• Author / Uploaded
• Prakhar Cm

• Categories
• Sistema no lineal
• Verdad lógica
• Ecuaciones
• Análisis matemático
• Objetos matemáticos

Citation preview

LECTURE # II Section 07 -08 •Separable Equations and applications •Implicit solutions and General Solutions •Homogeneous Equations •Equations reducible to homogeneous form •Exact equations

Linear vs. Nonlinear Differential Equations An ordinary differential equation is said to be linear if it is linear in all the “unknowns”

2

df d f f, , dt dt 2

etc…..

Thus, a general n-th order, linear, ordinary differential equation will be of the form

n

df ( t ) + ....... + a1 (t ) (t ) = a0 (t ). an (t ) n dt dt d f

It is important to note that the coefficient functions a0(t), a1(t),…,an(t) need not to be linear functions of t. The following example should convey the general idea.

Example: Nonlinear differential equation 3

dy 2 t +t +y =e 3 dt dt

d y

2

is a third-order, nonlinear, ordinary differential equation for y(t). The nonlinearity of the equation arises because of the presence of the term y2 (NOT of t2, NOT of et) which is a quadratic function of the unknown function y(t).

Separable Equation and Applications The first-order differential equation

dy = H ( x, y ) dx

(1)

is called separable provided that H(x,y) can be written as the product of a function of x and a function of y.

dy g (x ) = g ( x )h( y ) = dx f (y) where h(y) = 1/f(y). In this case the variable x and y can be separated-isolated on opposite sides on an equation – by writing informally the equation

f ( y )dy = g ( x )dx

which we understand to be concise notation for the differential equation

dy f ( y ) = g (x ) dx

(2)

It is easy to solve this special type of differential equation simply by integrating both sides with respect to x:

∫

dy f ( y ( x )) dx = ∫ g ( x )dx + C dx

equivalently,

∫

f ( y )dy = ∫ g ( x )dx + C

(3)

All that is required is that the antiderivatives

F ( y ) = ∫ f ( y )dy and G ( x ) = g ( x )dx

∫

can be found. To see that Eqs. (2) and (3) are equivalent, note the following consequence of the chain rule: dy D x [F ( y (x ))] = F ′( y ( x )) y ′(x ) = f ( y ) = g ( x ) = Dx [G ( x )], dx

which in turn is equivalent to

F ( y ( x )) = G ( x ) + C , because two functions have the same derivative on an interval if and only if they differ by a constant on that interval.

Example 6 Solve the initial value problem

dy = −6 xy, y (0 ) = 7 dx

Solution

Informally, we divide both sides of the differential equation by y and multiply each side by dx to get

dy = −6 xdx y

Hence

dy ( ) = − 6 x dx ; ∫y ∫

ln y = −3x + C 2

We see from the initial condition y(0) = 7 that y(x) is positive near x = 0, so we may delete the absolute value symbols.

ln y = −3x + C 2

and hence

y(x ) = e

−3 x 2 +C

=e

−3 x 2

e = Ae C

−3 x 2

Where A = eC. The condition y(0) = 7 yields A = 7, so the desired solution is

y ( x ) = 7e

−3 x 2

Implicit Solution and General Solutions In general, the equation K(x,y) = 0 is called an implicit solution of a differential equation if it is satisfied (on some interval) by some solution y = y(x) of the differential equation. But note that a particular solution y = y(x) of K(x,y) = 0 may or may not satisfy a given initial condition. For example, differentiation of x2+y2 = 4 yields.

dy =0 x+ y dx so x2+y2 = 4 is an implicit solution of the differential equation x + yy ′ = 0 . But only the first of the two explicit solutions

y ( x ) = + 4 − x and y ( x ) = − 4 − x 2

satisfied the initial condition y(0) = 2

2

Warning : Suppose, however, that we begin with the differential equation

dy = g ( x )h( y ) dx

(1)

and divide by h(y) to obtain the separated equation

1 dy = g (x ) h( y ) dx

(2)

If y0 is a root of the equation h(y) = 0 that is, if h(y0) = 0- then the constant function y(x) = y is clearly a solution of Eq (1), but may not be contained in the general solution of Eq. (2). Thus solutions of a differential equation may be lost upon division by a vanishing factor.

HOMOGENEOUS EQUATIONS A function f (x, y) is said to be a homogeneous function of its variables of degree n if the identity f (tx, ty) ≡ tn f(x,y)is valid. For instance, the function , f ( x, y) = x2 + y2 – xy is a homogeneous function of the second degree, since f ( tx, ty ) = (tx)2 + (ty)2 – (tx) (ty) = t2 ( x2 + y2 – xy) = t2 f ( x, y)

The differential equation M ( x, y )dx + N ( x, y )dy = 0

(1)

is said to be homogeneous, if M & N are homogeneous function of the same degree. Equation (1) can be put in the form

dy = f ( x, y ) dx Where,

− M ( x, y ) f ( x, y ) = N ( x, y )

(2)

Working Rule 1. Put the given equation in the form

dy ⎛ y⎞ = f⎜ ⎟ dx ⎝ x⎠ 2. Let z = y / x

(2)

⇒ y = zx

3. Differentiating (2) w.r.t. x,

dz dy = z+x dx dx

(3)

4. Using (2) & (3), (1) becomes

dz z + x = f (z ) dx or ,

Or,

dz x = f (z ) − z dx

dx dz = x f (z ) − z

A solution of DE usually contains one or more arbitrary constants equal in number to the order of the equation.

Example 1 Solve

(x

3

)

(

)

+ 3 xy dx + y + 3 x y dy = 0 2

3

2

Solution 2

⎛ y⎞ 1 + 3⎜ ⎟ 3 2 dy x + 3 xy x⎠ ⎝ =− 3 =− 2 3 dx y + 3x y ⎛ y⎞ ⎛ y⎞ ⎜ ⎟ + 3⎜ ⎟ ⎝ x⎠ ⎝ x⎠ y Put z = ⇒ y = zx (dividing by x3) x

(1)

dy dz ∴ = z+x dx dx

(2)

from (1) and (2)

dz 1 + 3z z+x =− 3 dx z + 3z 2

or

dz 1 + 3z 1 + 3z + z + 3z −z=− x =− 3 3 dx z + 3z z + 3z 2

2

dz z + 6z +1 x =− 3 dx z + 3z 4

or,

2

dx z + 3z dz =− 4 2 x z + 6z +1 3

4

2

or,

dx 4 z + 12 z 4∫ = − ∫ 4 dz 2 x z + 6z +1 3

or

(

)

4 log e x = − log e z + 6 z + 1 + log e C or

4

[ (

2

)]

log e x = log e C / z + 6 z + 1 4

4

2

or,

(

)

x z + 6z + 1 = C 4

4

2

⎡ ⎤ y y ⎛ ⎞ ⎛ ⎞ 4 x ⎢⎜ ⎟ + 6⎜ ⎟ + 1⎥ = C ⎝ x⎠ ⎢⎣⎝ x ⎠ ⎥⎦ 2 2 4 4 ⇒ 6x y + x + y = C 4

2

Equations reducible to homogeneous form Equation of the form

dy ax + by + c a b = , where ≠ dx a ′x + b′y + c′ a ′ b′ can be reduced to homogeneous form by letting

x = X + h⎫ ⎬. y =Y +k⎭ where h & k are constants and X and Y are new variables.

Problem 5 Solve

dy x + y + 4 = dx x − y − 6

(A)

Solution Take

x= X +h

dy dY ⇒ so that = y =Y +k dx dX ∴ (A) becomes

(1)

dY X + h + Y + k + 4 X + Y + (h + k + 4 ) = = dX X + h − Y − k − 6 X − Y + (h − k − 6 )

Choose h,k so that

h + k + 4 = 0⎫ ⎬ h − k − 6 = 0⎭ solving (3) we get, h = 1, k = -5

(3)

(2)

∴ X = x − h = x −1 &Y = y − k = y + 5

(4)

Using (3) in (2), we get, 1+ Y

dY X + Y X = = dX X − Y 1 − Y X

(5)

Let

Y = v ⇒ Y = vX X

dY dv =v + X ∴ dX dX

dv 1 + v = v+ X dX 1 − v

dv 1 + v 1+ v − v + v 1+ v X = = −v = dX 1 − v 1− v 1− v 2

dX v ⎤ ⎛ 1− v ⎞ ⎡ 1 ∴∫ dv = ∫ ⎢ dv − =∫⎜ 2 ⎟ 2 2⎥ X ⎣1 + v 1 + v ⎦ ⎝1+ v ⎠

1 2v ⇒ log e X = tan v − ∫ dv 2 2 1+ v −1

2

(

)

1 2 ⇒ log e X = tan v − log 1 + v + C 2 −1

(

−1

⇒ log e X = tan v − log 1 + v

(

⇒ log e X + log e 1 + v

)

1 2 2

)

1 2 2

+C

−1

= tan v + c

⎡ ⇒ log e ⎢ X 1 + v ⎣

(

)

1 2 2

⎤ −1 Y ⎥ = tan X + C ⎦

2 ⎤ ⎡ Y −1 ⎛ y + 5 ⎞ ⇒ log e ⎢( x − 1) 1 + 2 ⎥ = tan ⎜ ⎟+C X ⎥⎦ ⎝ x −1 ⎠ ⎢⎣

⎡ X 2 +Y 2 ⇒ log e ⎢( x − 1) 2 X ⎢⎣

⎤ −1 ⎛ y + 5 ⎞ ⎥ = tan ⎜ ⎟+C ⎝ x −1 ⎠ ⎥⎦

⎡ ⇒ log e ⎢( x − 1) ⎢⎣

(x − 1)2 + ( y + 5)2 ⎤⎥ = tan −1 ⎛ y + 5 ⎞ + c ⎜ ⎟ 2 (x − 1) ⎝ x −1 ⎠ ⎥⎦

⎛ y + 5⎞ ⎡ = log tan ⎜ ⎟ e ⎢ ⎣ ⎝ x −1 ⎠ −1

(x − 1)

2

+ ( y + 5) ⎤ + C ⎥⎦ 2

Q.5 (b) Solve dy x + y + 4 = dx x + y − 6

Solution If

a b = , a ′ b′ then discover a substitution which leads to variable separable.

dy ( x + y − 1) + 5 = dx ( x + y − 1) − 5

Let z = x + y – 1

dz dy ∴ = 1+ dx dx dy dz = −1 dx dx

dz z +5 −1 = dx z −5

dz z + 5 z +5+ z −5 2z = ∴ = +1 = dx z − 5 z −5 z −5 z −5 ∴ dz = dx 2z

1 5 dz ⇒ dz − = dx 2 2 z dz ⇒ dz − 5 = 2dx z

on integration,

z − 5 log e z = 2 x + c ⇒ x + y − 1 − 5 log e ( x + y − 1) = 2 x + c ⇒ y − x = 5 log e (x + y − 1) + c′

Problem 4(a) If

ae ≠ bd

show that the constants h & k can be chosen in such away that the substitution x = z-h, y = w-k reduce

⎛ ax + by + c ⎞ dy ⎟⎟ = F ⎜⎜ dx ⎝ dx + ey + f ⎠ to a homogeneous equation

Solution Now, x = z-h y = w-k

(1)

∴ dx = dz & dy = dw dy dw ∴ = dx dz

(2)

⎡ a ( z − h ) + b (w − k ) + c ⎤ dw = F⎢ ⎥ dz ⎣ d ( z − h ) + e (w − k ) + f ⎦

⎡ az + bw + ( − ah − bk + c ) ⎤ =F⎢ ⎥ ⎢⎣ dz + ew + ( − dh − ek + f ) ⎥⎦

(3)

In order to make (3) homogeneous, choose h & k so as to satisfy -ah-bk+c=0

and –dh+f-ek=0

(4)

Solving (4), we get,

ce − bf af − cd h= &k = ae − bd ae − bd given

(5)

ae ≠ bd ⇒ ae − bd ≠ 0

∴ h & k given by (5) are meaningful, & k will exist.

i.e., h

Exact Equations Let

f ( x, y ) = C

be a family of curves Then its differential equation df=0

(1)

∂f ∂f dx + dy = 0 ∂x ∂y

or,

(2)

Let us consider a differential equation

M (x, y)dx + N (x, y)dy = 0 If

∃ a function f ( x, y )

s.t.

(3)

∂f ∂f =M & =N ∂x ∂y then (3) can be written in the form

∂f ∂f dx + dy = 0 ∂x ∂y or

df = 0

(4)

and its general solution is f ( x, y ) = c Then Mdx+ Ndy is said to be differential & (3) is called differential equation.

Q

by calculus

∂ f ∂ f = ∂y∂x ∂x∂y 2

2

exact exact

∂M ∂N = ∂y ∂x

(5)

Then (3) is exact i.e. (5) is necessary condition for exactness.

∂f From (4) =M ∂x

(i)

and

∂f =N ∂y

(ii)

Integrating (i) w.r.t x,

f = ∫ Mdx + g ( y ) Then differentiating have,

(6)

(6) wrt y, we

∂ ⎡ ∂ f ⎤= Mdx + g y = N , from (4ii) ( ) ∫ ⎦ ∂y ∂y ⎣ ∂ g ′( y ) = N − ∫ Mdx or ∂y ⎡ ⎤ ∂ ∴ g ( y ) = ∫ ⎢ N − ∫ Mdx ⎥ dy ∂y ⎣ ⎦

(7)

Substitute (7) in (6) to get the solution.

Solve Q.1

Here

⎛ ⎜⎜ x + ⎝

2⎞ ⎟⎟dy + y dx = 0 y⎠

2 M = y, N = x + y

(1)

∂N ∂M = 1, =1 ∂y ∂x

∂M ∂N Q = ∂x ∂y

∴

(1) is exact

Hence

f = ∫ Mdx + g ( y )

= ∫ ydx + g ( y )

f = yx + g ( y ) ∂f ∴ = x + g ′( y ) ∂y

∂f ∴N = ∂y

(2)

2 ∴ x + = x + g ′( y ) y 2 ∴ g ′( y ) = y i.e. g ( y ) = ln y 2 + C From (2)

f = yx + ln y + C 2

i.e. yx + ln y = C 2

Solve Problem 18

xdx

(x

2

+y

Here,

+

) (x

3 2 2

ydy 2

+y

)

3 2 2

=0

(1)

M=

x

(x

2

+y

(

, N=

)

3 2 2

∂M − xy x + y = 3 2 2 ∂y x +y

(

and

)

1 2 2

2

)

y

(x

2

+y

)

3 2 2

(

∂N xy x + y =− 2 2 ∂x x +y

(

2

∂M ∂N Q = ∂y ∂x

∴

(1) is exact

f = ∫ Mdx + g ( y )

)

1 2 2

)

3

=∫

(x

x 2

+y

)

2⇒

3 2

dx + g ( y )

Let x2 + y2 = t

⇒ 2x dx = dt 1 dt 1 ∴ f = ∫ 3 + g(y) = − 1 + g(y) 2 (t ) 2 t2

f =

−1

(x

∂f ∴ = ∂y

2

+y

)

1 2 2

y

(x

2

+y

∂f QN = ∂y

)

3 2 2

+ g(y) + g ′( y )

∴

y

(x

2

+y

=

) (x

3 2 2

y 2

+y

)

3 2 2

⇒ g ′( y ) = 0 ⇒ g ( y ) = C ∴

1

(x

2

+y

)

1 2 2

1 = C

∴x + y = C 2

2

2

+ g ′( y )