 1 + p =0 x + 2qz 1 + q ⇒

p(2z – x) – (2z – y)q = (x – y) the desired partial differentiation equation.

Example 5: Form partial differential equation from the relation 1 2 (ii) z = f1(x + iy) + f2(x – iy). (i) z = y + 2 f x + logy

(

Solution: (i)

)

1 z = y2 + 2 f  + log y x 

…(1)



∂z 1   1 = 2 f ′  + log y ⋅  − 2  x   x  ∂x

and

∂z 1 = 2y + 2 f ′  + log y x  ∂y

 1 ⋅   y

1  On eliminating of 2 f'  + log y , we get (3) as x  ∂z ∂z 1 = 2y +  −x2  ∂y ∂ x y 

…(2) …(3)

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yq = 2y2 – x2p; (ii) Given

Similarly

∂z ∂z when ∂x = p and ∂y = q .

z = f1(x + iy) + f2(x – iy)

…(1)

∂z = f1′ (x + iy) + f2′ (x − iy) ∂x

…(2)

∂z = i f1′(x + iy) − i f2′(x − iy) ∂y

…(3)

∂2 z = f1′′ (x + iy) + f2′′ (x − iy) ∂x2

…(4)

∂2 z = i2 f1′′(x + iy) + i2 f2′′(x − iy) ∂y2

…(5)

∂2 z ∂2 z + = 0; where i2 = –1 ∂x2 ∂y2



Example 6: Form partial differential equations from the solutions (i) z = f(x) + ey g(x) (ii) z =

1 [F(r − at) + F(r + at)] r

[NIT Kurukshetra, 2008]

Solution: (i): Given z = f(x) + ey g(x)

∂z = ey g(x) , Keeping g(x) as constant. ∂y

∴ and Thus (ii) Given

∂2 z = ey g(x) , (On differentiating again with respect to y) ∂y2 ∂z ∂2 z = ∂y ∂y2 1 z = [F(r − at) + F(r + at)] r ∂z 1 =  F ′(r − at) ⋅ −a + F ′(r + at) ⋅ a  ∂t r  ∂2 z a2 = [ F ′′(r − at) + F ′′(r + at)] r ∂t2

1 ∂z 1 = [ F ′(r − at) + F ′(r + at)] − 2 [ F(r − at) + F(r + at)] r ∂r r ⇒

z ∂z 1 = [ F ′(r − at) + F ′(r + at)] − r ∂r r

…(1) …(2) …(3) …(4)

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1 ∂2 z 1 = [ F′′(r − at) + F′′(r + at)] − 2 [ F′(r − at) + F′(r + at)] r ∂ r2 r

− ⇒

1 2 F ′(r − at) + F ′(r + at)] + 3 [ F(r − at) + F(r + at)] 2[ r r

2 2 ∂2 z 1 = [ F ′′(r − at) + F ′′(r + at)] − 2 [ F ′(r − at) + F ′(r + at)] + 3 [ F(r − at) + F(r + at)] r r ∂r2 r …(5) On using (1), (3), (4) in (5), we get ∂2 z 1 ∂2 z 2  ∂z z  2 = − + + z ∂r2 a2 ∂t2 r  ∂r r  r2 ∂2 z 2 ∂z 1 ∂2 z + = ∂r2 r ∂r a2 ∂t2

or

a2 ∂  2 ∂z  ∂2 z is the desired p.d.e. r = r2 ∂r  ∂r  ∂t2

Example 7: (i) Find the differential equation of all planes which are at a constant distance ‘a’ from the origin. [NIT Kurukshetra, 2006] (ii) Find the differential equation of all spheres whose centre lies on the z-axis. (iii) Find the differential equation of all spheres of radius ‘d’ units having their centres in the xy-plane. Solution: (i) Equation of all planes is αx + βy + γz + δ = 0 …(1) Now perpendicular distance of P(0, 0, 0) from the plane (1) is given equal to ‘a’, i.e. α⋅0 +β⋅0 + γ ⋅0 + δ =a α2 + β2 + γ 2 ⇒

δ = a α2 + β2 + γ 2 Now on substituting the value of δ in equation (1), α x + β y + γ z + a α2 + β2 + γ 2 = 0

Taking partial derivative of equation (3) with respect to x, ∂z α+γ = 0 or α + γ p = 0 ∂x ∂z = 0 or β + γ q = 0 Likewise, β + γ ∂y On substituting values of α and β in terms of γ in equation (3), we get

…(2) …(3)

…(4) …(5)

−γ px − γ qy + γ z + a γ 2 p2 + γ 2q2 + γ 2 = 0



z = px + qy − a 1 + p2 + q2 , the desired partial differential equation.

(ii) Equation of spheres whose centre lies on z-axis is given by x2 + y2 + (z – c)2 = d2 (This represents a surface of revolution with axis OZ.)

…(1)

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First differentiating (1) partially with respect to x, we get ∂z =0 2x + 2(z − c) …(2) ∂x Likewise differentiating partially (1) with respect to y, we get ∂z =0 2y + 2(z − c) …(3) ∂y Now on eliminating (z – c) from equations (2) and (3), qx – py = 0, the desired p.d.e. (iii) Equation of all the spheres of radius ‘d’ whose centre lies in xy plane is given by …(1) (x – a)2 + (y – b)2 + z2 = d2 On differentiating (1) partially with respect to x, ∂z =0 2(x − a) + 2z …(2) ∂x ∂z =0 Likewise 2(y − b) + 2z …(3) ∂y Now on substituting values of (x – a) and (y – b) in equation (1) from equations (2) and (3) respectively, we get 2

2  −z ∂z  +  −z ∂z  + z2 = d2    ∂y  ∂x

i.e.

p2z2 + q2z2 + z2 = d2 or z2(p2 + q2 + 1) = d2

Note: Equation (x – a)2 + (y – b) 2 + z2 = d2 represents a paraboloid of revolution with vertex at (a, b, 0).

ASSIGNMENT 1 1. Form partial differential equations from the relations:

 x (i) z = f    y

(ii) z = emy φ(x – y)

y (iii) z = axe +

1 2 2y a e +b 2

2. Form the partial differential equation (by eliminating the arbitrary function) (i) xyz = φ(x + y + z) (ii) z = f1(x) f2(y) 3. Eliminate arbitrary constants a and b from the following relations: (ii) z = axy + b (i) z = ax + by + a2 + b2 2t − b (iv) ax2 + by2 + cz2 = 1 (iii) z = ae cos bx 4. If z = f(x + ct) + φ(x – ct), prove that

∂2 z ∂2 z = c2 2 2 ∂t ∂x

[J&K, 2001; KUK, 2008, 2009]

11.3 ABOUT SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS A solution of partial differential equation (p.d.e.) in some region R of the space of the independent variable is a function that has the partial derivatives appearing in the equation is some domain containing R and satisfies the equation everywhere in R (often one merely

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requires that the function is continuous on the boundary of R, has those derivatives in the interior of R, and satisfies the equation in the interior of R). In general, the totality of solutions of partial differential equation is very large. e.g. (i) u = x2 – y2, (iii) u = ex cos y (iii) u = log (x2 – y2) are three p.d.e. entirely 2 2 different from each other, still are the solution of ∂ u + ∂ u = 0 , as you may verify. We 2 ∂x ∂y2 shall see for unique solution of a p.d.e. corresponding to the given physical problem, will be obtainable by the use of additional conditions arising from the problem, for instance, the condition that the solution u assumes the given values on the boundary of the region considered (boundary conditions) or, when time t is one of the variables,

∂u or both prescribed at t = 0 (initial condition) ∂t We categorize the solution in the following sub-heads: that u or ut =

1. Complete Solution (Complete Integral) If we can obtain the relation F(x, y, z, a, b) = 0 which contains as many as arbitrary constants (viz., a and b) as there are independent variables in the partial differentiation equation f(x, y, z, p, q) = 0 is known as ‘Complete solution’. 2. Particular Solution (Particular Integral) Particular solution is obtained by giving particular values to the arbitrary constants or the arbitrary function in the complete solution. 3. General Solution (General Integral) If in the solution F(x, y, z, a, b) = 0, we put b = φ(a) and obtain the envelop of the family of surfaces F(x, y, z, a, φ (a)) = 0, we had a solution containing arbitrary function φ. This is called the general solution. 4. Singular Solution (Singular Integral) The envelop of family of surfaces F(x, y, z, a, b) = 0 obtained by elimination of arbitrary ∂F ∂F =0= constants a and b from F(x, y, z, a, b) = 0 and , is called singular solution. ∂a ∂b Remarks: A partial differential equation is said to be fully solved only if all the three types of integrals viz., complete integral, general integral and singular integrals are obtained.

Example 8: Show that if U1 and U2 be two solutions of linear homogeneous equation

∂2U ∂2U ∂2U ∂2U ∂U + 2 + 2 =a 2 +b , then C1U1 + C2U2 is also a solution. 2 ∂t ∂x ∂y ∂z ∂t Extend this result to a linear combination of n independent solutions. Will this result be true if n → ∞. Solution: As U1 and U2 are solutions of the given equation, therefore ∂2U1 ∂2U1 ∂2U1 ∂2U1 ∂U + + = a +b 1 2 2 2 2 ∂x ∂y ∂z ∂t ∂t

…(1)

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∂2U2 ∂2U2 ∂2U2 ∂2U ∂U + + = a 22 + b 2 2 2 2 ∂x ∂y ∂z ∂t ∂t

and

…(2)

Multiplying (1) by C1 and (2) by C2 and adding the two, we get

∂2 ∂2 ∂2 ( ( (C1U1 + C2U2 ) + + C U C C U C 2U2 ) + 2U2 ) + 1 1 1 1 ∂ x2 ∂ y2 ∂ z2 = a

∂2 ∂ (C1U1 + C2U2 ) + b (C1U1 + C2U2 ) ∂t2 ∂t

…(3)

Thus (C1U1 + C2U2) is also a solution of the given p.d.e. Generalisation: If U1, U2, … , Un are n independent solutions, then C1U1 + C2U2 + … + CnUn is also a solution. 2 2 Example 9: Verify that e− n t sin nx is a solution of the heat equation ∂U = ∂ U2 . Hence ∂t ∂x p

show that ∑ Cn e− n t sin nx; where C1, C2, … , Cp are all arbitrary constants, is a solution 2

n= 1

of this equation satisfying the boundary conditions U(0, t) = 0 and U(π π, t) = 0. Solution: Take U = e−n t sin nx, then we need to prove that U satisfies the heat equation. 2

Now ⇒ and

2 ∂U = e− n t n ⋅ cos nx ∂x

…(1)

2 ∂2U = e− n t (−n2 sin nx) 2 ∂x

…(2)

2 ∂U = e− n t (−n2 )sin nx ∂t

…(3)

Now (2) and (3), we have ∂2U ∂ U = ∂ x2 ∂t

Thus U = e−n2t sinnx is a solution of the given solution. Let n = 1, 2, … , p in U = e−n2t , then we get p different solutions. Hence by principle of superposition, we have U = C1 e −t sin x + C2 e−2 t sin 2x + … + Cp e−p t sin px 2

p

2

U(x, t) = ∑ Cn e−n t sin nx

or

2

n=1

is also a solution. Further,

…(4)

U(0, t) = 0 = U(p, t), since sin np = 0 for all integer values of n.

…(5)

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ASSIGNMENT 2

∂U ∂U 1. Show that U = f(x2 – y2) is a solution of y ∂x + x ∂y = 0. 2 nx − n2t sin   is a solution of the heat equation ∂ U = C2 ∂ U 2. Verify that e  C ∂t ∂ x2 N − n2t sin  nx  , where a1, a2, … , are arbitrary constants, is also a Hence show that ∑ an e  C n=1 solution satisfying the boundary conditions U(0, t) = 0 = U(πC, t).

11.4 EQUATIONS SOLVABLE BY DIRECT INTEGRATION Partial differential equations occuring with only one partial derivative can be solved directly by integration. However, in such cases, we must use arbitrary function of variable in place of constant of integration. Example 10: Solve

∂2 z ∂z + z = 0 given that when x = 0, z = ey and = 1. ∂x ∂x2

Solution: If z were a function of x alone, the solution would have been z = C1 cos x + C2 sin x …(1) where C1 and C2 are arbitrary constants. Since here z is a function of both x and y, therefore, C1 and C2 can be chosen arbitrary functions of y. Whence the solution of the given equation is z = f1(y)sin x + f2(y)cos x …(2)

∂z = f1(y)cos x − f2 (y)sin x ∂x When x = 0, z = ey ⇒ ey = f2(y) ∂z = 1 ⇒ 1 = f1(y) Also x = 0, ∂x Hence the required solution is z = sin x + ey cos x.



…(3) …(4) …(5)

∂z ∂2 z = − 2 sin y when x = 0, and z = 0, = sin x sin y , given that ∂y ∂ x∂ y π. when y is an odd multiple of 2 ∂2 z = sin x sin y with respect to x keeping y Solution: On integrating the given equation, ∂ x∂ y constant, ∂z = − cos x sin y + φ(y) …(1) ∂y

Example 11: Solve

∂z = −2 sin y ∂y φ(y) = – sin y

Given x = 0, or

implies

–2 siny = –sin y + φ(y) …(2)

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∂z = − cos x sin y − sin y ∂y Now, on integrating (3) with respect to y, we get z = cosx cos y + cos y + ψ(x) Thus (1) becomes

…(3) …(4)

π , z = 0 ⇒ ψ(x) = 0 2 z = (1 + cos x) cos y is the required solution.

Clearly, when y is an odd multiple of ∴

Example 12: Solve

∂ 3z = cos(2x + 3y). ∂ x2 ∂ y

∂3 z = cos(2x + 3y) ∂ x2∂ y On integrating (1) with respect to x, keeping y constant, we get sin(2x + 3y) ∂2 z = + f (y) 2 ∂ x∂ y Again integrating (2) with respect to x keeping y constant, we get cos(2x + 3y) ∂z =− + x f (y) + φ(y) ∂y 4 Now on integrating (3) with respect to y keeping x constant, we get

Solution: Given

sin(2 x + 3 y) + x ∫ f (y) dy + ∫ φ (y) dy + γ (x ) 12 sin(2x + 3y) z=− + x α(y) + β(y) + γ (x) 12

…(1)

…(2)

…(3)

z=−

or

…(4)

 ∂2 z   = (x + y) . Example 13: Solve log   ∂ x∂ y   ∂2 z 

log    ∂2 z  ∂ x∂ y  Solution: Given log  = e(x + y)  = (x + y) or e  ∂ x ∂ y   ∂2 z = e( x + y) ⇒ ∂ x∂ y On integrating (1) with respect to x keeping y constant, we get ∂z = e(x + y) + f (y) ∂y where f(y) is an arbitrary constant. Now, integrating (2) again with respect to y, keeping x constant z = e(x + y) + x f(y) + φ(x)

…(1)

…(2)

…(3)

Partial Differential Equations

Example 14: Solve

685

∂z ∂2z = e− x . = z, given that when y = 0; z = ex and ∂y ∂y2 [NIT Kurukshetra, 2010]

∂2 z = z , if we treat z as pure function of y only, we could solve it ∂ y2 like an ordinary differential equation with auxiliary equation as: D2 = 1 i.e., D = ±1 Solution: In the equation

so that z = Aey + Be–y …(2) Here z is a function of both x and y, since we are dealing in partial differential equations. Thus in z = Aey + Be– y, A and B are arbitrary constants, but are like A = φ(x) and B = ψ(x). Whence z = φ(x) ey + ψ(x) e– y …(3) Now, for y = 0, z = ex ⇒ ex = φ(x) e0 + ψ(x) e0 ex = φ(x) + ψ(x)

i.e.

Again for y = 0,

…(4)

∂z = e−x i.e., from equation (3), we get ∂y

e– x = [φ(x) ey – ψ(x) e– y]y=0 ⇒

e−x = φ(x) e0 − ψ(x)

1 = φ(x) – ψ(x) e°

…(5)

Now, on solving equations (4) and (5) for φ(x) and ψ(x), we get

 ex + e−x = cosh x  2   x −x and e −e ψ(x) = = sinh x   2 Therefore, z = (cosh x ey + sinh x e– y) is the desired solution. φ(x) =

ASSIGNMENT 3 1. Solve

∂2 z x = +a ∂ x∂ y y

3. Solve p.d.e.

4.

∂2 z = sin(xy) ∂ y2

∂z ∂2 z = a sin y and = a2 z , given that when x = 0, 2 ∂ x ∂x

∂2 z = e−t cos x ∂ x∂ t

6. Log s = x + y

2. Solve

5. xys = 1

 Hint: 

∂z = 0. ∂y

Rewrite as

  Hint: 

∂2 z 1 =  ∂ x∂ y xy 

Rewrite as,

 ∂2 z = ex + y  ∂ x∂ y 

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11.5 LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER A differential equation involving only first order partial differential coefficients p and q is called partial differential equation of first order. Further, if the degrees of p and q are unity only then it is termed as linear p.d.e. of first order. If each term of such an equation contains either the dependent variable or one of the derivatives, the equation is said to be homogeneous, otherwise non-homogeneous. Some important partial differential equations of second order are as follows: 1.

2 ∂2 u 2 ∂ u c = , one dimensional wave equation (hyperbolic) ∂t2 ∂ x2

2 2. ∂ u = c2 ∂ u , one dimensional heat equation (parabolic) ∂t ∂x2 2 2 3. ∂ u + ∂ u = 0 , two dimensional Laplace equation (elliptic) ∂ x2 ∂ y2

4.

∂2u ∂2u + = f (x, y) , two dimensional Poisson equation ∂ x2 ∂ y2

2 2 2 5. ∂ u2 = c2 ∂ u2 + ∂ u2 , two dimensional wave equation ∂t ∂x ∂y

6.

∂2u ∂2u ∂2u + + = 0 , three dimensional Laplace equation ∂ x2 ∂ y2 ∂ z2

Here c is a constant, t is time, x, y, z are Cartesian co-ordinates Equations other than (4), all are homogeneous. Lagrange’s Linear Equation Ist order linear partial differential equation in its standard form Pp + Qq = R

…(1)

where P, Q, R are functions of x, y, z is called Lagrange’s Linear Equation. This equation is obtained by eliminating arbitrary function f from f(u, v) = 0 …(2) where u, v are functions of x, y, z. Here we show that its solution depends on the solution of the equations dx dy dz = = …(3) P Q R Differentiating (2) partially with respect to x and y respectively, we get

∂f ∂u ∂f ∂v  + = 0 ∂u ∂x ∂v ∂x   (as (2) is an implicit relation) ∂f ∂u ∂f ∂v + = 0 ,  ∂u ∂y ∂v ∂y

Partial Differential Equations

687

More precisely,

∂f  ∂u ∂u  ∂f  ∂v ∂v   + p + + p = 0     ∂u ∂x ∂z ∂v ∂x ∂z   , ∂ f  ∂u ∂u  ∂ f  ∂v ∂v  q + q = 0  + +    ∂u  ∂y ∂z  ∂v  ∂y ∂z  

…(4)

∂f ∂x (as ∂ x = 0 = ∂ y , x and y being two independent variables.) From above equations, on eliminating

implying

∂u ∂u + p ∂x ∂z

∂v ∂v + p ∂x ∂z

∂u ∂u + q ∂y ∂z

∂v ∂v + q ∂y ∂z

∂f ∂f and , we have ∂u ∂v

=0

 ∂u ∂v ∂u ∂v  ∂u ∂v ∂u ∂v  ∂u ∂v ∂u ∂v   ∂y ∂z − ∂z ∂y  p +  ∂z ∂x − ∂x ∂z  q = ∂x ∂y − ∂y ∂x

…(5)

…(6)

which is the same as equation (1) with  ∂u ∂v ∂u ∂v  P= −  ∂ y ∂ z ∂ z ∂ y 

 ∂u ∂v ∂u ∂v  Q= −  ∂z ∂x ∂x ∂z   ∂u ∂v ∂u ∂v  R= −  ∂ x ∂ y ∂ y ∂ x  Now in order to find u and v, let u = a and v = b, where a and b are two arbitrary constants, so that

∂u ∂u ∂u  dx + dy + dz ∂x ∂y ∂ z   and ∂v ∂v ∂v  dx + dy + dz 0 = dv = ∂x ∂y ∂z  From above simultaneous equations, we get 0 = du =

dx

∂u ∂v ∂u ∂v ⋅ − ⋅ ∂y ∂z ∂z ∂y or

=

dy

∂u ∂v ∂u ∂v − ∂z ∂x ∂x ∂z

=

dz ∂u ∂v ∂u ∂v − ∂x ∂y ∂y ∂x

dx dy dz = = P Q R Solution of above differential equation are u = a and v = b.

…(7)

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whence the solution of Lagrange’s Linear equation Pp + Qq = R is f(u, v) = 0 or f(a, b) = 0. Working Rule for Solving Lagrange’s Equations (i) Corresponding to Lagrange’s Equation (linear partial differential equation) Pp + Qq = R. Form the auxiliary equations dx dy dz = = P Q R (ii) Solve these auxiliary equations by the method of grouping or the method of the multiplier or both for getting two independent integrals, say, u = a and v = b. Then the general integral of the given equation will be f(u, v) = 0 or u = f(v), where f is an arbitrary function. Note: In case of linear equation with n independent variables, say, P1p1 + P2p2 + … + Pnpn = R, where

pj =

∂z j = 1, 2, … , n; , Pl, P2, … , Pn and R are functions of x1, x2, … , xn and z. ∂ xj

The subsidiary equation is

dx1 dx2 dx dz = =…= n = P1 P2 Pn R and solution is f(u1, u2, … , un) = 0 where u1 = const., u2 = const. so on. un = constant, are the solutions of the subsidiary equations.

Geometrical Interpretation of Lagrange’s Equation Lagrange’s linear equation Pp + Qq = R …(1) may be written as Pp + Qq + (– 1)R = 0

n p:q:–1

Let the solution of (1) be f(x, y, z) = 0 …(2) representing a surface, the normal to which at any point has direction cosines proportional to ∂f ∂f ∂f : : ∂x ∂y ∂z

∂f − ∂f ∂x : ∂y : −1 ∂f ∂f − − ∂z ∂z

− or

or

∂z ∂z : : −1 ∂x ∂y

or p : q : –1

P:Q:R

f(x, y, z) = 0

dx dy dz = = P Q R

Fig. 11.1

Partial Differential Equations

689

Further, the simultaneous equations

dx dy dz = = P Q R

…(3)

represent a family of curves such that the tangent to which at any point has the direction cosines proportional to P, Q, R and that f(u, v) = 0 represents a surface through such curves where u = constant, v = constant (say a and b respectively), are two particular integrals of (3). Hence, the geometrical interpretation of equation (1) is that ‘‘the normal to the surface (2) is perpendicular to a line (say u = a or v = b = f(a)) whose direction cosines are proportional to P, Q, R and so that the sum of their respective product is Pp + Qq + R(–1) = 0 or Pp + Qq = R. Or in other words, the equation (1) states that normal to the surface (2) at any point is perpendicular to the members of the family (3) through that point and which is true for every point on the surface (1). Thus, the equation (1), Pp + Qq = R and the equation (3),

dx dy dz defines the same = = P Q R

set of surfaces and hence equivalent. Example 15: Solve the following equations (i) (ii) (iii) (iv)

(z2 – 2yz – y2)p + (xy + xz)q = (xy – zx) p tanx + q tany = tanz px – qy = (y2 – x2) y2p – xyq = x(z – 2y)

(KUK, 2000) (KUK, 2008)

Solution: (i) The subsidiary equations are given by

x dx + y dy + z dz dy dx dz = = = 2 (z − 2yz − y ) (xy + xz) (xy − zx) 0 2

I

Taking

II

III

…(1)

IV

dy dz = (y + z) (y − z)

Which on simplification results to (y dy – z dz) + (z dy + y dz) = 0 ⇒

y dy z dz − + d(yz) = 0 2 2 On integrating, we have

(y2 – z2 – 2yz) = C1 Also from (1), we have

…(2)

x dx + y dy + z dz dx = 2 (z − 2xz − y ) 0 2

i.e.

x dx + y dy + z dz = 0

⇒ x2 + y2 + z2 = C2

…(3)

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∴ The desired solution is f(C1, C2) = 0 or f(y2 – z2 – 2yz, x2 + y2 + z2) = 0

dy dx dz (ii) tan x = tan y = tan z I

II

III

dy dx On taking I & II, tan x = tan y

⇒ or

∫ cot x dx = ∫ cot y dy  sin y  C1 =   sin x 

…(1)

Likewise taking II & III, or ⇒

or log sinx = log siny – log C1

dy dz = ⇒ tan y tan z

∫ cot y dy = ∫ cot z dz

log sin y = log sin z – logC2  sin z  C2 =   sin y 

…(2)

 sin y sin z  , ∴ The desired solution is f(C1, C2) = 0 = f   sin x sin y  (iii) The subsidiary equations are

x dx + y dy + dz dx dy dz = = 2 = 2 0 x −y y − x I

II

III

IV

dx dy On taking I & II, we have x = −y ⇒

log(xy) = log C1 ⇒ Taking I & IV,

or ∴

⇒ logx = – log y + logC1

xy = C1

…(1)

dx (x dx + y dy + dz) = x 0

x dx + y dy + dz = 0 ⇒ x2 + y2 + 2z = C2

…(2)

f(C1, C2) = 0 or f(xy, x2 + y2 + 2z) = 0, the desired solution.

(iv) Here subsidiary equations are

dy dx dz = = 2 y −xy x(z − 2y)

…(1)

Partial Differential Equations

dy dx On taking I & II, y2 = −xy

⇒ x dx + y dy = 0, i.e. (x2 + y2) = C1

691

…(2)

Likewise, on taking

dy dz − dy dz = = −xy xz − 2xy xz − xy dy dz − dy = −y z−y

i.e.,

…(3)

On simplifying, we get z dy + y dz = 2y dy, i.e. d(yz) = d(y2) or y2 – yz = C2 Alternately, ∴

…(4)

dy dz − dy f ′(x) is of the form = f (x) z−y −y

–log y = log(z – y) – log C2 or y(z – y) = C2

…(5)

Hence the solution is f(x2 + y2, y2 – yz) = 0 Example 16: Solve Ist order linear partial differential equation p(x2 – y2 – z2) + q(2xy) = 2xz Solution: Here we have,

x dx + y dy + z dz dy dx dz = = = 2 2 2 2xy 2xz x(x − y2 − z2 ) + y(2xy) + z(2xz) x −y −z 2

I

II

Taking II and III,

Taking III and IV, ⇒ ⇒ ⇒

III

IV

dy dz y = ⇒ = C1 y z z dz (x dx + y dy + z dz) = 2xz x(x2 + y2 + z2 )

dz d(x2 + y2 + z2 ) = 2 , z (x + y2 + z2 )

f ′(x)    which is of the form f (x) 

log z + log C2 = log(x2 + y2 + z2)

C2 =

(x2 + y2 + z2 ) z

Hence the desired solution is f(C1, C2) = 0 or

…(1)

 y x2 + y2 + z2  f ,  = 0 z z

…(2)

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Example 17: Solve the following Ist order linear partial differential equations (i) y2zp + x2zq = y2x (ii) p – q = log(x + y) (iii) pyz + qzx = xy

[NIT Kurukshetra, 2007]

Solution: (i) Here subsidiary equations are

dy dx dz = 2 = 2 2 yz xz yx On taking I and II, we get

dy dx = y2 z x2 z

or x2dx = y2dy

or (x3 – y3) = C1

…(1)

or x dx = z dz ⇒ (x2 – y2) = C2

…(2)

Likewise, taking I & III, we get

dy dx = 2 2 yz yz

Hence the desired solution is f(x3 – y3, x2 – y2) = 0. (ii) The subsidiary equations are

dx dy dz = = 1 − 1 log(x + y)

I

II

III

On taking I and II,

dx dy = 1 −1

⇒ (x + y) = a (say)

…(1)

Now on taking I & III, we get

dx dz = 1 log(x + y)



dx dz = 1 log a

⇒ (log a) · dx = dz

On integrating, (log a)x = z + b (say) ⇒ ⇒

x · log(x + y) = z + b

…(2)

On using (1)

x log (x + y) = z + φ(a) as b = φ(a)

⇒ x log(x + y) – z = φ(x + y) (iii) Here the subsidiary equations are

…(3)

dx dy dz = = yz zx xy I

II

III

On taking I and II, we get dx dy ⇒ x dx = ydy = yz zx

⇒ (x2 – y2) = C1

…(1)

Partial Differential Equations

693

On taking II and III, we get

dy dz ⇒ y dy = z dz ⇒ (y2 – z2) = C = 2 zx xy Hence the desired solution is f(C1, C2) = 0 or f(x2 – y2, y2 – z2) = 0.

…(2)

Example 18: Solve the following Lagrange’s Linear partial differential equations (i) (x2 – yz)p + (y2 – zx)q = (z2 – xy) [KUK, 2009] [KUK, 2004-05] (ii) x2(y – z)p + y2(z – x)q = z2(x – y) [NIT Jalandhar, 2006; KUK, 2003-04] (iii) x(y2 – z2)p + y(z2 – x2)q – z(x2 – y2) = 0 Solution: (i) Here the auxiliary equations are

dy dx dz = = x2 − yz y2 − zx z2 − xy I

or

II

III

dy dx dz = 2 = 2 (x − yz) (y − zx) (z − xy) dx − dy dy − dz dz − dx = 2 = 2 = 2 2 2 (x − yz) − (y − zx) (y − zx) − (z − xy) (z − xy) − (x2 − yz) 2

A

B

C

On taking expressions A and B,

dx − dy dy − dz = 2 2 ( x − yz) − (y − zx) (y − zx) − (z2 − xy) On simplification we get, 2

dx − dy dy − dz = (x − y)(x + y + z) (y − z)(x + y + z) or

f ′(x) dx − dy dy − dz , which is of the form = f (x) (x − y) (y − z) On integration, log(x – y) = log(y – z) + logC1 ⇒ Likewise, on taking II and III,

 x − y C1 =   y − z 

…(1)

dy − dz dz − dx = (y − z) (z − x)

or log(y – z) = log(z – x) + logC2 ⇒

 y − z C2 =   z − x 

 x − y y − z Hence, the desired solution of the given p.d.e. is f  , = 0.  y − z z − x 

…(2)

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Engineering Mathematics through Applications

(ii) Here, the subsidiary equations are

dy dx dz = 2 = x (y − z) y (z − x) z2 (x − y)

…(1)

2

I

II

III



1 1 1 dx + dy + dz dy x y z dx dz = 2 = 2 = 2 1 1 1 x (y − z) y (z − x) z (x − y) x2(y − z) + y2(z − x) + z2(x − y) x y z



1 1 1 dx + dy + dz dy x y z dx dz = 2 = 2 = 2 x (y − z) y (z − x) z (x − y) 0

1 1 1 dx + dy + dz = 0 x y z

⇒ ⇒

log xyz = log C1 ⇒ xyz = C1

…(2)

Likewise, (1) also becomes 1 1 1 dx + 2 dy + 2 dz dy x2 y z dx dz = = = x2 (y − z) y2 (z − x) z2 (x − y) (y − z) + (z − x ) + (x − y)

1 1 1 dx + 2 dy + 2 dz = 0 2 x y z

or

On integration of each term with respective variable, we get ⇒



1 1 1 1 or C2 = ( x−1 + y−1 + z−1 ) − − =− x y z C2

∴ The desired solution, f (xyz, x–1 + y–1 + z–1) = 0. (iii) The subsidiary equations are

1 1 1 dx + dy + dz dy x dx + y dy + z dz x y z dx dz = = = = =β 2 2 2 2 2 2 x(y − z ) y(z − x ) z(x − y ) 0 0 I

II

III

IV

…(1)

V

From above, expression IV gives x dx + y dy + z dz = 0 ⇒

x2 y2 z2 + + =c 2 2 2

⇒ x2 + y2 + z2 = C1

…(2)

Partial Differential Equations

695

From expression V, we get

1 1 1 dx + dy + dz = 0 x y z log x + log y + log z = log C2 ⇒ xyz = C2 Hence, the desired solution is

f(x2

+

y2

+

z2,

…(3)

xyz) = 0.

Example 19: Solve the following partial differential equations (i) x(y – z)p + y(z – x)q = z(x – y) (ii) (y + z)p + (z + x)q = (x + y).

[KUK, 2002-03]

Solution: (i) Here the subsidiary equations are

dy dx dz = = x(y − z) y(z − x) z(x − y) 1 1 1 , , , we get x y z

On using the multiplier

1 1 1 dx + dy + dz dy x y z dx + dy + dz dx dz − = = = = x(y − z) y(z − x) z(x − y) 0 0 I

II

III

From expression IV, we get

IV

V

1 1 1 dx + dy + dz = 0 x y z

log xyz = log C1 ⇒ xyz = C1 From expression V, dx + dy + dz = 0 ⇒ (x + y + z) = C2 Hence the desired solution is f (xyz, x + y + z) = 0. (ii) Here in this case, the auxiliary equations are

dy dx dz = = (y + z) (z + x) (x + y)

…(1)

The relation (1) is extended to,

dy dx + dy + dz dx − dy dy − dz dx dz = = = = = (y + z) (z + x) (x + y) 2(x + y + z) −(x − y) −(y − z) I

II

III

IV

V

VI

From IV and V, we get (dx + dy + dz) (dx − dy) f ′(x) , which is of the form = 2(x + y + z) −(x − y) f (x)

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1 log(x + y + z) + log(x − y) = log C 2 log C1 = log(x + y + z)(x – y)2

or C1 = (x + y + z)(x – y)2

…(1)

From expressions V and VI,

(dx − dy) (dy − dz) = (x − y) (y − z) ⇒ log(x – y) = log(y – z) + log C2

 x − y ⇒ log  = log C2  y − z 

or

 x − y  y − z  = C2

…(2)

  x − y Hence, the desired solution is f  (x + y + z)(x − y)2 ,   = 0.  y − z    Example 20: Solve the linear partial differential equation (i) x2x3p1 + x3x1p2 + x1x2p3 = – x1x2x3

(ii) –p1 + p2 + p3 = 1

∂u ∂u ∂u (iii) x ∂ x + y ∂ y + z ∂ z = xyz

[CDLU, 2004]

Solution: (i) Comparing the given equation x2x3p1 + x3x1p2 + x1x2p3 = – x1x2x3

…(1)

with the equation P1p1 + P2p2 + P3p3 + … + Pnpn = R

…(2)

we get the auxiliary equations as:

dx1 dx2 dx3 dx = = = …= n = R P1 P2 P3 Pn ∴

dx1 dx dx dz = 2 = 3 = x2 x3 x3 x1 x1x2 −x1x2 x3

…(3) …(4)

On taking I and IV, we get x1dx1 + dz = 0 ⇒ x12 + 2z = C1

…(5)

Likewise, from I & II and I & III, we get x12 – x22 = C2 and x12 – x32 = C3

…(6)

Hence the general integral is f (x12 + 2z, x12 – x22, x12 – x32) = 0. (ii) As explained above, the corresponding auxiliary equations in this case are

dx1 dx2 dx3 dz = = = 1 1 1 −1 I

II

III

IV

…(1)

Partial Differential Equations

Taking I and IV, dx1 + dz = 0 ⇒ x1 + z = C1 Likewise, from I and II and I and III, we get x1 + x2 = C2 and

x1 + x3 = C3

697

…(2) …(3) …(4)

Hence, the desired solution is f(x1 + z, x1 + x2, x1 + x3) = 0. (iii) Here in this case when u is a function of three independent variables x, y and z, the desired auxiliary equations are

dx dy dz du = = = x y z xyz I

II

III

On taking I and II, ⇒

…(1)

IV

dx dy = x y

logx = log y + logC ⇒

x = C1 y

Similarly taking II and III, we get

y = C2 z

yz dx + zx dy + xy dz du = 3xyz xyz yz dx + zx dy + xy dz = 3du or d(xyz) = 3du or xyz – 3u = C3

…(2) …(3)

Again ⇒

…(4)

x y  Hence, the desired solution is f  , , xyz − 3u . y z  Example 21: Solve p + 5q = 9z + tan(y – 5x). Solution: Here the auxiliary equations are

dx dy dz = = 1 5 9z + tan(y − 5x)

…(1)

Taking I and II, we get

dx dy = ⇒ (y – 5x) = C1 1 5 On taking I and III, we get dx dz = 1 9z + tan C1 ∴

x=

log(9z + tan C1 ) − log C2 9

9x = log(9z + tan(y – 5x)) – logC2

…(2)

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⇒ ∴

log C2 = log[9z + tan(y – 5x)] – 9x C2 = e–9x[9z + tan(y – 5x)] Hence, the desired solution is

(

…(3)

)

f e−9x [ 9z + tan(y − 5x)] , y − 5x = 0. Example 22: Solve p cos(x + y) + q sin(x + y) = z. Solution: The subsidiary equations are

dy (dx + dy) dx − dy dx dz = = = = cos(x + y) sin(x + y) z cos(x + y) + sin(x + y) cos(x + y) − sin(x + y) I

II

III

(x + y) = u ⇒ (dx + dy) = du

Let

From III and IV,

dz du = = z cos u + sin u

log z =



IV

du π 1 , since sin  u +  = (sin u + cos u)   π 4 2 2 sin  u +   4

1 log[cosec U − cot U ] + log C1, 2

1 1 cos U  z log  , = − 2 C1  sin U sin U 



log



 2 sin2 U   1 − cos U  z  2  log = log  = log  U U C1  sin U   2 sin cos  2 2 



log



2 log

1 z U log  tan  = 2  2 C1  z u π  π = log tan  +   , as U = u +   2 8  C1 4 



z 2 = C3 u π  tan  +  2 8



C3 = z Further,

2

u π cot  +  = z 2 8

2

V

x + y π ⋅ cot  +  8  2

dx + dy dx − dy = cos(x + y) + sin (x + y) cos(x + y) − sin (x + y)

U =u+

π 4

Partial Differential Equations



cos(x + y) − sin (x + y)

∫ cos(x + y) + sin (x + y)(dx + dy) = ∫ (dx − dy)

Here LHS is comparable to ∴



f ´(x) dx = log f (x) f (x)

log[cos(x + y) + sin(x+ y)] = (x – y) + logC2

Rewrite as ⇒

699

log[cos(x + y) + sin(x + y)] = log e(x – y) + logC2

e–(x – y) [cos(x + y) + sin(x + y)] = C2 .

 Hence f  z 

2

 x + y π  ( y − x) cos( + ) + sin( + )  x y } = 0 cot  + , e { x y  2  8

Example 23: Solve the equation z − xp − yq = a x2 + y2 + z2 . Solution: The subsidiary equations are as follows:

x dx + y dy + z dz dx dy dz = = = 2 2 2 2 2 x y z−a x +y +z (x + y + z2 ) − az x2 + y2 + z2 Putting (x2 + y2 + z2) = u2 so that (xdx + y dy + zdz) = u du

u du dx dy dz du = = = 2 = 2 2 2 x y u − azu u − az z−a x +y +z



dx dy dz du du + dz = = = = x y z − au u − az (1 − a)(u + z) Taking

dx du + dz dx du + dz = or (1 − a) = x (1 − a)(u + z) x (u + z)

Integrating (1 – a)log x = log (u + z) + log C1

{

x(1 − a) = C1(u + z) = C1 z + x2 + y2 + z2

⇒ Again, ⇒



}

…(1)

dx dy = x y

x = C2 y Therefore, general solution is f(C1, C2) = 0 log x = log y + logC2 or

…(2)

 x x(1 − a) = z + x2 + y2 + z2 φ    y

{

}

Example 24: Solve the Lagrange’s Linear differential equation px(z – 2y2) = (z – qy)(z – y2 – 2x3)

[KUK, 2007, 2010]

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Engineering Mathematics through Applications

Solution: On rewriting the given equation, we have px(z – 2y2) + qy(z – y2 – 2x3) = z(z – y2 – 2x3) On comparing with Pp + Qq = R, we have  P = x(z – 2y2) Q = y(z – y2 – 2x3) y2

2x3)

R = z(z – – Here subsidiary equations are

   

…(1)

dx dy dz = = P Q R dy dx dz = = 2 x(z − 2y ) y(z − y2 − 2x3 ) z(z − y2 − 2x3 )

or

I

II

…(2)

III

On considering II and III, we get

dy dz = y z

or y = az

…(3)

Now, consider I and III and make use of expression (3), we have

dx dz = x(z − 2a2 z2 ) z(z − a2 z2 − 2x3 ) dx dz = x(1 − 2a2 z) (z − a2 z2 − 2x3 )

or ⇒

(x dz – z dx) + 2x3 dx = 2a2 xz dz – a2 z2 dx

Divide throughout by x2,

(x dz − z dx) a2 (2xz dz − z2 dx) 2 + x dx = x2 x2 z  z2  d   + d(x2 ) = a2 d    x  x 2 z 2 2 z   + x − a  = b2 or x x

y2 z + x2 − =b x x

y z y2  Hence the required solution is, f  , + x2 −  = 0 x  z x Example 25: Solve Lagrange’s Linear differential equation (i)

dx dy dz = = 2 x y z − a x + y2 + z2

(ii) p(y2 + z2 + yz) + q(z2 + zx + x2) = (x2 + xy + y2).

Partial Differential Equations

701

Solution: The corresponding subsidiary equations are (i)

x dx + y dy + z dz dx dy dz = = = 2 2 2 2 2 x y z−a x +y +z (x + y + z2 ) − az x2 + y2 + z2 Let x2 + y2 + z2 = t2 so that (x dx + y dy + z dz) = t dt On using (2), relation (1) becomes

t dt dx dy dz = = = x y z − at t2 − azt

…(1) …(2)

…(3)

Now from (3), we have dx dy = ⇒ log x = log C1y ⇒ x y

x = C1 y

…(4)

dx dz dt dt + dz = = = x z − at t − az (t − az) + (z − at)

and ⇒

(dt + dz) dx = x (1 − a)(t + z)



log(t + z) = (1 – a)log x + logC2

(t + z) = C2 x1− a





x2 + y2 + z2 + z = C2 x1− a

 x x2 + y2 + z2 + z  Here the solution is f(C1, C2) = 0 = f  y ,  x1− a (ii) The subsidiary equations are

(dy − dx) (dz − dy) dy dx dz = 2 = 2 = = 2 2 2 (x − y)(x + y + z) (y − z)(x + y + z) y + yz + z z + zx + x x + xy + y 2

⇒ ⇒ ⇒

(dy − dx) (dz − dy) = (x − y) (y − z) log(y – x) = log(y – z) + logC1 (y – x) = C1(y – z)

Similarly ⇒ ⇒ ⇒

…(1)

dy − dx dz − dx = (x − y)(x + y + z) (x − y)(x + y + z) (dy − dx) (dz − dx) = ( x − y) (x − y) log(x – y) = log(x – z) + logC2 (x – y) = C2(x – z)

∴ The desired solution is f(C1, C2) = 0

…(2) or

 x − y y − x f = 0. ,  x − z y − z 

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Engineering Mathematics through Applications

Example 26: x

xy ∂u ∂u ∂u . +y +z = au + ∂x ∂y ∂z z

Solution: Here the subsidiary equations are

dx dy dz du = = = xy x y z au + z I

II

III

…(1)

IV

Taking I and II, we get y dx dy = ⇒ = C1 x y x Likewise from I & III, we have z = C2 x Again from I and IV, we get

dx du = xy x au + z y du a − u= dx x z



…(2)

…(3)

du ⇒ = dx

or

au +

xy z

x

C du a − u= 1 dx x C2

(on using (2) and (3))

…(4)

Which is an ordinary linear differential equation with Integrating Factor, a − ∫ dx x

I.F. = e

∴ or or

= e− a log x =

1 xa



C 1 1 = 1 dx + C3 a x C2 xa y x−a + 1 + C3 ux−a = ⋅ z −a + 1 y x1 − a = C3 ux−a − ⋅ z 1− a

u⋅

…(5)

y x1 − a  y z , ux−a − ⋅  =0 Hence the general integral, f  , x x z 1 − a

ASSIGNMENT 4 1. Solve z(xp – yq) = (y2 – x2) 2. Solve (y3x – 2x4) + (2y4 – x3y)q = 9z(x3 – y3) 3. Solve Lagrange’s equation

y2 z p + zxq = y2 . x

…(6)

Partial Differential Equations

703

4. Solve linear partial differential equation (mz – ny)p + (nx – lz)q = (ly – mx). 5. Find the surface whose tangent planes cut of an intercept of constant length k from the z-axis. [Hint: Equation of the tangent plane at (x, y, z) is (Z – z) = (X – x)p + (Y – y)q.] 6. Solve

(y + z + u)

∂u ∂u ∂u + (z + x + u) + (x + y + u) = (x + y + z) . ∂x ∂y ∂z

11.6 NON-LINEAR EQUATIONS OF FIRST ORDER As already defined, when p and q occur other than in the first degree, the equation is a nonlinear one and its general solution contains only two arbitrary constants (viz equal to the number of independent variable i.e., x and y). These equations are discussed under following four standard forms. I. When Equation Contains p and q Only (i.e., no x, y, z.) – First Standard Form Let the equation be f(p, q) = 0 The obvious complete solution for this equation is z = ax + by + c viz replacement of p, q by two arbitrary constants a, b respectively as

p=

∂z ∂z = a and q = =b ∂x ∂y

…(1) …(2)

…(3)

whence a and b are related by the relation f(a, b) = 0 …(4) Further (4) gives b = f(a) and with this, the complete solution (2) may be written as z = ax + f(a)y + c …(5) Example 27: Solve (i) pq + p + q = 0

(ii) p3 – q3 = 0.

Solution: (i) As the given equation falls under the Ist category i.e., f(p, q) = 0 whence f(a, b) = ab + a + b = 0 ⇒

(a + 1)b + a = 0 or

a  b = −   a + 1 

…(1) …(2)

Hence the desired solution, z = ax + by + C z = ax + f(a)y + C, where b = f(a) a z = ax − y+C (a + 1) (ii) Here, f(a, b) = (a3 – b3) = 0 ⇒ (a – b)(a2 + b2 + ab) = 0 i.e.

either a = b or a2 + b2 + ab = 0

…(3)

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Engineering Mathematics through Applications

On using, b = a in the obvious solution, z = ax + by + C we get z = a(x + y) + C. Example 28: Solve x2p2 + y2q2 = z2. Solution: On rewriting the given equation as 2

2  x ⋅ ∂ z  +  y ∂z  = 1    z ∂y  z ∂x 

…(1)

dy dx dz = dX, = dY, = dZ. x y z so that X = log x, Y = log y, Z = log z Further (2) implies Now let

∂Z x ∂z = ∂X z ∂x

and

…(2) …(3)

∂Z y ∂z = ⋅ ∂Y z ∂y

…(4)

or P2 + Q2 = 1

…(5)

On using (4), (1) becomes 2

2

 ∂Z  +  ∂Z  = 1     ∂X  ∂Y 

where f(p, q) = 0 has complete integral as Z = aX + bY + C Here, ∴ or ⇒

f(a, b) = 0 ⇒ a2 + b2 = 1 or

b = ± 1 − a2

Z = aX ± 1 − a2 Y + log C log z = a log x ± 1 − a2 log y + log C, z = Cxa ⋅ y

1− a2

Using (3)

is the required solution.

II. Equation Containing p, q and z (i.e., no x and y) – Second Standard Form The equation is of the form f(p, q, z) = 0 Let us assume its solution be z = φ(u) where u = x + ay

∂z dz ∂u dz = = ∂x du ∂x du ∂z dz ∂u dz q= = =a and ∂y du ∂y du Whence the equation (1) reduces to with

…(6)

…(1) …(2)

p=

…(3)

dz dz f  z, , a  = 0  du du 

…(4)

Partial Differential Equations

705

dz i.e. a first order ordinary linear differential equation, and du hence solved by variable seperable method. which is clearly a relation in z,

Note: Sometimes the equation in its given form is not of the form f(p, q, z) but after certain transformation or substitution it reduces to f(p, q, z) = 0.

Example 29: Obtain the complete solution of following equations: (i) z = p2 + q2

(ii) p(1 + q2) = q(z – a).

Solution: (i) Let u = x + ay, so that

∂z dz ∂u dz  = = , ∂x du ∂x du   ∂z dz ∂u dz q= = =a  du  ∂y du ∂y p=

With the above values of p and q, the given equation reduces to 2

dz dz z =   +  a   du   du 

dz z = (1 + a)2    du 

⇒ ⇒ ⇒

2

2

or du = (1 + a)1/2

1 1 z2

dz

u + b = 2(1 + a2)1/2 z1/2, b is an arbitary constant (x + ay + b)2 = 4(1 + a2)z

which is the desired solution. dz dz , q=b , the given equation becomes (ii) With u = x + by and p = du du 2 dz   dz   dz 1+ = b   ( z − a) du   du    du  ⇒

2

dz 1 +   = b(z − a)  du  2

 dz  = b(z − a) − 1   du

⇒ ⇒ ⇒ ⇒

dz = ± du b(z − a) − 1 2⋅

bz − ba − 1 = ± (u + c) b

bz − ab − 1 = ±

b (x + by + c) is the desired solution. 2

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Engineering Mathematics through Applications

Example 30: Solve z2(p2 + q2 + 1) = C2.

Solution: Let u = x + ay

∂z dz ∂u dz  = = ∂x du ∂x du   ∂z dz ∂u dz q= = =a  du  ∂y du ∂y p=

so that

Now given equation reduces to 2  dz 2  dz z2   +  a  + 1 = C2  du   du    2



dz z2 (1 + a2 )   = (C2 − z2 )  du  1

(1 + a2 )2



dz C2 − z2 =± du z

z du dz = ± 2 C −z 1 + a2

or

2

On integrating both sides, we get

− (C2 − z2 ) = ±

u +b 1 + a2

or

C2 − z2 = m

Alternately: Put z dz = dZ

…(1)

2

z = Z or z2 = 2Z 2

⇒ Now

(x + ay) −b 1 + a2

dZ dZ dz = = zp dx dz dx

(using (1))

…(2) …(3)

dZ dZ dz = = zq dy dz dx Therefore, the given equation reduces to 2

2

 ∂Z  +  ∂Z  + 2Z = C2,  ∂x   ∂y  which is clearly of the form f(p, q, z) = 0 Now, let Z = f(x + ay) = f(u), when u = x + ay dZ dZ du dZ dZ  and = = ⋅1= dx du ∂x du du   dZ dZ du dZ dZ = = ⋅a= a  dy du dy du du 

…(4)

…(5)

Partial Differential Equations

707

Therefore (4) reduces to 2

 dZ  (1 + a2 ) + 2Z = C2   du 



(1 + a2 )1/2

dZ = (C2 − 2Z)1/2 du



(1 + a2 )1/2

dZ = du (C − 2Z)1/2 2



− 1 + a2 C2 − 2Z = u + b



(1 + a2)(C2 – 2z) = (x + ay + b2) which is the desired solution.

Example 31: Solve p(p2 + 1) + (b – z)q = 0.

[KUK, 2005]

Solution: The equation p(p2 + 1) + (b – z)q = 0 falls under non-linear partial differential equation of the type f(p, q, z) = 0 Take

u = (x + ay), so that

∂u ∂u = 1, =a ∂x ∂y

therefore

p=

∂z dz ∂u dz = = ∂x du ∂x du

and

q=

∂z dz ∂u dz = = a ∂y du ∂y du

…(1)

…(2)

Whence the given equation reduces to 2  dz  dz  dz =0   + 1 + a(b − z) du  du du 

or

2  dz  dz      + 1 + a(b − z) = 0 du  du 

I

II

From II,

dz = a(z − b) − 1 du



a(z − b) − 1 =u+c 1 a⋅ 2

or



1 2[a(z – b) – 1] 2 = (au + c)



4[a(z – b) – 1] = [a(x + ay) + c]2

Example 32: Solve z2 = l + p2+ q2.

dz = du a(z − b) − 1

…(3)

708

Engineering Mathematics through Applications

∂z dz ∂u dz  = = , ∂x du ∂x du   ∂z dz ∂u dz q= = =a  du  ∂y du ∂y

Solution: Let u = (x + ay) so that p =

With the above values of p and q, the given equation reduces to 2

dz dz z2 = 1 +   +  a   du   du  dz (z2 − 1) = (1 + a2 )    du 

or ⇒

1 + a2





2

dz = z2 − 1 du

dz = z2 − 1

cosh−1 z =



2

1 ∫ du 1 + a2 u+c 1 + a2

 x + ay + c  . z = cosh   1 + a2 

or

III. Variable Separable Form or f1(x, p) = f2(y, q) – Third Standard Form Let each side of this equation be equal to an arbitrary constant i.e., f1(x, p) = f2(y, q) = a, (say) Solve above relations for p and q, p = F1(x) and q = F2(y),

dz =

then

∂z ∂z dx + dy becomes ∂x ∂y

dz = F1(x)dx + F2(y)dy which on integration results in

z = ∫ F1(x) dx + ∫ F2 (y) dy as the required complete solution. Example 33: Solve (i) yp + xq + pq = 0

[MDU, 2009]

(ii) yp = 2xy + log q.

Solution: (i) The given equation can be written like yp + xq = – pq ⇒

or

 y   q  =  −1 −

y x + = −1 q p

x = a (say) p 

…(1)

Partial Differential Equations

709

So this clearly falls under category f1(x, p) = f2(y, q)

y y   = a, q= ,  q a  ⇒ x  x  p= −1− =a  −1 − a  p 

whence

…(2)

Now we know that for z(x, y),

y ∂z ∂z x  dx + dy or dz =  dx + dy   −1 − a  ∂x ∂y a

dz =

y2 x2 + +c 2(−1 − a) 2a

z=



⇒ 2z = −

y2 x2 + +b (−1 + a) a

(ii) The given equation can be written as

p = 2x + ⇒

1 log q y

(p − 2x) =

1 log q = a y

(say)

…(1)

which is clearly of the form f1(x, p) = f2(y, q) ∴

p − 2x = a,  1 log q = a  ⇒  y

Whence

or or

p = (a + 2x),  q = eay 

…(2)

∂z ∂z dx + dy becomes ∂x ∂y dz = (a + x)dx + eay dy dz =

1 ay e +b a az = ax2 + a2x + eay + ab, the desired solution of given equation. z = ax + x2 +

Example 34: Solve (i) p + q = sin x + sin y, Solution: (i) Rewrite p + q = sin x + sin y as: (p – sin x) = (sin y – q) = a (say), where a is an arbitrary constant. Now from (1), we have

p − sin x = a, sin y − q = a

}

∂z = (a + sin x), x ∂ ⇒ ∂z  = (sin y − a)  ∂y 

(ii)

p+ q =x+y …(1)

…(2)

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Engineering Mathematics through Applications

Now we know that dz = On using (2), or

∂z ∂z dx + dy ∂x ∂y

…(3)

dz = (a + sin x) dx + (sin y – a) dy

z = a (x – y) – (cos x + cos y + b)

as the desired solution. (ii) The given equation can be written as:

p − x = y − q = a (say) p − x = a  y − q = a 



Now

dz =

∂z = (a + x)2  a + x = p  ∂x ⇒  ⇒ ∂z  (y − a) = q  = (y − a)2  ∂y 

…(4)

∂z ∂z dx + dy, on using (4) becomes ∂x ∂y

dz = (a + x)2 dx + (y – a)2 dy ⇒

3z = (a + x)3 + (y – a)3 + b

which is the desired solution. Example 35: Solve z(p2 – q2) = (x – y). Solution: The given equation can be written as: 2

Putting

2  z ∂z  −  z ∂z  = (x − y)    ∂x  ∂y 

…(1)

2 3 z dz = dZ so that Z = z2 3

…(2) 2

2  ∂Z   ∂Z  Thus the given equation reduces to   −   = (x − y)  ∂x   ∂y 

or

(P2 – Q2) = (x – y); where P =

∂Z ∂Z , Q= ∂x ∂y

or

(P2 – x) = (Q2 – y) = a (say), Third Standard Form

…(3)

so that

P= a+ x (P2 − x) = a  ⇒ (Q2 − y) = a  Q= a+y

…(4)

whence

dZ = Pdx + Q dy

Partial Differential Equations

i.e.,

dZ = a + x dx + a + y dy Z=

or

711

…(5)

2 2 (a + x)3/2 + (a + y) + b 3 3

z3/2 = (a + x)3/2 + (a + y)3/2 + c,

or

where a and c are two arbitrary constants. IV. Fourth Standard Form: Equation of the Form z = px + qy + f(p, q) The solution of this equation is z = ax + by + f(a, b) which is obtained by replacing p and q by arbitrary constants a and b respectively in the

  ∂z ∂z = a and q = = b for z(x, y) given equation.  i.e., p = ∂x ∂y   This equation is analogous to Clairut’s ordinary differential equation y = px + f(p), where

p=

dy and has solution y = ax + f(a), i.e. replace p by ‘a’. dx

Example 36: Solve z = px + qy − 2 pq . Solution: As the given equation is Fourth standard Form Whence complete integral of the given equation is

z = ax + by − 2 ab

for z(x, y)

…(1)

For singular Integral, differentiate (1) partially with respect to a and b, we have

2 ⋅ b, ab   2 ⋅a  0=y− ab  0=x−



b  , a   a y= b

x=

On eliminating a and b, the singular solution is xy = 1. Miscellaneous Problems Example 37: Obtain the complete solution of the equation (x – y)(px – qy) = (p – q)2. Solution: Let (x + y) = u and xy = v so that ∂z ∂z ∂u ∂z ∂v p= = + ∂x ∂u ∂x ∂v ∂x

∂z ∂z  = ⋅1 + ⋅y  ∂u ∂v 

as

∂u ∂v = 1, =y ∂x ∂x

…(2)

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Engineering Mathematics through Applications

Similarly q =

∂z ∂z ∂u ∂z ∂v = + ∂y ∂u ∂y ∂v ∂y ∂z ∂z  ∂u ∂v =  1+ = 1, =x x as  ∂u ∂v  ∂y ∂y

Now on substituting the values of

∂z ∂z in the given equation, we get and ∂x ∂y

 ∂z ∂z ∂z ∂z    (x − y)  x  + y  − y  + x   =  ∂z + y ∂z  −  ∂z + x ∂z    ∂u   ∂u ∂v  ∂v   ∂v   ∂u ∂v     ∂u

 ∂z ∂z ∂z ∂z ∂z ∂z  + xy −y − xy  =  y − x   (x − y)  x  ∂u ∂v ∂u ∂v    ∂v ∂v  



(x − y)(x − y)



∂z  ∂z  =  ∂u  ∂v 

⇒ ⇒

∂z ∂z = (x − y)2    ∂u ∂v 

2

2

2

2

∂z ∂z , Q= ∂u ∂v This relationship is comparable to Ist category, where f(p, q) = 0 Therefore in finding its solution, replace P by a and Q by b in it i.e. P − Q2 = 0

where P =

a – b2 = 0 or a = b2 Whence

z = au + bv + c z = a(x + y) + bxy + c z = b2(x + y) + b(xy) + c.

Example 38: Solve pq = xmynzl. Solution: In the given equation, on putting

xm + 1 = X, m+1

yn + 1 =Y n+1

…(1)

we get

∂z ∂z ∂X ∂z  = ⋅ = xm ∂x ∂X ∂x ∂X   ∂z ∂z ∂Y ∂z  = ⋅ = yn q= ∂y ∂Y ∂y ∂ Y  p=

and

…(2)

Partial Differential Equations

713

On substituting (2), the given equation reduces to

∂z ∂z = zl ∂X ∂Y

i.e. P · Q = zl

which is of the form f(z, p, q) = 0 ∴ Putting z = f(u) where u = (X + aY) ∂z dz ∂u dz so that = = ∂ Y du ∂ X du ∂z dz ∂u dz = =a and ∂ Y du ∂ Y du Equation (3) becomes, 2

dz a   = zl  du 



l

or z 2 dz =

…(3)

…(4)

1 du a

which on integration implies l − +1 2

z

l − +1 2 l − +1 2

z or



l +1 2

=

u +b a

=

yn + 1  1  xm + 1 +a +b  a m+1 n + 1

…(5)

Example 39: Solve q2y2 = z(z – px). Solution: On rewriting the given equation as 2

 ∂z  ∂z    y ∂ y  = z  z − x ∂ x  dx = dX   x  so that X = log x  Y = log y dy = dY   y ∂z ∂z ∂Y ∂z 1 = = ∂y ∂Y ∂y ∂Y y

Let

Therefore,

∂z ∂z  = ∂ y ∂ Y   ⇒ ∂z ∂z  x = ∂ x ∂ X  Likewise, on using (3), (1) becomes

…(1) …(2)

y

…(3)

2

 ∂z  = z  z − ∂z      ∂Y  ∂X  which is of the form f(p, q, z) = 0.

…(4)

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Engineering Mathematics through Applications

The obvious solution for this is z = f(u), u = X + aY

…(5)

∂z dz ∂ u dz = = ⋅1, ∂ X du ∂ X du  

Therefore

∂z dz ∂u dz = = a ∂ Y du ∂ Y du On using (6), (4) becomes

 

…(6)

2

dz dz a2   + z − z2 = 0  du  du

which is a quadratic in

dz and therefore, du

dz − z ± z2 + 4a2 z2 z = = 2  − 1 ± 1 + 4a2  2a2 2a du 2a2



dz  = − 1 ± 1 + 4a2  du z 



2a2 log z =  − 1 ± 1 + 4a2  (u + log c)



2a2 log z =  − 1 ± 1 + 4a2  (X + aY + log c) as u = X + aY



2a2 log z =  − 1 ± 1 + 4a2  (log x + a log y + log c) 2 log z2 a =  − 1 ± 1 + 4a2  (log xy ac)



z2 a = ( xyac )



− 1± 1+ 4a2

2

as the desired solution.

Example 40: Solve z2(p2 + q2) = (x2 + y2). Solution: On re-writing the above equation as 2

2  z ∂ z  +  z ∂ z  = (x2 + y2 )    ∂y  ∂x

Let z dz = dZ Now

z2 2 ∂Z ∂Z ∂z ∂z = =z = P (say) ∂x ∂z ∂x ∂x so that Z =

∂Z ∂Z ∂z ∂z = =z = Q (say) ∂y ∂z ∂y ∂y Whence the given equation reduces to P2 + Q2 = x2 + y2 ⇒ (P2 – x2) = (y2 – Q2) = a (say) which clearly falls under the category

…(1) …(2) …(3)

and

…(4)

Partial Differential Equations

715

f1(x, p) = f2(y, q) = a (say) ∴ From (4), we get and

P2 − x2 = a  P2 = a + x2   ⇒  2 2 y −Q = a Q2 = y2 − a 



dZ = P dx + Q dy,



dZ = x2 + a dx + y2 − a dy



x 2 a x + a + log  x + x2 + a  2 2

Z=

+



…(5)

y 2 a y − a − log  y + y2 − a  + b 2 2

 x + x2 + a  z2 = x x2 + a + y y2 − a + a log   + 2b 2  y + y − a

(

)

which is the desired solution. Example 41: Solve z = px + qy + c (1 + p2 + q2 ) . Solution: Clearly this equation is of the form z = px + qy + f(p, q) analogous to Clairaut’s form. ∴ Complete solution is z = ax + by + c (1 + a2 + b2 )

…(1)

Singular Integral: Differentiating (1) partially with respect to ‘a’ and ‘b’ respectively,



or

0=x+

ac (1 + a2 + b2 )

…(2)

0=y+

bc (1 + a2 + b2 )

…(3)

(x2 + y2 ) =

c2 (a2 + b2 ) (1 + a2 + b2 )

(x2 + y2 ) =

c2 + c2(a2 + b2 ) − c2 (1 + a2 + b2 )

(x2 + y2 ) =

c2 + (1 + a2 + b2 ) − c2 (1 + a2 + b2 )

(c2 − x2 − y2 ) =

c2 (1 + a2 + b2 )

716

or

Engineering Mathematics through Applications

(1 + a2 + b2 ) 1 = 2 (c − x2 − y2 ) c2 Also from (2) and (3), a = −

b=

…(4)

x 1 + a2 + b2 = c

x c − x2 − y2 2

−y 1 + a2 + b2 y =− 2 (using (4)) c c − x2 − y2

Putting these values of ‘a’ and ‘b’ in (1), the singular integral is

z=

= or

− x2 − c2 − x2 − y2

y2 + c2 − x2 − y2

c2 c2 − x2 − y2

− c 2 − x 2 − y2 = c2 − x2 − y2 c 2 − x 2 − y2

(x2 + y2 + z2) = c2.

ASSIGNMENT 5 Solve 1. q2 = z2p2(1 – p2) 3. q(p – cos x) = cos y

2. z = px + qy + p2 q2 4. (pa – p – q)(z – px – qy) = pq.

11.7 CHARPIT’S METHOD It is a general method due to Charpit for solving non-linear equations of first order. When it is difficult to solve such equations under any of the standard forms (as discussed in previous article) then this method is employed to find the complete integrals. Let the given equation be f(x, y, z, p, q) = 0 …(1) If we succeed to find another relation F(x, y, z, p, q) = 0 …(2) satisfied by p and q, then we can solve equation (1) and (2) for p and q Since z consists of two independent variables x and y. ∴ dz = p dx + q dy, For determining F, differentiate (1) and (2) with respect to x and y respectively giving

∂f ∂f ∂f ∂p ∂f ∂q  + p+ + =0  ∂x ∂z ∂p ∂x ∂q ∂x   ∂F ∂F ∂F ∂p ∂F ∂q + p+ + = 0  ∂x ∂z ∂p ∂x ∂q ∂x 

…(4)

Partial Differential Equations

∂f ∂f ∂f ∂p ∂f ∂q  + q+ + = 0 ∂y ∂z ∂p ∂y ∂q ∂y   ∂F ∂F ∂ F ∂p ∂ F ∂q  q+ + + =0  ∂y ∂z ∂p ∂y ∂q ∂y  Eliminating

717

…(5)

∂p from the first pair viz (4), we get ∂x

 ∂f ∂F ∂F ∂f   ∂f ∂F ∂F ∂f   ∂f ∂F ∂F ∂f  ∂q  ∂ x ∂ p − ∂ x ∂ p  +  ∂ z ∂ p − ∂ z ∂ p  p +  ∂ q ∂ p − ∂ q ∂ p  ∂ x = 0

…(6)

∂q Similarly on eliminating ∂ y from the second pair viz (5), we get  ∂f ∂F ∂F ∂f   ∂f ∂F ∂F ∂f   ∂f ∂F ∂F ∂f  ∂p  ∂y ∂q − ∂y ∂q  +  ∂z ∂q − ∂z ∂q  q +  ∂p ∂q − ∂p ∂q  ∂y = 0       Since

…(7)

∂q ∂p ∂2 z , = = ∂ x ∂ x∂ y ∂ y

whence the last terms in (6) and (7) are the same with opposite signs. Adding (6) and (7), we get

∂f  ∂F  ∂f ∂f  ∂F  ∂f ∂f  ∂F  ∂f  ∂F  ∂f  ∂F  ∂f +p  + +q  +  −p −q  + −  + −  =0  ∂z  ∂p  ∂y ∂z  ∂q  ∂p ∂q  ∂z  ∂p  ∂x  ∂q  ∂y  ∂x

…(8)

Clearly this is Lagrange’s equation (linear equation of first order) with x, y, z, p, q as independent variables and F as dependent variable. Thus, identical to Article 4.4, its solution will depend on solution of the subsidiary equations dp dq dy dz dx dF …(10) = = = = = ∂f ∂f ∂f ∂f ∂f ∂f ∂f ∂f 0 +p +q −p −q − − ∂x ∂z ∂y ∂z ∂p ∂q ∂p ∂q An integral of the above equations (10) which involves p or q or both may be taken as assumed relation (2). The more simple the integrals involving p or q or both derived from (10), the more easy to solve them for p and q and the given equation (1). Example 42: Using Charpit’s method find complete integral of pxy + pq + qy = yz. [KUK, 2002-03] Solution: Here f(x, y, z, p, q) = pxy + pq + qy – yz = 0 then by Charpit’s method, the auxiliary equation is

dp dq dy dφ dx dz = = = = = − fp − fq − pfp − qfq 0 fx + pfz fy + qfz

…(1)

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Engineering Mathematics through Applications

dp dy dx dz = = = = dz − (xy + q) − (p + y) − p(xy + q) − q(p + y) py + p( − y)

or

=

dq (px + q − z) + q (− y)

implying dp = 0 or p = a Putting p = a in (1), axy + aq + qy = yz or q(a + y) = y(z – ax)

q=



y(z − ax) (a + y)

Also we know that for z(x, y), dz = p dx + q dy On substituting the values of p and q from (2) and (3),

dz = a dx +

…(2)

…(3) …(4)

y(z − ax) dy (a + y)

dz − a dx  a  = 1 − dy (z − ax) a + y  

or

Integrating, log(z – ax) = y – a log(a + y) + log b (z – ax) = b ey(y + a)–a.

or

Example 43: Find complete integral of the equation p2x + q2y = z. [NIT Kurukshetra, 2008] Solution: Here f(x, y, z, p, q) = p2x + q2y – z = 0 By Charpit’s method, auxiliary equations are

…(1)

dp dq dy dφ dx dz = = = = = − fp − fq − pfp − qfq 0 fx + pfz fy + qfz i.e.

dp dq dy dx dz = = = = 2 2 2 − 2px − 2qy − 2(p x + q y) − p + p − q + q2

From above, we have

(p2 dx + 2px dp) (q2 dy + 2py dq) = p2 x q2 y

or

d(p2 x) d(q2 y) = 2 p2 x qy

On integration x = aq2y where a is an arbitrary constant.

…(2)

Partial Differential Equations

719

Putting value of p2 x from (2) in (1), 1/2

aq2y

+

q2y

= z or

 z  q=   (1 + a)y 

…(3)

1/2

 az  p=   (1 + a)x 

∴ From (1),

…(4) 1/2

 az  dz = p dx + d dy =    (1 + a)x 

Thus,

(1 + a)

or

1/2

 z  dx +    (1 + a)y 

dy

dy dz dx = a 1/2 + 1/2 1/2 z x y

(1 + a) z = ax + y + b

or

which is required complete integral. Example 44: Solve the equation z2 = pqxy.

[KUK, 2003-04, 2010]

Solution: z2 = pqxy may be written as z2 – pqxy = 0 …(1) then by Charpit’s method of solving non-linear equations, we have auxiliary equations as: dp dq dy dx dz = = = = qxy pxy pqxy + qpxy − pqy + p2z − pqx + q2z

From above, we get =

x dp + p dx y dq + q dy = (− pqxy + 2pzx) + pqxy (− pqxy + 2qzy) + qpxy

x dp + p dx y dq + q dy = 2pzx 2qzy

i.e.

…(2)

On integrating both sides,

∫ ⇒ ⇒

(x dp + p dx) = px



(y dq + q dy) qy

log px = log qyc

or px = qyc

y x Substituting this in equation (1), we get p = cq

 y 1  z z2 −  cq  q dx = 0 or q2 =    x c  y

2

…(3)

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Engineering Mathematics through Applications

 z q = b   y

or

…(4)

Again, putting this value of q in (3) for getting p,

 z z2 − p  b  xy = 0  y z2 – bxzp = 0 or

p=

1  z   b  x 1z z dx + b dy bx y

dz = p dx + q dy =

Now

…(5)

dy 1 1 dx dz = +b 2 b x y

or

log z = log x1/b + log yb + log a

or

z = ax1/ b yb = axc

1/ 2

or

× yc

–1/ 2

where b2 =

1 c

Example 45: Solve the equation q + xp = p2. Solution: The given equation is f(x, y, z, p, q) = (q + xp – p2) = 0 Its subsidiary equations are given by

…(1)

dp dq dy dφ dx dz = = = = = − fp − fq − pfp − qfq 0 fq + pfz fy + qfz dp dq dφ dy dx dz = = = = = 2p − x − 1 2p2 − xp − q p + 0 0 0

or Taking ∴

p2

i.e. Thus,

dq = 0 ⇒ q = c (constant) – xp – q = 0 becomes p2 – xp – c = 0

p=

dz = (p dx + q dy)

dz = ⇒

x ± x2 + 4c 2 x ± x2 + 4c dx + c dy, using (2) and (3) 2

x 1 2 z=  ± x + (2c1/2 )2  dx + c ∫ dy + d 2 2 



…(2)

…(3)

Partial Differential Equations

z=

721

2c1/2 x2  x 2 x  sinh−1 1/2  + cy + d ± x + 4c + 4 2 2 2c  Q 

a2 + x2 dx =

…(4)

x 2 a x a + x2 + sinh–1  2 2 a

Alternatively: By another possible combination, dp dy ⇒ log p = – y + log b or p = b e–y = p −1 Implying q = p2 – xp = b2 e–2y – x be–y, on using the above value of p.

Also



…(5) …(6)

dz = p dx + q dy, using (5) and (6) dz = be–y dx + (b2 e–2y – x be–y) dy dz = b(e–y dx – xe–y dy) + b2 e–2y dy

dz = b ⋅ d ( xe−y ) − Implying,

z = b xe− y −

b2 ⋅ d ( e− 2y ) 2

b2 − 2y +a e 2

Example 46: Solve p(p2 + 1) + (b – z)q = 0. Solution: Subsidiary equations under Charpit Method are

dp dq dy dx dz dF = = = = = − fp − fq − pfp − qfq 0 fx + pfz fy + qfz where

fp = 3p2 + 1  fq = (b – z)   fx = 0   fy = 0  fz = – q 

…(1)

…(2)

On using (2), equation (1) becomes

dp dq dy dx dz dF = = = = = 2 2 2 0 − (3p + 1) − (b − z) − p(3p + 1) − q(b − z) − pq − q ⇒

⇒ ⇒

p = aq On using (3) in the given equation, aq(a2q2 + 1) + (b – z)q = 0 q[q2 a3 + {–z + (a + b)}] = 0 either q = 0

or q2 a3 = (z – (a + b))

…(3)

722

or

Engineering Mathematics through Applications

q=

z − ( a + b) , p= a a

z − (a + b) a



dz = pdx + qdy



dz =

z − ( a + b) dx + a

…(4)

z − ( a + b) dy a a

dy dz dx = + z − (a + b) a a a

or

On integrating, we get

y (z − (a + b))1/2 x = + +b 1/2 a a a 2 z − (a + b) = [(ax + y) + C] / a a Example 47: Solve 1 + p2 = qz by Charpit’s method. Solution: Given (1 + p2 – qz) = 0 Here subsidiary equations are

…(1)

dp dq dy dx dz = = = = − 2p + z − 2p2 + qz 0 + p(−q) 0 + q(−q)

or

Taking or

dp dq = − pq − q2 log p = log qa



p = qa Putting this value of p in (1), we have a2q2 – z q + 1 = 0 which is a quadratic equation in q and on solving,

q= Now

…(3)

1  z ± z2 − 4a2 , p= z ± z2 − 4a2  2a2 2a 

dz = p dx + q dy  z ± z2 − 4a2 dz = a  2 a2 





…(2)

2a2

(

(

dz = (a dx + dy) z ± z2 − 4a2

)

)

z ± z2 − 4a2  dx dy +  2 a2

Partial Differential Equations

2a2 ⋅





2

2a

(z m

(z ±

z2 − 4a2 2

4a z m 



(z m z2 − 4a2

)

z2 − 4a2

)( z m

z2 − 4a2

) dz = (a dx + dy)

)

dz = (a dx + dy) rationalization

 taking ± both signs for the term ( z ± z2 − 4a1 )  

z2 − 4 a2  dz = 2(a dx + dy)

(

On integration (taking ± both signs for the term z m z2 − 4a2

{

723

(

z2 z 2 ± z − 4a2 − 2a2 log z + z2 − 4a2 2 2

)} = 2(ax + y) + b.

)

Example 48: Solve the equation p(q2 + 1) + (b – z)q = 0. Solution: Here subsidiary equations are

dp dq dy dx dz dF = = = = = − fp − fq − pfp − qfq 0 fx + pfz fy + qfz dp dq dy dx dz dF = = = = = 2 2 − (q + 1) − 2 pq + z − [p(q + 1) + q(2 pq − z)] − pq − q2 0

⇒ ⇒

dp dq = − pq − q2

or

log p = log q + log c

…(1)



p = cq On using (2), given equation becomes cq(q2 + 1) + (b – z)q = 0 or q[cq2 + c + b – z] = 0

…(2)

⇒ either cq2 + (b + c – z) = 0 or q = 0 Considering cq2 + (b + c – z) = 0, we get

…(3)

q=± Now

or

dz = p dx + q dy

dz = c



c

z − (b + c) z − (b + c) , p=±c c c z − (b + c) dx + c

dz = (c dx + dy) z − (b + c)

z − (b + c) dy c

…(4)

724



Engineering Mathematics through Applications

c

z − (b + c) = (cx + y + a) 1/ 2

4(cz – bc – c2) = (cx + y + a)2.

or

ASSIGNMENT 6 Solve 1. (p2 + q2)y = qz 3. 2(z + xp + yp) = yp2

2. 2zx – px2 – 2pxy + pq = 0

4. x2p2 + y2p2 = z.M 5. z2(p2 + q2) = x2 + y2.

[Hint: Put X = log x, Y = log y, Z = z ] [Hint: Put z2 = Z]

11.7 HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS An equation of the form …(1) (Dn + k1 Dn– 1 D' 1 + k2 Dn–2 D'2 + … + kn D'n)z = F(x, y) where D =

∂ ∂ and k , … , k all constants, is termed as a homogenous and D′ = 1 n linear ∂x ∂y

partial differential equation of nth order with constant coefficients. Alike to an ordinary linear differential equation, we can rewrite (1) as f(D, D´)Z = F(x, y) …(2) and its complete solution consists of two parts: one is called complementary function and the other is called particular integral. Complementary function is the solution of the equation f(D, D´)Z = 0 …(3) and particular integral is the particular solution of f(D, D')Z = F(x, y) obtained by giving particular values to the arbitrary constants in the general solution due to F(x, y). To Find Complementary Functions Take a simple case of 2nd order homogenous linear equation for finding complementary function and then extend it to higher order. Let

∂2 z ∂2 z ∂2 z + K + K =0 1 2 ∂x2 ∂x∂y ∂y2

be a second order equation which can be written in its simplified form as (D2 + K1 DD' + K2D'2) = 0 with its auxiliary equation as (D2 + K1 DD' + K2 D'2) = 0 giving D/D' = m1, m2 (say) as two of its roots. Case I: When the roots are real and distinct Equation (2) may be written as (D – m1D') + (D – m2D')z = 0

…(1) …(2) …(3)

…(4)

Partial Differential Equations

725

The solution of (D – m1D')z = 0 will satisfy equation (4), Now (D – m, D')z = 0 i.e., p – m1 q = 0, Lagrange’s linear equation with auxiliary equation, is dy dx dz = = 1 − m1 0

giving (y + m1 x) = a and z = b ∴ z = φ(y + m1 x) Similarly (4) will also be satisfied by (4) (D – m2D')z = 0 i.e., z = ψ(y + m2x) Hence in this case, the complete solution of (2) is z = φ(y + m1x) + ψ(y + m2x) Case II: When the roots are equal (repeated): Take m1 = m2 = m(say), so equation (2) becomes (D – mD')(D – mD')z = 0 …(5) Let (D – mD')z = u, then the above equation reduces to (D – mD')u = 0 which is again a Lagrange’s linear equation and has solution u = φ(y + mx) Now with this value of u, equation (D – mD')z = u becomes, p – mq = φ(y + mx)

dy dx dz Its auxiliary equation is 1 = − m = φ (y + mx) I

II

III

from which on considering I and II, we get (y + mx) = a and on considering I and II, we get dz = φ(a)dx i.e., z = φ(a)x + b or

z = x φ(y + mx) + φ1(y + mx) which is the complete solution.

Example 49: Solve the following homogenous linear partial differential equations (i) (D4 – 2D3D' + 2DD'3 – D'4)z = 0. (iii) 25r – 40s + 16t = 0.

∂4z ∂4z − = 0. ∂x4 ∂y4 (iv) r = a2 t.

(ii)

Solution: (i) Its auxiliary equation is m4 – 2m3 + 2m – 1 = 0, where D/D' = m ⇒

(m + 1)(m – 1)3 = 0 or m = – 1, 1, 1, 1 Hence the solution (the complementary function) is z = φ1(y – x) + φ2(y + x) + x φ3(y + x) + x2φ4(y + x) (ii) Here the given equation in its symbolic form is written as (D4 – D'4)z = 0 Its auxiliary equation is

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Engineering Mathematics through Applications

(m4 – 1) = 0

or

(m – 1)(m + 1)(m2 + 1) = 0

i.e., m = – 1, 1, i, – i ∴ The solution z = φ1(y + x) + φ2(y – x) + φ3(y + ix) + φ4(y – ix). (iii) The given equation is symbolic form is as follows: 25D2z – 40DD'z + 16D'2 z = 0 Its auxiliary equation is 25(D/D')2 – 40(D/D') + 16 = 0 i.e., 25m2 – 40 m + 16 = 0 where D/D' = m 4 4 or (5m – 4)2 = 0 ⇒ m = 5 , 5 which is a case of repeated roots ∴ The solution is z = φ1(5y + 4x) + x φ2(5y + 4x). (iv) The given equation r = a2t in its symbolic form is D2z – a2D'2 z = 0 Its auxiliary equation is m2 – a2 = 0 or m = ± a, which is a case of distinct roots. ∴ z = φ1(y + ax) + φ(y – ax). To Find Particular Integral Consider the symbolic form of the equation as f(D, D')z = F(x, y) For this, Particular Integral (P. I.) =

1 F(x, y) f (D, D′)

Case I: When F(x, y) = eax + by 1 1 P.I. = eax +by = eax +by , f(a, b) ≠ 0. f (D, D′) f (a, b) (i.e., replace D by a, D' by b)

…(1) …(2)

…(3)

Case II: When F(x, y) = sin (ax + by) or cos (ax + by),

P.I. = =

1 sin (ax + by) f (D, D′) 1 sin(ax + by), Provided f(–a2, –ab, –b2) ≠ 0. f (− a , − ab, − b2 ) 2

…(4)

(i.e., replace D2 = – a2, D'2 –b2, DD´= – ab) Case III: When F(x, y) = xm yn, where m, n are positive integer

1 xm yn = [ f (D, D′)] −1 xm yn …(5) f (D, D′) (i.e., expand [f(D, D')]–1 in ascending powers of D/D' and operate on xmyn term by term.) P.I. =

Partial Differential Equations

727

Case IV: When F(x, y) is any function of x and y.

P.I. = and

1 1 F(x, y) = F(x, y) (D − m1D′)(D − m2D′) … f (D, D′)

1 F(x, y) = ∫ F(x, c − mx)dx, where c = y + mx. D − mD′

Case V: When F(x, y) = eax + by U(x, y),

P.I. =

1 1 eax + by ⋅ U(x, y) = eax + by U(x, y) f (D, D′) f (D + a, D′ + b)

(i.e., replace D by D + a and D' by D' + b in f(D, D')) 2

3

Example 50: Solve (D3 + D2 D' – DD' – D' )z = ex cos 2y.

[KUK, 2000]

Solution: The given equation (D3 + D2 D' – DD'2 – D'3)z = ex cos 2y has auxiliary equation as (m3 + m2 – m – 1) = 0 or (m – 1)(m + 1)2 = 0 or m = 1, – 1, – 1 whence complementary function z = φ1(y + x) + φ2(y – x) + xφ3(y – x) Now

P.I. =

1 ex cos 2y D3 + D2D′ − DD′2 − D′3

=

 e2iy + e− 2i0ta  1 ex   2 3  2 D + D D ′ − DD ′ − D ′

=

1 1 {ex + 2iy + ex − 2iy } 3 2 2 D + D D′ − DD′2 − D′3

3

2

I

II

  1 eax + by ax + by = , f (a, b) ≠ 0  Discuss under, P.I. = f (D, D′) e f (a, b)   =

1 1 1 1 ex + 2iy + ex − 2iy 2 (1 + 2i + 4 + 8i) 2 (1 − 2i + 4 − 8i) (as in Ist, a = 1, b = 2i; in II, a = 1, b = – 2i)

=

1 1 1 1 ⋅ ex + 2iy + ⋅ ex − 2iy 2 5(1 + 2i) 2 5(1 − 2i)

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Engineering Mathematics through Applications

=

(1 + 2i) 1 1 (1 − 2i) ex + 2iy 1 ⋅ + ⋅ ex − 2iy 2 5 5(1 + 2i)(1 − 2i) 2 5(1 − 2i)(1 + 2i)

=

1  (1 − 2i) ex + 2iy (1 + 2i) ex − 2iy  +  25  2 2 

=

ex  e2iy + e− 2iy e2iy − e− 2iy  − 2i   25  2 2 

P.I. =

ex [cos 2y + 2 sin 2 y] 25

Therefore complete solution

z = φ1(y + x) + φ2 (y − x) + x φ3 (y − x) +

Example 51:

ex [cos 2y + 2 sin 2y] 25

2 ∂2 y 2 ∂ y = E sin pt. 2 −a ∂t ∂x2

Solution: The given equation in its symbolic form is written as (D2 – a2 D'2)y = E sin pt Corresponding A.E. is m2 – a2 = 0 i.e., m = ± a ∴

yC.F. = φ(t + ax) + φ2(t – ax) Now

P.I. =

…(1)

…(2)

1 E sin pt D − a2 D′ 2 2

1  using, P.I. = sin(ax + by), replacing D2 by − a2 , DD’ by provided 2 2  f (D , DD′, D′ )   2 2 f ( a , ab , b ) ≠ 0) − − −   i.e.,

P.I. =

E sin pt − p2

∴ Complete solution y = φ1(t + ax) + φ2(t − ax) +

…(3)

E sin pt (− p2 )

y = φ1(t + ax) + φ2(t – ax) – E sin pt/p2 2 2 Example 52: Solve ∂ z − ∂ z = E cos x ⋅ cos 2y . 2 ∂x ∂x∂y

Solution: The given equation can be expressed as (D2 – DD')z = cosx · cos2y

[NIT Kurukshetra, 2003]

Partial Differential Equations

729

Corresponding A.E. is m2 – m = 0, when D/D' = m ⇒ m = 0, m = 1 whence, C.F. = φ1(y) + φ2(y + x) 1 P.I. = cos x cos 2 y 2 Now (D − DD′ )

=

1 1 [cos(x + 2y) + cos(x − 2y)] 2 2 D − DD′ I

=

II

1 1 1 cos(x + 2y) + cos(x − 2y)  −1 − 2  2  −1 + 2

( When cos (ax + by) replace D2 by – a2, D'2 by – b2, DD' = – ab) P.I. =



1 1  cos(x + 2y) − cos(x − 2y)  2 3 

Therefore, complete solution,

z = φ1(y) + φ2 (y + x) +

1 1 cos(x + 2y) − cos(x − 2y) 2 6

2 2 Example 53: Solve ∂ z2 + ∂ z2 = cos mx cos ny . ∂x ∂y

Find complete solution if F(x, y) = cos x cos 2y. Solution: The given equation in its symbolic form is (D2 + D'2)z = cos mx cos ny =

1 [cos(mx + ny) + cos(mx − ny)] 2

Corresponding auxiliary equation is m2 + 1 = 0 i.e., m = ± i whence C.F. = φ1(y + ix) + φ2(y – ix) Now Replace i.e.,

P.I. =

D2 by – m2, DD' by – mn, D'2 by – n2;

P.I. = =



1 1 [cos(mx + ny) + cos(mx − ny)] 2 ( D2 + D′2 )

P.I. =

1 1 cos(mx + ny) + cos2 (mx − ny) − 2(m2 + n2 ) − 2(m2 + n2 ) 1 [cos(mx + ny) + cos(mx − ny)] − 2(m2 + n2 ) −1 cos mx cos ny (m2 + n2 )

…(1)

…(2)

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Engineering Mathematics through Applications

Therefore complete solution is

z = φ1(y + ix) + φ2 (y − ix) +

−1 cos mx cos ny (m + n2 ) 2

Further if m = 1, n = 2 as F(x, y) = cos x cos2y, we have

P.I. = −

1 1 cos x cos 2y = − cos x cos 2y 2 (1 + 2 ) 5 2

Hence complete solution in this case is

z = φ1(y + ix) + φ2 (y − ix) −

Example 54: Solve

1 cos x cos 2y 5

∂3 z ∂3 z ∂3 z –4 2 +4 = 4sin(2x + y). 3 ∂x ∂x ∂y ∂ x∂ y2

Solution: The symbolic form of the above equation is (D3 – 4D2D' + 4DD'2)z = 4sin (2x + y) Corresponding A.E. is m3 – 4m2 + 4m = 0 or m(m – 2)2 = 0 i.e., m = 0, 2, 2 whence C.F. = φ1(y) + φ2(y + 2x) + xφ3(y + 2x) Now

P.I. =

1 4 1 4 sin (2x + y)a = ⋅ sin (2x + y) 2 D(D − 4DD′ + 4D ′ ) D (− 4 + 8 − 4) 2

4 1 ⋅ sin (2x + y) D 0 Hence method fails. =

PI =



1 1 ⋅ 4 sin (2x + y) 2 (D + 2D′) D

(Here, differentiate f(D, D') twice with respect to D and multiply the numerator twice by x) i.e.,

P.I. =

x2 {−2 cos(2x + y)} = −x2 cos (2x + y) 2

∴ The complete solution is z = φ1(y) + φ2(y + 2x) + xφ3(y + 2x) – x2 cos (2x + y). Example 55: Solve

∂2z ∂2z ∂2z 6 + − = cos (2x + y). ∂x2 ∂x∂y ∂y2

Solution: The given equation is a homogenous linear partial differential equation of 2nd order and its symbolic form is (D2 + DD' – 6D'2)z = cos (2x + y) …(1) Auxiliary equation is

Partial Differential Equations

731

(m2 + m – 6) = 0 or m = 2, – 3. whence complementary function (C.F.) = φ1(y + 2x) + φ2(y – 3x)

…(2) …(3)

D2 = −4  1 P.I. = 2 cos(2x + y), DD´= −2 2 (D + DD′ − 6D′2 ) D´ = − 1 (i.e., on replacing D2 by a2, DD' by –ab, D'2 = – b2 provided φ(–a2, –ab, –b2) ≠ 0) Clearly it is a case of failure as φ(–a2, –ab, –b2) = 0. Therefore, we discuss it under category IV i.e., factorise φ(D, D') and apply these factors turn by turn 1 P.I. = cos(2x + y) ∴ (D − 2D′)(D + 3D′)

=

1 D + 3D′



cos(2x + c1 − 2x) dx as here y + 2x = c1

D − 2D ′



=

1 (cos c1) dx D + 3D′

=

1 1 (x cos c1 ) = x cos(y + 2x) (D + 3D′) D + 3D′



=

D + 3D’



= x cos(5x + c2 ) dx = =



x cos(y + 2x) dx = x cos(c2 + 3x + 2x) dx

³ y – 3x = c2

x sin(5x + c2 ) 1 cos(5x + c2 ) + 5 5 5

x 1 sin (2x + y) + cos(2x + y) 5 25

…(4)

∴ Complete Integral z = C.I. + P.I.

= φ1(y + 2x) + φ2 (y − 3x) +

1 x sin(2x + y) + cos(2x + y). 5 25

Alternately

P.I. = =

Now

1 1 F(x, y) = cos(2x + y) f (D, D′) (D − 2D′(D + 3D′)

…(5)

1 cos(2 x + y) D D     − 2 D′ +3 D´  D′   D′ 

…(6)

D 1 1 =m = where ′ D D D     ( m − 2)( m + 3)´ − 2 + 3   D′   D′ 

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Engineering Mathematics through Applications

1 1 1  −  5  (m − 2) (m + 3) 

=

(By Partial Fraction)

   1 1 1 =  −  5   D − 2  D + 3    D′     D′  D′  1 1 −  5  (D − 2D′) (D + 3D′) 

=

…(7)

Therefore on substituting (7) in (6), we get

P.I. =

 1  1 1 cos(2x + y) −  5D′  (D − 2D′) (D + 3D′)  1 5D′

=



cos(2x + y) dx −

(D − 2D´)

1 1 5 D´



cos(2x + y)dx

D + 3D´



=

1 1 cos(2x + c1 − 2x) dx − ∫ cos(2x + c2 + 3x) dx 5D′ 5D′

=

1 1 sin (5x + c2 ) x cos c1 − 5D′ 5D′ 5

= P.I. =

Example 56: Solve



x 1 1 cos(2x + y) − sin(2x + y) 5 D′ 25D′

replacing

c1 = y + 2x c2 = y − 3x

x 1 sin (2x + y) + cos(2x + y) 5 25

∂ 2z ∂ 2z ∂ 2z −4 + 3 2 = x + 3y . 2 ∂x ∂y ∂x ∂y

Solution: The given equation which is an homogenous linear partial differential equation of 2nd order can be written in its symbolic form as …(1) (D2 – 4DD' + 3D'2)z = (x + 3y)1/2 Its auxiliary equation is as (m2 – 4m + 3) = 0 or m = 1, 3 whence C.F. = φ1(y + x) + φ2(y + 3x) For Particular Integral,

P.I. =

1 1 (x + 3y)1/2 F(x, y) = (D − D′)(D − 3D′) f (D, D′)

…(2)

Partial Differential Equations

=

1 (D − 3D′)



733

(x + 3y)1/2 dx

D − D´

=

1 1  x + 3 ( c1 − x ) 2 dx (D − 3D′)

=

1 (D − 3D′)



since y + x = c1 for m = 1

1

∫ ( 3c1 − 2x)2 dx

   (3c1 − 2x)3/2  1 = (D − 3D′)  − 2 × 3    2   =

1  − 1 (x + 3y)3/2  ,  (D − 3D′)  3

=−

=− =−

=

1 3



On replacing c1 = y + x

(x + 3y)3/2 dx

D − 3D´



1 (x + 3c2 − 9x)3/2 dx, 3

as (y + 3x) = c2 for m = 3

1 (3c2 − 8 x)5/2 3 −8× 5 2

(3c2 − 8x)5/2 (x + 3y)5/2 , = 60 60

replacing c2 = y + 3x

…(3)

Here, complete solution

z = φ1(y + x) + φ2 (y + 3x) + Alternately: P.I. = =

Considering

(

(x + 3y)5/2 60

1 (x + 3y)1/2 (D − D′)(D − 3D′) 1 1 (x + 3 y)1/2 D ′2  D − 1  D − 3  D′   D′ 

1 1 D , = =m (m − 1)(m − 3) when D′ D D −1 −3 D′ D′

)(

)

…(4)

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Engineering Mathematics through Applications

=

1 1 1  − 2  (m − 3) (m − 1) 

   1 1 1 =  −  2   D − 3  D − 1    D′     D′ D′  1 1  − 2  D − 3D′ D − D′  Therefore using (5), we have =

P.I. = =

1  1 1  (x + 3y)1/2 −  2D′  D − 3D′ D − D′  1 2D′



(x + 3y)1/2 dx −

D − 3D´

1 2D′



(x + 3y)1/2 dx

D − D´



=

1 1 (x + 3c2 − 9x)1/2 dx − ∫ (x + 3c1 − 3x)1/2 dx 2D′ 2D′

=

1 1 (3c2 − 8x)1/2 dx − 2D′ 2D′

=



P.I. = − = =

P.I. =

∫ (3c − 2x) 1

1/2

dx

1 (3c2 − 8x)3/2 1 (3c1 − 2x)3/2 − 2D ′ − 8 × 3 2D ′ − 2 × 3 2 2

=−



…(5)

1 1 1 1 (3c2 − 8x)3/2 + (3c1 − 2x)3/2 24 D′ 6 D′ 1 1 1 1 (x + 3y)3/2 + (x + 3y)3/2 , 24 D′ 6 D′

3 1 (x + 3y)3/2 24 D′ 3 3 (x + 3y)5/2 (x + 3 y)3/2 dy = 24 24 3 × 5 2



(x + 3y)5/2 60

Hence the complete solution

z = φ1(y + x) + φ2(y + 3x) +

(x + 3y)5/2 . 60

(as c2 = y + 3x, c1 = y + x)

Partial Differential Equations

735

Example 57: Solve the equation 4r + 12s + 9t = e3x – 2y. Solution: For z(x, y), we know that r =

∂2 z ∂2 z ∂2 z , s = , t = for z as a function of x and y. ∂ x2 ∂ x∂ y ∂ y2

So the given equation becomes

4

∂2 z ∂2 z ∂2 z + 12 + 9 2 = e(3x − 2y) 2 ∂x ∂ x∂ y ∂y

(4D2 + 12DD' + 9D'2)z = e(3x – 2y)

or where

D=

…(1)

∂ ∂ and D′ = ∂x ∂y

Now for equation (1) A.E. is as follows: 4m2 + 12m + 9 = 0, where D/D' = m or

…(2)

3 3 m = − ,− 2 2

(2m + 3)2 = 0 i.e.,

whence C.F. = φ1(2y – 3x) + xφ2(2y – 3x) Now for obtaining particular integral, we have P.I.(z) =

[³ y + mx = c]

1 eax + by , provided f(a, b) ≠ 0 eax + by = f (D, D ′) f (a, b)

1 e(3x − 2y) is clearly a case of failure as D = a = 3 and 4D2 + 12DD′ + 9D′2 D' = b = – 2, f(a, b) = 0 for Therefore,

Here z =

P.I. =

=

=

1 3x − 2y 2 e 3   4 D + D´  2  1 3  4 D + D´  2 

2c

2

e3 x − 2 y dx

 D + 3 D´   2 

∫e 3 4  D + D′ 1



=



dx as y = c −



1 3 4  D + D′  2 

x e2c =

1 3 4  D + D´  2 

3 x 2

x e3 x − 2 y

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Engineering Mathematics through Applications

=

= ⇒

1 4



x e3x − 2y dx

 D + 3 D´   2 



1 1 x2 x e2c dx = ⋅ ⋅ e2c 4 4 2

3 x2 3x − 2y ; replace, e c =  y + x  8 2  Therefore the complete solution x2 z = φ1(2y − 3x) + x φ2 (2y − 3x) + e3x − 2y 8 P.I. =

Example 58: Solve

∂ 3z ∂ 3z ∂ 3z ∂ 3z −4 2 +5 − 2 3 = e2 x + y . 3 2 ∂x ∂x ∂y ∂ x∂ y ∂y

Solution: The symbolic form of the above equation is (D3 – 4D2D' + 5DD'2 – 2D'3)z = e2x + y The corresponding A.E. is m3 – 4m2 + 5m – 2 = 0 i.e., (m – 2)(m2 – 2m + 1) = 0 or

(m – 2)(m – 1)(m – 1) = 0 i.e., m = 1, 1, 2



C.F. = φ1(y + x) + x φ2(y + x) + φ3(y + 2x)

1 e2x + y D3 − 4D2 D′ + 5DD′2 − 2D′3 It is a case of failure as f(a, b) = 0 for D = a = 2 and D' = b = 1 Therefore, Now

P.I. =

P.I. =

1 e2x + y (D − 2D′)(D − D′)2

P.I. =

1 1 ⋅ e2 x + y D − 2D′ (2 − 1)2

=

1 e2x + y D − 2D′

= ∫ e2x + c − 2x dx, as y + 2x = c = ∫ ec dx = x ec = x ey + 2x ⇒

P.I. = x e2x + y Therefore the complete solution, z = φ1(y + x) + x φ2(y + x) + φ3(y + 2x) + x e2x + y.

Partial Differential Equations

737

Example 59: Solve (D2 – DD' – 2D'2)z = (y – 1)ex. Solution: Corresponding A.E. is (m2 – m – 2) = 0 i.e., (m – 2)(m + 1) = 0

…(1)

or whence

…(2)

Now

m = 2, – 1 C.F. = φ1(y + 2x) + φ2(y – x) 1 P.I. = 2 (y − 1) ex D − DD′ − 2D′2 1 (y − 1) ex = (D − 2D ′)(D + D′)

=

1 (D − 2D′)



(y − 1) ex dx

D + D′

Corresponding to the factor D + D', y = c1 + x ⇒

P.I. =



1 (c1 + x − 1) ex dx (D − 2D ′)

∫ ((c − 1) e

+ x ex ) dx

=

1 (D − 2D′)

=

1 (c1 − 1) ex + (x − 1) ex  D − 2D′ 

=

1 (y − 2) ex  ; − D 2D′ 

P.I. =



1

x

replacing, c1 = (y – x)

(y − 2) e x dx

D − 2D′

Expressing (y – 2) in terms of x as y + 2x = c2 corresponding to the factor (D – 2D') ⇒

P.I. =

∫ (c

2

− 2x − 2 ) ex dx =

∫ ((c

2

− 2) − 2x ) ex dx

= (c2 – 2) ex – 2 (x – 1) ex P.I. = yex, on replacing c2 by (y + 2x) Therefore complete solution z = φ1(y + 2x) + φ2(y – x) + yex. Example 60: Solve (r + s – 6t) = y cos x. Solution: The given equation can be written as (D2 + DD' – 6D'2)z = y cosx Corresponding A.E. is (m2 + m – 6) = 0 ⇒ (m + 3)(m – 2) = 0 i.e., m = – 3, + 2 ∴

C.F. = φ1(y – 3x) + φ2(y + 2x)

…(1)

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Engineering Mathematics through Applications

Now

P.I. =

1 y cos x D + DD′ − 6D′2 2

1 y cos x (D + 3D′)(D − 2D′) Now apply these operators viz., (D + 3D') and (D – 2D') turn by turn on y cosx as a function of x only =

1 (D + 3D ′)

=



y cos x dx

D − 2D′



1 (c1 − 2x)cos x dx , Here y + 2x = c1 (D + 3D′) 1 = [(c1 − 2x)sin x − 2 cos x ] replace c1 = y + 2x (D + 3D′) =

=

1 (y sin x − 2 cos x) (D + 3D ′)

=



y sin x dx −





(c2 + 3x) sin x dx − 2

D + 3D′

=

D + 3D′

D + 3D′

2 cos x dx





D + 3D´

cos x dx y – 3x = c2



= − (c2 + 3x)cos x − 3 (− cos x) dx − 2 cos x dx



P.I. = − y cos x + cos x dx = − y cos x + sin x,

replacing (c2 + 3x) by y

z = φ1(y – 3x) + φ2(y + 2x) – y cos x + sin x as complete solution.



Example 61: Solve (D2 + 3DD' + 2D'2)z = 24 xy. Solution: Here Auxiliary Equation (A.E.) is (m2 + 3m + 2) = 0 where m = D/D' ⇒ ⇒

…(1)

m2 + 2m + m + 2 = 0 (m + 2)(m + 1) = 0

∴ Now

i.e., m = – 1, – 2

…(2)

C.F. = φ1(y – x) + φ2(y – 2x)

…(3)

P.I. = =

1 24xy D + 3DD′ + 2D′2 2

1 24xy  3  D ′ 2D′2   2 D 1 +  + 2  D    D

Partial Differential Equations

−1

=

1 D2

  3D′ D′2   1 +  D + 2 D2   24xy  

=

1 D2

1 + 1 − 3 D′  24xy  D  D′ since in the term xy, y is in power one only) D

(On leaving higher order terms of

= ⇒

739

1 D′ 24xy − 3 ⋅ 3 24xy 2 D D

P.I. = 4x3y – 3x4 Therefore the complete solution is z = φ1(y – x) + φ2(y – 2x) + 4x3y – 3x4.

3 3 Example 62: Solve ∂ z − ∂ z = x3y3. ∂x3 ∂y3

[NIT Kurukshetra, 2007]

Solution: The given equation can be written as (D3 – D'3)z = x3y3 Corresponding auxiliary equation is m3 – 1 = 0 i.e., (m – 1)(m2 + m + 1) = 0 (m – 1)(m – w)(m – w2) = 0,

or

for w to be the cube root of unity

C.F. = φ1(y + x) + φ2(y + wx) + φ3(y +



w2x)

…(1)

…(2) …(3)

Now P.I. =

1 x3 y3 = D3 − D ′3

1 x3 y3 3   D′ D3 1 −     D  

−1

  D′  3  3 3 D ′3  3 3 1  1 −    x y = 3 1 + 3 + … x y D D D    

=

1 D3

=

1 3 3 D′3 3 3 xy + 6 xy D3 D

D′ to the powers to which y is appearing in F(x, y) = xnym i.e. neglecting terms D containing powers more than 3 in D´) (On taking

=

x6 y3 6x9 + 4⋅5⋅6 4⋅5⋅6⋅7⋅8⋅9

740

Engineering Mathematics through Applications

x6 y3 x9 + 120 10080 Hence complete solution P.I. =

z = φ1(y + x) + φ2 (y + wx) + φ3 (y + w2 x) +

Example 63: Solve

x6 y3 x9 + 120 10080

∂ 2z ∂ 2z ∂ 2z + − 6 2 = x2sin (x + y). 2 ∂x ∂y ∂x ∂y

Solution: The given equation in its symbolic form is written as (D2 + DD' – 6D'2)z = x2 sin (x + y) sin(x + y) = imaginary part of

ei(x + y);

since we know that

eiθ

…(1)

= cos θ + i sin θ Real Part

Imaginary Part



f(D, D')z = Im. ei(x + y) · x2 Its auxiliary equation is m2 + m – 6 = 0 i.e., m = 2, – 3 ∴ C.F. = φ1(y + 2x) + φ2(y – 3x) Now

P.I. = Im⋅

…(2)

1 ei(x + y) ⋅ x2 D + DD′ − 6D′2 2

= Im⋅ ei(x + y) ⋅ x2 = Im⋅ ei( x + y) x2

1 x2 (D + i) + (D + i)(D′ + i) − 6(D′ + i)2 2

1 x2 D2 + 3iD + DD′ − 11iD′ − 6D′2 + 4 −1

P.I. = Im⋅ ei( x + y)

1   D2 3iD DD′ 11 6  1+  + − iD′ − D′2   x2  4   4 4 4 4 4

= Im⋅ ei(x + y)

1   D2 3i 6 9 2 2  2 DD′ 11 + D+ − iD′ − D′2 + 1− i D x   4  4 4 4 4 4 16

= Im⋅ ei(x + y)

1  2 1 3i 9 x − − 2x −  4  2 4 8

= Im⋅

1  2 13  3i   x −  − x  [ cos(x + y) + i sin (x + y)] 4  8 2 

1  13 3 P.I. =   x2 −  sin (x + y) − x cos(x + y)   8 8 4  Hence the complete solution

z = φ1(y + 2x) + φ2(y − 3x) +

1  2 13   3  x −  sin(x + y) −   x cos(x + y) 4 8 8

…(3)

Partial Differential Equations

741

MISCELLANEOUS PROBLEMS Example 64: Solve 4r – 4s + t = 16 log (x + 2y). Solution: The symbolic form of the above equation is (4D2 – 4DD' + D'2)z = 16 log (x + 2y) Corresponding A.E. is 4m2 – 4m + 1 = 0 i.e., ∴

…(1)

1 1 , 2 2 C.F. = φ1(2y + x) + xφ2(2y + x)

(2m – 1)2 = 0 or

P.I. =

m=

1 2 ⋅ 16 log (x + 2y) 1   4 D − D′  2 



1 log 2 c dx as 1 D − D′ 2 1 = 4⋅ ⋅ x log (x + 2y) 1 D − D′ 2 = 4 ⋅ ∫ x log 2c dx

= 4⋅

= 4⋅

( 2y + x ) = c

x2 ⋅ log 2c 2

= 2x2 log 2c(x + 2y) Hence the complete solution, z = φ1(y + 2x) + x φ2(y + 2x) + 2x2 log (x + 2y). Example 65: Find a real function V of x and y reducing to zero when y = 0 and satisfying

∂2V ∂2V + + 4π(x2 + y2 ) = 0 . ∂x2 ∂y2 Solution: Here the given function V(x, y) will be obtainable from the Particular Integral only, since C.F. = φ1(y + ix) + φ2(y – ix) is simply an imaginary function, if y = 0. 1 P.I. = 2 { − 4π(x2 + y2 )} Now D + D′2

=

=

1 { − 4π(x2 + y2 )} 2 D  ′  D2  1 + 2   D  −1 D′2   2 −4π  1 +   (x + y2 )   D2  D2  

742

Engineering Mathematics through Applications

=

− 4π  D′2  2 1 − (x + y2 ), D2  D2 

On neglecting higher powers of

D′2 D2

 1 D′2  = − 4π  2 (x2 + y2 ) − 4 (x2 + y2 ) D D     x4  y2 x2  x4 = − π 4  + −4×2×   1 ⋅ 2 ⋅ 3 ⋅ 4    3 ⋅ 4 1 ⋅ 2  =−

π x4 x4 − 2π x2 y2 + π 3 3

P.I. = – 2πx2y2 ∴

V(x, y) = – 2πx2y2.

Example 66: Solve r – t = tan3 x tany – tanx tan3 y. Solution: The given equation can be written in symbolic form as (D2 – D'2)z = tan3 x tan y – tan x tan3 y Corresponding auxiliary equation is (m2 – 1) = 0 i.e., m = 1, – 1

…(1)

C.F. = φ1(y + x) + φ2(y – x)

∴ Now

P.I. =

…(2)

1 (tan3 x tan y − tan x tan3 y) D2 − D′2

=

1 (tan3 x tan y − tan x tan3 y) (D + D′)(D − D′)

=

1 (D + D′)



(tan3 x tan(c1 − x) − tan x tan3 (c1 − x)) dx

here (y + x) = c1

D − D′

=

1 (sec2 x − 1)tan x tan (c1 − x) − tan x tan (c − x) {sec2 (c1 − x) − 1}  dx (D + D′) 

=

1  tan (c1 − x)tan x sec2 x − tan x tan (c1 − x)sec2 (c1 − x) dx (D + D′) 





=





1 1 tan (c1 − x)tan x sec2 x dx − tan x tan (c1 − x)sec2(c1 − x) dx D + D′ D + D′ I

[Integration by parts

II

I

viz., Ist function Int. II – ∫ (Diff. Ist) (Int. IInd) dx]

II

Partial Differential Equations

=

743

1  tan2 x 1  tan (c1 − x) sec2 (c1 − x) tan2 x dx  +  2 2 D + D′  



+

 tan2(c1 − x) 1 1  − tan x sec2 x tan2(c1 − x) dx   D + D′  2 2 



   tan2 (c1 − x)   tan2 x  2 = tan(c1 − x)sec2 (c1 − x)(−1) since d   = tan x sec x and d   2 2     ⇒

P.I. =

1 tan2 x tan(c1 − x) + tan x tan2 (c1 − x) 2(D + D′) 

∫ {(sec x − 1)sec (c

+

2

2

1

}

− x) − sec2 x (sec2 (c1 − x) − 1) dx  

∫ {sec x − sec (c − x)} dx ,

=

1  tan2 x ⋅ tan(c − x) + tan x ⋅ tan2(c − x) + 1 1 2(D + D′) 

=

1 tan2 x ⋅ tan y + tan x ⋅ tan2 y + (tan x + tan y)  , 2(D + D′) 

2

2

1

(on replacing (c1 – x) by y in all terms)

=

1  tan2 x ⋅ tan y + tan y + tan x + tan x ⋅ tan2 y  2(D + D′) 

=

1  tan y sec2 x + tan x sec2 y  2(D + D′) 

=

1 2

∫ {tan (c

2

+ x)sec2 x + tan x sec2 (c2 + x)} dx

D + D′

(As here y – x = c2 for the factor D + D' in f(D, D')) =





1 tan (c2 + x) sec2 x dx + tan x sec2 (c2 + x) dx  ,  2  I II (i.e. taking up 2nd integral only)





=

1 tan (c2 + x) sec2 x dx + tan x tan(c2 + x) − sec2 x tan(c2 + x) dx   2 

=

1 tan x tan (c2 + x) 2

1 tan x ⋅ tan y (replacing c2 + x, by y) 2 Therefore complete solution, P.I. =

744

Engineering Mathematics through Applications

z = φ1 (y + x) + φ2(y − x) +

1 tan x ⋅ tan y . 2

Example 67: A surface is drawn satisfying r + t = 0 and touching x2 + z2 = 1 along its section by y = 0. Obtain its equation in the form z2(x2 + z2 – 1) = y2(x2 + z2). Solution: The symbolic form of the given equation is (D2 + D'2)z = 0 or (D + iD')(D – iD')z = 0

…(1)



…(2)

C.F. (z) = φ1(y + ix) + φ2(y – ix) Now we know that

and

p=

∂z = i φ1′(y + ix) − i φ2′(y − ix) ∂x

…(3)

q=

∂z = φ1′(y + ix) + φ2′(y − ix) ∂y

…(4)

Also from the given x2 + z2 = 1 or

z = 1 − x2

∂z x =− ∂x 1 − x2 ∂z q= =0 and ∂y Now under the given condition that at y = 0, the surface touches x2 + z2 = 1 ∂z ∂z (i.e., at y = 0, the tangents ∂ x = p, ∂ y = q must be equal) Precisely, Thus,

p=

[i φ1′(y + ix) − i φ2′ (y − ix)]y = 0 = − and

…(5)

[φ1′(y + ix) + φ2′ (y − ix)]y = 0 = 0

or

2 i φ1′ (y + ix) =

or

φ1′(y + ix) =

−x 1 − x2

or φ1′(y + ix) =

y + ix 2 1 + (y + ix)2

x  1 − x2   

…(6)

…(7)

x , y=0 2 1 + (ix)2

(no matter, since y = 0)

…(8)

Integrating both sides, ⇒

1 1 + (y + ix)2 + a 2 Now from (7), we get φ'(y – ix) = – φ1'(y + ix) at y = 0 2 φ1(y + ix) =

=−

(y + ix) 1 1 ix  = − 2 2 1 + (y + ix) 2 1 − x2  at y = 0

…(9)

Partial Differential Equations

φ2′(y − ix) =

(y − ix) 1 2 1 + (y − ix)2

at y = 0

On integrating both sides, we get 1 1 + (y − ix)2 + b φ2 (y − ix) = 2 1 1 + (y + ix)2 + 1 + (y − ix)2 + c z= whence 2 where c = (a + b) Now on equating the two values of z as in (5) and (10), we get

{

}

{

745

…(10)

}

1 1 − x2 =  2 1 − x2 at y = 0 + c ⇒ c = 0 2 y = 0 1 1 + (y + ix)2 + 1 + (y − ix)2 z= Therefore 2

{



}

2z − 1 + (y + ix)2 = 1 + (y − ix)2 Squaring on both sides, we get 4z2 + 1 + (y + ix)2 − 4z 1 + (y + ix)2 = 1 + (y − ix)2 (z2 + ixy) = z 1 + (y + ix)2

or

Again squaring on both sides, we get z4 + 2ixyz2 + i2x2y2 = z2{1 + (y + ix)2} ⇒

z4 + 2ixyz2 – x2y2 = z2 + z2y2 – z2x2 + 2z2xyi



z2(z2 + x2 – 1) = y2(z2 + x2)

Example 68: Find a surface passing through the two lines z = x = 0, z – 1 = x – y = 0 satisfying r – 4s + t = 0. Solution: The given equation may be written as (D2 – 4DD' + 4D'2)z = 0 Its auxiliary equation is m2 – 4m + 4 = 0 or (m – 2)2 = 0 i.e., m = 2, 2 ∴

C.F. (z) = φ1(y + 2x) + x φ2(y + 2x) Since the above surface (2) passes through the lines z = x = 0 and z – 1 = x – y = 0 ∴ z = 0 = φ1(y + 2x) + 0 · φ2(y + 2x) i.e., and ⇒

φ1(y + 2x) = 0 z – 1 = 0 = x – y i.e., z = 1 and x = y

1 = φ1(y + 2x) + x φ2(y + 2x) On using (4), we get 1 1 3 3 3 or φ2 (y + 2x) = = φ2 (y + 2x) = = = x x 3x 2x + x 2x + y

…(1) …(2) …(3)

…(4)

746



Engineering Mathematics through Applications

3 2x + y Hence the required solution is 3 z=x or z(2x + y) = 3x 2x + y φ2 (y + 2x) =

…(5)

11.9 NON-HOMOGENOUS LINEAR EQUATIONS If the differential coefficients involving in the partial differential equation f(D, D') = F(x, y) are not of the same order than it is called ‘non-homogenous linear partial differential equation’. Alike homogenous linear p.d.e. its solution also consists of two parts viz. complementary function and particular integral. To Find Complementary Function For Complementary Function, factorise f(D, D') into factors of the form (D – mD' – α), say, so that (D – mD' – α)z = 0 or p – mq = az is solved for z. The subsidiary equations for above equation are dy dx dz = = 1 − m αz I

II

III

On considering I & II, we get dy dx i.e., (y + mx) = c1 = 1 −m From I and III, we get dx dz dz = or α dx = i.e., z = c2 eαx αz 1 z or z = eαx φ0(c1) = eαx φ(y + mx) Likewise, find solutions for other such factors and add them to get complementary function.

Example 69: Solve (D – D' – 1)(D – D' – 2)z = e2x – y + x. Solution: Here f(D, D') = (D – D' – 1)(D – D' – 2) On comparing it with f(D, D') = (D – m1D' – α1)(D – m2D' – α2) we get ∴

}

}

m1 = 1 m2 = 1 , α1 = 1 α2 = 2

C.F. = eα1x φ1(y + m1x) + eα 2 xφ2(y + m2 x)

Now

P.I.1 = =

1 1 F(x, y) = e2 x − y (D − D′ − 1)(D − D′ − 2) f (D, D′) 1 ⋅ e2 x − y (2 + 1 − 1)(2 + 1 − 2) (Replace D by a = 2 and D' by b = – 1, provided f(a, b) ≠ 0)

Partial Differential Equations

= P.I. 2 = =

1 2x − y e 2 1 x (D − D′ − 1)(D − D′ − 2) 1 1 −  1 − (D − D′) × −2  1 − (D − D′)   2

x

−1

=

1 1 [1 − (D′ − D)]−1 1 − (D − D′) x 2 2  

=

1 [1 + (D′ − D) + …] 1 + 12 (D − D′) + … x 2

1 D 1 13 x 3  = − 1+ D + +… x = x +  2 2 2 22 2 4 Therefore complete solution, =

z = ex φ1(y + x) + e2x φ2(y + x) +

e2x − y x 3 + + 2 2 4

Example 70: Solve the following equation (D + D' – 1)(D + 2D' – 3)z = (4 + 3x + 6y) Solution: Here f(D, D') = (D + D' – 1)(D + 2D' – 3) is non linear and comparable to (D – m1 D' – c1)(D – m2 D' – c2) where

}

m1 = −1 c1 = 1

and

}

m2 = − 2 c2 = 3

From (D – m1 D' – c1)z = 0, we have z = φ(y + m1x) ec1x = φ((y − x) e x

and from (D – m2D' – c2)z = 0, we get z = φ(y + m2 x) ec2 x = φ(y − 2x) e3 x

∴ Complementary Function (C.F.) = φ(y – x)ex + φ(y – 2x) e3x Now for Particular Integral

or

P.I.( z) =

1 1 (4 + 3x + 6y) F(x, y) = 2 2 (D + 2D′ + 3DD′ − 4D − 5D′ + 3) f (D, D′)

P.I.( z) =

1 (4 + 3x + 6y) 1   2 2 3 1 + (D + 2D′ + 3DD′ − 4D − 5D′) 3  

747

748

Engineering Mathematics through Applications

=

−1

1 1 1 + (D2 + 2D ′2 + 3DD ′ − 4D − 5D ′) (4 + 3x + 6y) 3  3 

1 1 1 − (− 4D − 5D′) + … (4 + 3x + 6y)  3 3  (Taking higher order terms as zero 1 4 5 = (4 + 3x + 6y) + D (3x) + D′ (6y) 3 3 3  =

=

1 4 + 3x + 6y + 4 + 10 3

18 + 3x + 6y = (x + 2 y + 6) 3 ∴ Complete solution = φ(y – x)ex + φ(y – 2x)e–2x + (x + 2y + 6). =

Example 71: Solve (2DD' + D'2 – 3D')z = 3 cos (3x – 2y). Solution: Here f(D, D') = (2DD' + D'2 – 3D') = D'(2D + D' – 3) ∴ Complementary function = e0y φ1(x) + e3y φ2(2y – x) Now

P.I. = =

1 3 cos(3x − 2y) 2DD′ + D ′2 − 3D′ (− 1) 3 cos(3x − 2y) (3D′ − 8)



2

f(D2, DD', D' ) = f(–a2, –ab, –b2) ≠ 0]

On replacing D2 = –a2 = 9, DD' = –ab = 6,

= −3 =

D'2 = –b2 = – 4

(3D′ + 8) cos(3x − 2y) 9D′2 − 64

3 (3D′ + 8)cos(3x − 2y) 100

3 [3 sin (3x − 2y) + 4 cos(3x − 2y)] 50 Therefore complete solution, =

z = φ1 (x) + e3y φ2 (2y − x) +

3 [4 cos(3x − 2y) + 3 sin(3x − 2y)] 50

Example 72: Solve r – s + p = 1. Solution: The given equation can be expressed in symbolic form as (D2 – DD' + D)z = 1 For complementary function, we write it as D(D – D' + 1)z = 0

Partial Differential Equations



749

C.F. = e0x φ1(y) + e–x φ2(y + x) Now

P.I. = =

1 ⋅1 D(D − D′ + 1) 1 1 1 (1 + D − D′)−1 ⋅ 1 = [1 − D + …]1 =  − 1 + … ⋅ 1 D D D 

=x–1 ∴ Complete solution is z = φ1(y) + e–xφ2(y + x) + x – 1. Example 73: Solve (D2 – D'2 – 3D + 3D')z = ex + 2y + xy. Solution: Rewriting the given equation as (D – D')(D + D' – 3)z = xy + ex + 2y comparable to (D – m1 D' – α1)(D – m2D' – α2)z = F(x, y) C.F. = eα1xφ1(y + m1x) + eα 2xφ2 (y + m2 x)

or

C.F. = e0x φ1(y + x) + e3x φ2(y – x)

³ m1 = 1; m2 = – 1

c1 = 0; c2 = 3 Now Particular Integrals corresponding to xy and ex + 2y are as follows:

P.I.1 =

=

1 1 xy = xy D′   D D′  (D − D′)(D + D ′ − 3)  1− − −3D 1 −  3 3 D −1  D′  1− 3D  D

−1

D D′  ⋅  1 − −   3 3

−1

xy

1 1 1 2 D′ 1 1 1 D′ = −  + + + D + D′ + DD′ + 2 + … xy 3 D 3 3 D 9 3 9 D  (On taking terms of D, D', DD' to the power 1 since in F(x, y), x and y are in power 1 only)

1 1 1 1 1 1 1 1 P.I.1 = −  x2 y + xy + x2 + y + x + + x3  3 2 3 3 9 3 9 6  P.I. 2 =

1 ex + 2y , (D − D′)(D + D′ − 3)

=

1 ex + 2 y (1 − 2)(D + D′ − 3)

=

1 ex + 2 y D + D′ − 3

= − ex + 2y

1 .1 [D + 1 + D′ + 2 − 3]

Replacing D by 1 and D´ by 2

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Engineering Mathematics through Applications

= − ex + 2y ⋅

1 .1 D + D′ −1

1 D′  ⋅1 1+ D D = – x ex+ 2y. Therefore complete solution z = φ1(y + x) + e3xφ2(y – x) –x ex + 2y. = − ex + 2 y ⋅

Example 74: Solve (D – 3D' – 2)2z = 2e2x tan (y + 3x). Solution: Here symbolic form of the equation (D – 3D' – 2)z = 2e2x tan (y + 3x) is comparable to f(D, D') = (D – 3D' – 2)2z = 2e2x tan(y + 3x) a non-homogenous linear equation with repeated factors (D – mD' – a). whence C.F. = e2x φ1(y + 3x) + xe2x φ2(y + 3x) Now

P.I. =

1 2e2x tan(y + 3x), (D − 3D′ − 2)

= 2e2 x

1 tan (y + 3x), (D + 2 − 3D′ − 2)

= 2e2x

1 tan (y + 3x) (D − 3D′)2



= 2e2 x ⋅

1 ⋅ tan c dx, D − 3D′

= 2e2 x ⋅

1 ⋅ x tan c = 2e2 x D − 3D′

= 2e2 x ⋅

(Replace D by (D + 2))

c = (y + 3x)



x tan c dx

D − 3D´

x2 tan (y + 3x) 2

P.I. = x2 e2x tan(y + 3x) Therefore the complete solution z = e2x φ1(y + 3x) + xe2x φ2(y + 3x) + x2e2x tan (y + 3x). Example 75: Solve (D2 + 2DD' + D'2 – 2D – 2D')z = sin (x + 2y). Solution: The symbolic form of the given equation is (D2 + 2DD' + D´2 – 2D – 2D')z = sin(x + 2y) Thus f(D, D') = (D + D')(D + D' – 2) is comparable to (D – m1D' – α)(D – m2D' – α2) so that C.F. = eα1x φ1(y + m1x) + eα2x φ2 (y + m2 x)

Partial Differential Equations

= e0x φ1(y – x) + e2x φ2(y – x)

751

³ m1 = 1, α1 = 0 m2 = 1, α2 = 2

Now

P.I. = =

1 sin(x + 2y) (D + 2DD′ + D2 − 2D − 2D′) 2

1 sin (x + 2y) (−1 + 2(− 2) + (− 4) − 2D − 2D′) (Replacing D2 = – 1, DD' = – 2, D'2 = – 4)

=

−1 sin (x + 2y) 2(D + D′) + 9

=

− 1[ 2(D + D′) − 9] sin (x + 2y) [ 2(D + D′) + 9][ 2(D + D′) − 9]

=

−1[ 2(D − D′) − 9] sin (x + 2y) 4[−1 + 2(−2) − 4] − 81

=

1  2D sin (x + 2y) − 2D′ sin (x + 2y) − 9 sin (x + 2y) 117 

−1  2 cos(x + 2y) − 9sin(x + 2y) 117  Hence the complete solution P.I. =

1 2 cos(x + 2y) − 9sin (x + 2y) 117 

z = φ1(y − x) + e2 x φ2 (y − x) −

Example 76: Solve

∂ 2z ∂ 2z ∂z − + = x2 + y2 . ∂x2 ∂x∂y ∂ x

Solution: The symbolic form of the above equation is (D2 – DD' + D)z = (x2 + y2) Here with So that Now

D, D') = D(D – D' + 1) comparable to = (D – m1D' – α1)(D – m2D' – α2), m1 = 0, α1 = 0; m2 = 1, α2 = – 1 C.F. = φ1(y) + e–x φ2(y + x)

P.I. = =

1 (x2 + y2 ) D(D − D′ + 1) 1 (x2 + y2 ) D [1 + (D − D′)]

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Engineering Mathematics through Applications

= =

1 [1 + (D − D′)]− 1 (x2 + y2 ) D

2 3 1 1 − (D − D′) + (D2 + D ′2 − 2DD ′) + (D3 − 3D2D′ + 3DD′ − D ′ ) (x2 + y2 )  D

=

1 1 D′ D′2  −1+ +D+ − 2D′ + D2 + 3D′2  (x2 + y2 )  D D D D 

 x3  =  + xy2 − x2 − y2 + 2yx + 2x − 2y − 2 − 6  3  =

x3 + xy2 − x2 − y2 + 2xy + 4x − 2y + 8 3

Therefore complete solution

z = φ (y) + e− x φ2(y + x) +

x3 + xy2 − x2 + 2xy + 4x − y2 − 2y + 8 3

Example 77: Solve (D2 – DD' + D' – 1)z = cos(x + 2y) + ey. Solution: Here f(D, D) = (D2 – DD' + D' – 1) = (D – 1)(D – D' + 1) On comparing it with (D – m1D' – α1)(D – m2D' – α2) we see m1 = 0, α1 = 0 m2 = 1, α2 = – 1; whence C.F. = ex φ1(y) + e–x φ2(y + x). Now P.I. = P.I.1 + P.I.2, where P.I.1 Particular Integral corresponding to cos (x + 2y) and P.I.2, Particular Integral corresponding to ey

P.I.1 =



1 cos(x + 2y) (D − DD′ + D′ − 1) 2

=

1 cos(x + 2y) , −1 − (−2) + D′ − 1

=

1 cos(x + 2y) D′

=

1 sin (x + 2y) 2

(Replace D2 = – 1, DD' – 2, D'2 = – 4)

1 ey D2 − DD′ + D′ − 1 It is a case of failure as f(D, D') = f(a, b) = 0 for a = 0 and b = 1 in ey. Therefore differentiate f(D, D') with respect to D and multiply F(x, y) by x.

and



P.I. 2 =

1 ey 2D − D′ Now replace D by a (= 0) and D' by b (= 1) P.I.2 = x ⋅

Partial Differential Equations

P.I. 2 =



1 x ey = −xey 0−1

z = φ1(y) + e− x φ2 (y + x) +

Therefore,

753

1 sin(x + 2y) − xey 2

Example 78: Solve (D3 – 3DD' + D' + 4)z = e 2x + y Solution: Here f(D, D')z = (D3 – 3DD' + D' + 4)z which can not be factorised in terms of D and D'. Let its solution be z = eax + by Implies (D3 – 3DD´ + D´ + 4)z = φ(a, b)e ax + by = (a3 – 3ab + b + 4)e ax + by (D3 – 3DD' + D' + 4)z = 0, if (a3 – 3ab + b + 4) = 0

then Whence

C.F. (z) = ∑ λ e ax + by , where (a3 – 3ab + b + 4) = 0

Now

1 e 2x + y D3 − 3DD′ + D′ + 4 e2x + y = 3 2 − 3(2) + 1 + 4 as here a = 2, b = 1

P.I. =

=

1 2x + y e 7

∴ Complete solution (z) = ∑ λ e ax + by +

1 2 x + y where (a3 – 3ab + b + 4) = 0 , e 7

ASSIGNMENT 7 Solve the following equations: 1.

∂2 z ∂2 z ∂z + + − z = e− x 2 ∂x ∂ x∂ y ∂ y

2. DD'(D + 2D' + 1)z = 0.

3. (D2 – DD' + D' – 1)z = cos (x + 2y) + ey.

4. (D2 – D')z = 2y – x2.

ANSWERS Assignment 1 1. (i) px + qy = 0,

(ii) p + q = mz,

(iii) q = px + p2

∂2 z ∂ z ∂z − =0 ∂ x∂ y ∂ x ∂ y

2. (i) x(y – z)p + y(z – x)q = z(x – y)

(ii) z

3. (i) z = px + qy + p2 + q2

(ii) px – qy = 0

754

Engineering Mathematics through Applications

(iii)

∂2 z ∂ z = ∂ x2 ∂ t

(iv) z(px + qy) = z2 – 1

Assignment 3

1 sin xy + y f (x) + φ (x) x2 4. 4 = e–t sin x + φ(x) + φ(y) 6. z = ex + y + φ(y) + φ(x)

x2 log y + axy + φ (x) + ψ(y) 2 3. z = sin x + eycos x 5. z = log x · log y φ(x) + φ(y)

2. z = −

1. z =

Assignment 4 1. f(xy, x2 + y2 + z2) = 0

x  y  2. f  2 + 2 , xyz1/3  = 0 x  z

3. f(x3 – y3, x2 – z2) = 0

4. f(lx + my + nz, x2 + y2 + z2) = 0

 y z − k 5. f  ,  =0 x x 

x−y z−u 1/3  , , ( z − u) ( x + y + z + u )  = 0 6. f  y−z y−z 

Assignment 5 1.

z2 = a2 + (x + ay + b)2 or z = b

3. z = ax + sin x +

2. z = ax + by + a2b2

1 sin y + c a

4. z = ax + by +

ab  ab − a − b 

Assignment 6 1. z2 = (a + bx)2 + a2y2

2. z = ay + b(x2 – a)

ax a2 b 3. z = y2 − 4y3 + y .

4. 4z = (a log x + 6 log y + c)2 wehre a2 + b2 = 1

(

{

}

{

5. z2 = b + x x2 + a2 + a log x + x2 + a2 + y y2 − a2 + a log y y + y2 − a2

})

Assignment 7 −x 1. z = e−x φ1(y) + ex φ2 (y − x) − xe 2

3. z = exφ1(y) + e−x φ2 (y + x) −

1 sin (x + 2y) − xey 2

2. z = φ1(y) + φ2(–x) + e–x φ3(y – 2x) 4 4. z = Σλ e ax + by − y2 − x 12