• Author / Uploaded
• Ryan Mills

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Differential Equation – An equation containing an independent variable and / or dependent variable and differential coefficient of dependent variable with respect to independent variable is called a differential equation. e.g.

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Step- 2 Put y = vx and

equation in step 1 and cancel out x from the right hand side, the equation reduces to the form

dy d2 y dy -5 + 2xy = x3 and + 6y = x2 dx dx 2 dx

Order of a Differential Equation – The order of a differential equation is the order of the highest order derivative appearing in the equation. Degree of a Differential Equation – The degree of a differential equation is the power of the highest order derivative when differential coefficients are made free from radicals and fractions. Solution of a Differential Equation – The solution of a differential equation is a relation between the variables involved, not involving the differential coefficients, such that this relation and derivatives obtained from it satisfy the given differential equation. General Solution – The solution which contains as many as arbitrary constants as the order of the differential equation is called the general solution of the differential equation. Particular Solution – Solution obtained by giving particular values to the arbitrary constants in the general solution of a differential equation is called a particular solution. Formation of Differential Equations – Formation of a differential from a given equation representing a family of curves means finding a differential equation whose solution is the given equation. If an equation representing a family of curves, contains n arbitrary constants, then we differentiate the given equation n times to obtain n more equations. Using all these equations, we eliminate the constants. The equation so obtained is the differential equation of order n for the family of given curves. Homogeneous Differential Equation –

dy A differential equation of the form = F (x, y) is dx said to be homogeneous if F (x, y) is a homogeneous function of degree zero. Algorithm may be used to solve a homogeneous differential equation. Algorithm:Step- 1 Put the differential equation in the form

dy f ( x, y ) = dx Y ( x, y )

dy dv = v+x in the dx dx

v+x

dv = f (v) dx

Step- 3 Shift v on RHS and separate the variable v and x. Step- 4 Integrate both sides to obtain the solution in terms of v and x. Step- 5 Reduce v by 9.

y in the solution obtained in x

step (4) to obtain the solution in terms of x and y. Linear Differential Equation – A differential equation is known as first order linear differential equation, if the dependent variable (y) and its derivative appear in first degree. The general form is

dy + Py = Q, where dx

P and Q are constant or functions of x. Algorithm used to solve a linear differential equation: (i) Write the given differential equation in the form

dy + Py = Q and obtain P and Q. dx Pdx (ii) Find integrating factor, I.F. = e ò (iii) Multiply both sides of equation in (i) by I.F. (iv) Integrate both sides of the equation obtained in (iii) w.r.t. x to obtain y (I.F.) = ò Q.(I.F.) dx + C This gives the required solution. In case, the first order linear differential equation is in the form

dx + P1 x = Q1, where , P1 and Q1 are dy

constants or functions of y only. Then I.F. = e ò P1dy and the solution of the differential equation is given by x . (I. F) = ò(Q1 × I.F.) dy + C 10. Differential Equations of the type

d2 y dx 2

= f (x)

(i) Integrate both sides of the differential equation in (i) with respect to x to obtain a first order first degree differential equation. (ii) Again integrate both sides of the first order differential equation obtained in (i) with respect to x.

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E X E R C I S E S Determinate order and degree (if defined) of the differential equations given in exercises 1 to 10. d4 y + (sin y¢¢¢) = 0 dx 4 Sol. Order of the equation is 4 It is not a polynomial in derivatives so that it has not degree. 2. y’ + 5y = 0 Sol. y’ + 5y = 0 It is a D.E. of order one and degree one.

1.

æ d 2s ö æ ds ö + 3s çè ÷ø çè dt 2 ÷ø = 0 dt Sol. Order of the equation is 2. Degree of the equation is 1. 4

3.

2

æ d 2 yö æ dy ö 4. ç 2 ÷ + cos çè ÷ø = 0 dx è dx ø 2 2 æ d yö æ dy ö Sol. ç 2 ÷ + cos çè ÷ø = 0 dx è dx ø It is a D.E. of order 2 and degree undefined.

5.

d2 y dx 2

= cos3x + sin 3x

d2 y

= cos3x + sin 3x dx 2 It is a D.E. of order 2 and degree 1. 6. (y¢¢¢)2 + (y¢¢)3 + (y¢)4 + y5 = 0 Sol. Order of the equation is 3 Degree of the equation is 2 y '''+ 2 y ''+ y ' = 0 7. Sol.

Sol. y '''+ 2 y ''+ y ' = 0 The highest order derivative is y ''' . Thus the order of the D.E. is 3. The degree of D.E is1 8.

y '+ y = e x

Sol. y '+ y = e x The order of the D. E. = 1 (highest order derivative) The degree of the D.E. = 1.

9. y ''+ ( y ')2 + 2 y = 0 2 Sol. y ''+ ( y ') + 2 y = 0 The highest derivative is 2. \ Order of the D.E. = 1. Degree of the D.E = 1 10. y¢¢ + 2y¢ + sin y = 0 Sol. Order of the equation is 2 Degree of the equation is 1 11. The degree of the differential equation 3

æ d 2 y ö æ dy ö 2 æ dy ö ç 2 ÷ + çè ÷ø + sin çè ÷ø + 1 = 0 is dx dx è dx ø (a) 3 (b) 2 (c) 1 (d) not defined. 3

æ d 2 y ö æ dy ö 2 æ dy ö Sol. ç 2 ÷ + çè ÷ø + sin çè ÷ø + 1 = 0 dx dx è dx ø The degree not defined. Hence option (d) is correct. 12. The order of the differential equation d2y dy 2 x 2 2 – 3 + y = 0 is dx dx (a) 2 (b) 1 (c) 0 (d) not defined. 2 dy 2d y –3 + y =0 Sol. 2 x 2 dx dx Thus order of the D.E. = 2 Hence option (a) is correct.

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation 1. y = e x + 1: y ''– y ' = 0 Sol. y = ex+1 Þ y ' = ex Þ y '' = ex Now L.H.S. = y '' – y ' = ex – ex = 0 Hence y = ex + 1 is a solution of y '' – y ' = 0. 2. y = x 2 + 2 x + c : y '– 2 x – 2 = 0 Sol. y = x 2 + 2 x + c \ y ' = 2 x + 2 Þ y '– 2 x – 2 = 0 Hence y = x 2 + 2 x + c is a solution of y '– 2 x – 2 = 0. 3. y = cos x + c : y '+ sin x = 0 Sol. y = cos x + c Þ y ' = – sin x Þ y '+ sin x = 0 Hence y = cos x + c is a solution of y '+ sin x = 0.

3 4.

y = 1 + x2 : y ' =

tan–1 y

9. x+y= Sol. x + y = tan–1 y

xy

1+ x 1 2 –1/ 2 ´ 2x Sol. y = 1 + x 2 Þ y ' = (1 + x ) 2 x y' = = L.H.S. 1 + x2 xy x R.H.S. = Thus y ' = 1+ x2 1 + x2 2

Hence y = 1 + x 2 is a solution of y ' = 5. y = Ax : xy ' = y ( x ¹ 0) Sol. y = Ax Þ y ' = A L.H.S. = xy ' = xA = y = R.H.S. Hence y = Ax is a solution of xy ' = y.

xy 1+ x

10.

. 2

y = x sin x ; xy¢ = y +x x 2 - y2 (x ¹ 0 and x > y or x < – y) Sol. y = x sin x ...(i) Differential w.r.t. x y¢ = 1. sin x + x cos x y¢ = sinx + x cos x [Q sin2 x + cos2 x = 1] y from (i) sin x = x ... (ii) y¢ = sin x + x 1 - sin 2 x from (i) & (ii) 6.

y¢ =

y y2 y x 2 - y2 + x 1- 2 = + x x x x x2

Multiplying by x, xy¢ = y + x x 2 - y2 is the reqd. differential equation.

which

y2 (xy ¹1) 1 - xy Sol. xy = log y + c Differentiating w.r.t. x 1 1 × y + xy¢ = × y ¢ or y2 + xyy¢ = y¢ y

7.

xy = log y + c; y¢ =

y2 or y2 = y¢ – xyy¢ = y¢ (1 – xy) \ y¢ = 1 - xy This is the reqd. differential equation. y - cos y = x : ( y sin y + cos y + x ) y ' = y 8. Sol. y - cos y = x Þ y '+ sin yy ' = 1 1 y '(1 + sin y ) = 1 Þ y ' = 1 + sin y L.H.S. = ( y sin y + cos y + x ) y '

= ( y sin y + y ) y '

1 = y = R.H.S. 1 + sin y Here y – cos y = x is a solution of ( y sin y + cos y + x ) y ' = y. = (1 + sin y) y.

; y2

y¢ + y2

+1=0

y¢ Differentiating w.r.t. x 1 + y¢ = 1 + y2 (1 + y2) + y¢ (1 + y2) = y¢ or 1 + y2 + y¢ (1 + y2 – 1) = 0 or y2 y¢ + y2 + 1 = 0 which is the reqd. differential equation. dy y = a 2 - x 2 x Î (– a, a) ; x + y = 0, (y ¹ 0) dx

Sol. y = a 2 - x 2 Squaring both sides y2 = a2 – x2 or x2 + y2 = a2 Differentiating w.r.t. x 2x + 2yy¢ = 0

dy ö æ Þ 2 çè x + y ÷ø = 0 is the required solution. dx 11. The number of arbitrary constants in the general solution of a differential equation of fourth order are : (a) 0 (b) 2 (c) 3 (d) 4 Sol. The general solution of a differential equation of fourth order has 4 arbitrary constants. Hence option (d) is correct. 12. The number of arbitrary constants in the particular solution of a differential equation of third order are : (a) 3 (b) 2 (c) 1 (d) 0 Sol. Number of arbitrary constants = 0 Hence option (d) is correct.

In each of the following, Q. 1 to 5 form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. x y + =1 1. a b x y Sol. Given + = 1 .............(i) a b where a and b are two arbitrary constants, Since (i) contains two arbitrary we differentiate it two times w.r.t. x and the differential equation will be of second order. Differentiating (i) w.r.t. x, we get 1 1 + y' = 0 .............(ii) a b 1 y '' = 0 Þ y '' = 0 Þ b which is the required differential equation.

4 2. y2 = a (b2 – x2) Sol. y2 = a (b2 – x2) ...(i) Differentiating w.r.t x 2yy¢ = – 2ax ... (ii) Again differentiating 2(y¢2 + yy¢¢) = – 2a ... (iii) Dividing (iii) by (ii) 2 (y ¢2 + yy¢¢) - 2a = , x (y¢2 + yy¢¢) = yy¢ 2yy ¢ - 2ax i.e. the differential equation 2

æ d2 y ö dy æ dy ö =0 xy ç 2 ÷ + x ç ÷ - y dx è dx ø è dx ø 3 y = ae3x + be–2x Sol. y = ae3x + be–2x ... (i) Differentiating w.r.t. x y¢ = 3ae3x – 2be–2x ... (ii) Again differentiating y¢¢ = 9 ae3x + 4be–2x ... (iii) Multiply equation (i) by 2 and add with (ii) i.e., 2y = 2ae3x + 2be -2x y¢ = 3ae3x - 2 be -2x or ae3x = 2y + y¢ 5 2y + y¢ = 5a e3x Multiply (i) by 3 and subtract (ii) from it 3y = 3ae3x + 3be -2x 3y - y¢ y¢ = 3ae3x - 2be -2x or be–2x = 2x 5 3y - y ¢ = 5 be Putting the values of a and b in (iii) æ 2y + y¢ ö æ 3y - y¢ ö +4ç y¢¢ = 9 ç ÷ ÷ è 5 ø è 5 ø 5y¢¢ = 30y + 5y¢ or y¢¢ – y¢ – 6y = 0 Required. differential equation is d 2 y dy – 6y = 0 dx 2 dx 4. y = e 2 x (a + bx) Sol. y = e 2 x a + bxe 2 x .............(i) 2x 2x y ' = 2e (a + bx) + be .............(ii) y ' = 2 y + be 2 x 2x .............(iii) y '' = 2 y '+ 2be From (ii) and (iii), we get 2 y ' = 4 y + 2be 2 x

y¢ = ex (a cos x + b sin x) + ex (– a sin x + b cos x) y¢¢= ex [(a + b) cos x – (a – b) sin x] ... (ii) y¢¢ = ex [(a + b) cos x – (a – b) sin x] + ex [– (a + b) sin x – (a – b) cos x] = ex [2b cos x – 2a sin x] = 2ex [b cos x – a sin x] y¢¢ x = e (b cos x – a sin x) ... (iii) or 2 Adding (ii) and (iii) y¢¢ y+ = ex [(a + b) cos x – (a – b) sin x] = y¢ 2 or 2y + y¢¢ = 2y¢ Þ y¢¢ – 2y¢ + 2y = 0 d2 y dy -2 + 2y = 0 Hence the diff equn is 2 dx dx 6. Form the differential equation of the family of circles touching the y axis at origin Sol. The equation of the circle with centre (a, 0) and radius a, which touches y – axis at origin y

O

(a, 0)

x

(x – a)2 + y2 = a2 or x2 + y2 = 2ax ... (i) Differentiating w.r.t. x 2x + 2yy¢ = 2a or x + yy¢ = a Putting value of a in (i) x2 + y2 = 2x (x + yy¢) = 2x2 + 2xyy¢ dy \ Reqd. diff. equation is 2xy + x2 – y2 = 0. dx 7. Form the differential equation of the family of parabolas having vertex at origin and axis along positive y–axis Sol. The equation of parabola having vertex at the origin and axis along positive y–axis is y

y '' = 2 y '+ 2be 2 x – – 2 y '– y '' = 4 y – 2 y ' Þ y ''– 4 y '+ 4 y = 0 5. y = ex (a cos x + b sin x) Sol. The curve is y = ex (a cos x + b sin x) Differentiating w.r.t. x

x¢

x

0

... (i) y¢

5 x2 = 4ay

... (i)

Differentiating w.r.t x 2x = 4ay¢ Dividing (ii) by (i) 2x 4ay ¢ y¢ 2 = Þ = 2 x 4ay y x \ xy¢ = 2y

... (ii)

dy – 2y = 0 dx 8. Form the differential equation of family of ellipses having foci on y-axis and centre at origin. Sol. The equation of family ellipses having foci at y-axis is

Reqd. diff. equation is

x

y

a 0 b

x2 y2 + = 1, a > b b2 a2

x

Again differentiating, we get b2 = yy ''+ ( y ')2 a2 Thus yy ' = xyy ''+ x( y ')2 which is the required equation of the hyperbola. 10. Form the differential equation of the family of circles having centre on y-axis and radius 3 units. Sol. Let centre be (0, a) and r = 3 Equation of circle is .............(i) x 2 + ( y – a)2 = 9 Differentiating both sides, we get dy ( y – a) = –x dx –x .............(ii) Þ y–a= y' From (i) and (ii), we get x2 = 9 Þ x 2 [( y ')2 + 1] = 9( y ') 2 x2 + ( y ') 2 which is the required equation. 11. Which of the following differential equation has y = c1e x + c2 e – x as the general solution ? d2y d2y y + = 0 – y=0 (a) (b) dx 2 dx 2

... (i)

2x 2yy¢ + 2 =0 Differentiating w.r.t. x b2 a x yy¢ or 2 + 2 = 0 ...(ii) b a 1 1 Again Differentiating 2 + 2 (y¢2 + yy¢¢) = 0 b a 1 y ¢2 + yy ¢¢ \ 2 =b a2 Putting this value in (ii) x (y¢2 + y y¢¢) yy¢ + 2 = 0 or x (y¢2 + yy¢¢) = yy¢ – a2 a or xyy¢¢ + xy¢2 – yy¢ = 0 \Differential equation is 2

æ d2 y ö æ dy ö æ dy ö xy ç 2 ÷ + x ç ÷ - y ç ÷ = 0 dx dx è ø è dx ø è ø 9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin. x2 y 2 Sol. Equation of the hyperbola is 2 – 2 = 1 a b Differentiating both sides w.r. to x

12.

+1 = 0

(d)

d2y

d2y

– y=0 dx 2 Option (b) is correct. Which of the following differential equation has y = x as one of its particular solution ? d2y dy – x2 + xy = x (a) 2 dx dx d2y dy + x + xy = x (b) 2 dx dx Þ

(c)

x

y dy – 2 =0 Þ 2 a b dx

d2y

–1 = 0 dx 2 dx 2 Sol. y = c1e x + c2 e – x dy = c1e x – c2 e – x Þ dx d2y d2y x –x c =y = + e c e Þ 1 2 dx 2 dx 2 (c)

(d)

d2y dx

2

d2y dx

2

– x2 +x

dy + xy = 0 dx

dy + xy = 0 dx

6

d2y dy =0 =1, dx dx 2 d2y 2 dy + xy = 0 Now 2 – x dx dx Hence option (c) is correct.

Sol. y = x Þ

5. (ex + e– x) dy – (ex – e– x) dx = 0 Sol. We have (ex + e– x) dy = (ex – e– x) dx , e x - e- x dy = e x + e - x dx Integrating both sides, we get

ò For each of the following D.E in Q. 1 to 10 find the general solution: 1.

dy 1– cos x = dx 1 + cos x

dy 1– cos x dy = Sol. = Þ dx 1 + cos x dx

æ xö 2sin 2 ç ÷ è 2ø æ xö 2cos 2 ç ÷ è 2ø

dy æ xö = tan 2 ç ÷ Þ è 2ø dx 2 x dx Þ ò dy = ò tan 2 æ 2x ö –1÷ dx Þ y = ò çè sec 2 ø x Þ y = 2 tan – x + C (Required solution) 2 dy 2 = 4 – y (–2 < y < 2) 2. dx dy dy = 4 – y2 Þ ò Sol. = ò dx dx 4 – y2 –1 y = x + C (Required solution) Þ sin 2 y = sin( x + C ) 2 Hence, y = 2sin( x + C ). dy + y = 1 ( y ¹ 1) 3. dx dy dy = – ò dx + y =1 Þ ò Sol. y –1 dx Þ log( y –1) = – x + C Þ y = 1 + e – x .eC Hence y = 1 + Ae – x . which is the required solution. 4. sec2 x tany dx + sec2 y tan x dy = 0 Sol. We have sec2 x tan y dx + sec2 y tan x dy = 0 sec 2 y sec 2 x Þò dy = -ò dx tan y tan x

Þ log ( tan y ) = - log ( tan x ) + log c Þ log |tan x tan y| = log c Þ tan x tan y = c

dy =

ò

ex - e-x dx Þ e x + e- x

ò dy = ò t

dt

[Put ex + e–x = t so that (ex – e–x) dx = dt] Þ y = log |t| + c Þ y – c = log |ex + e–x | (x Î R) or y = log (ex + e–x) + c Q ex, e–x > 0 6.

dy = (1 + x 2 ) (1 + y 2 ) dx

dy

Sol.

1 + y2

= (1 + x 2 ) dx

Integrating both sides, we get 1 tan –1 y = x + x3 + C 3 which is the required solution. 7. y log y dx – xdy = 0 Sol. y log y dx – x dy = 0 or y log y dx = x dy

or

ò

dy = dx Integrating, we get y log y 1 dy dx = ò Put log y = t so that y dy = dt y log y x

òt

dt

= log x + log c or

log t = log x + log c log (log y) = log cx \ log y = cx Þ y = ecx. x5

dy = – y5 dx

5 Sol. x

dy = – y5 dx

8.

Þ

òy

–5

dy = – ò x –5 dx

1 1 Þ – 4 = 4 + 4C y x 1 1 Þ – 4 = 4 + K where K = 4 C. y x Þ x –4 + y –4 = K .

7 9.

Integrating, we get

Solve the following : dy = sin -1 x dx dy = sin -1 x Þ ò dy = ò sin -1 x dx Sol. dx x dx Þ y = x sin–1 x – ò 1 - x2 Let 1 – x2 = t Þ – 2x dx = dt 1 1 dt Þ y = x sin–1 x + ò 2 t Þ y = x sin–1 x +

3

2x 2 + x dx = + x 2 + x +1

ò

2x 2 + x dx 2 + 1)

ò (x +1) (x

2x 2 + x A Bx + C = + 2 Let 2 (x + 1) (x + 1) x + 1 x +1 2x2 + x = A (x2 + 1) + (Bx + C) (x + 1) = A (x2 + 1) + B (x2 + x) + C (x + 1) 1 Put x = – 1, 2 – 1 = A (1 + 1) Þ A = 2 Comparing the coefficient of x2 and x 2 = A + B and 1 = B + C, B = 2 – A = 2 – 3 1 C = 1 –B= 1– = 2 2

=

1 2

ò x + 1 dx + 2 ò x

1 2

1

1

ò x +1 dx + 4 ò x 1

3

3x - 1 dx 2 +1

2x 1 dx 2 +1 2

òx

dx 2 +1

1 3 1 log (x + 1) + log (x2 + 1) – tan–1 x + c 2 4 2 1 3 1 1 = log 1 + log 1 – tan–1 0 + c 2 4 2 [ Q y = 1 when x = 0] 1 = 0 + c Þ c = 1 Thus the solution is 1 3 1 y = log (x + 1) + log (x2 + 1) – tan–1 x + 1 2 4 2 1 1 y = log (x + 1)2 (x2 + 1)3 – tan–1 x + 1 4 2 dy 2 = 1, y = 0 when x = 2 12. x( x –1) dx dy 2 =1 Sol. x( x –1) dx dx Þ ò dy = ò x( x + 1) ( x –1) 1 A B C Let x( x + 1) ( x –1) = x + x + 1 + x –1

=

Þ y = x sin –1 x + 1 - x 2 + C (Required solution) 10. ex tan y dx + (1– ex) sec2 y dy = 0 Sol. ex tan y dx + (1– ex) Sec2 y dy = 0 or (1–ex) sec2 y dy = – ex tan y dx Dividing by (1 – ex) tan y sec 2 y -e x dx dy = tan y 1 - ex sec2 y -e x dy = dx Integrating, we get tan y 1 - ex \ log tan y = log (1 – ex) + log c log tan y = log c (1– ex) Þ tan y = c (1 – ex) Find a particular solution satisfying the given condition for the following differential equation in Q.11 to 14. dy = 2x2 + x; y = 1, when x = 0 11. (x3 + x2 + x + 1) dx dy Sol. (x3 + x2 + x + 1) = 2x2 + x or (x3 + x2 + x + 1) dy dx 2x 2 + x = (2x2 + x) dx or dy = x 3 + x 2 + x + 1 dx Integrating, we get

ò dy = ò x

ò

2x 2 + x dx (x + 1) (x 2 + 1)

=

t +C

ò

y=

1 3 = 2 2

Þ 1 = A(x + 1) (x – 1) + Bx(x – 1) + Cx(x + 1) Let x = 0 Þ A = –1 1 Let x – 1 = 0 Þ x = 1, C = 2 1 Let x + 1 = 0 Þ x = –1, B = 2 é –1 1/ 2 1/ 2 ù \ y = ò ê x + x + 1 + x –1 ú dx ë û

1 1 = – log x + log( x + 1) + log( x –1) + C 2 2 2 æ ö 1 x –1 y = log ç 2 ÷ + log C 2 è x ø 1 3 Now x = 2, y = 0 Þ log C = – log 2 4 Hence particular solution is :

æ x 2 –1ö 1 1 3 y = log ç 2 ÷ – log . 2 2 4 è x ø

8 13.

æ dy ö cos ç ÷ = a, (a Î R), y = 1 when x = 0 è dx ø

æ dy ö dy Sol. cos ç ÷ = a \ = cos–1a dx è dx ø or dy = (cos–1 a) dx Integrating òdy = cos–1a òdx or y = (cos–1 a) x + c we have y = 2 when x = 0 Þc=2 \ Solution is y = x (cos–1a) + 2 y-2 y-2 is the reqd. solu. or cos–1 a = or a = cos x x dy = y tan x, y = 1, when x = 0 14. dx dy dy = tan x dx = y tan x Þ ò Sol. y ò dx Þ log y = log sec x + C When x = 0, y = 1 Þ log1 = log sec 0 + C Þ 0 = log1 + C Þ C=0 \ log y = log sec x Hence y = sec x. 15. Find the equation of the curve passing through the point (0,0) and whose differential equation y¢ = ex sin x. Sol. Differential equation is y¢ = ex sin x dy = ex sin x \ dy = ex sin x dx or dx Integrating, we get ò dy = ò ex sin x dx y = ex (– cos x) – ò ex (–cos x) dx = – ex cos x + ò ex cos x dx Again integrating by parts taking ex as first function = – ex cos x + ex sin x – ò ex sinx dx or 2y = – ex cos x + ex sec x + c ex \ y= [– cos x + sin x] + c 2 1 1 Put x = 0, y = 0 Þ 0 = – +c \ c= 2 2 1 ex \ Solution is y = (sin x – cos x) + 2 2 dy 16. For the differential equation xy = (x + 2) dx (y + 2) find the solution curve passing through the point (1, – 1) dy Sol. The differential equation is xy = (x + 2) (y + 2) dx or xy dy = (x + 2) (y + 2) dx

y x+2 dy = dx Integrating, we get y+2 x dy x+2 y = dx y+2 x

ò ò

ò

y+2-2 2ö æ dy = ç 1 + ÷ dx è y+2 xø

ò

æ

ö

æ

2ö

ò èç1 - y + 2 ø÷ dy = ò çè1 + x ÷ø dx 2

y – 2 log (y + 2) = x + 2 log x + c The curve passes through (1, –1) \ – 1 – 2 log 1 = 1 + 2 log 1 + c (log 1 = 0) Þ – 1 = 1 + c Þ c = – 2 Putting c = – 2 Þ y – 2 log (y + 2) = x + 2 log x – 2 or y – x = 2 [log (y + 2) + log x] – 2 = 2 log x (y + 2) – 2 Solution is y = x + 2 log x (y + 2) – 2. 17. Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve the product of the slope of its tangent and y – coordinate of the point is equal to the x – coordinate of the point. Sol. According to the question dy y =x dx Þ ò y dy = ò x dx y2 x2 = +C 2 2 (0, – 2) lies on it. Þ C = 2 \ Equation of the curve is x2 – y2 + 4 = 0. 18. At any point (x, y) of a curve the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3) find the equation of the curve given that it passes through (– 2, 1). dy Sol. Slope of the tangent to the curve = slope of dx the line joining (x, y) and (– 4, – 3) Þ

y + 3 dy æy + 3ö =2ç , ÷ x + 4 dx èx + 4ø 2 (y + 3) dy 2 dx Þ dx = \ dy = x+4 y+3 x+ 4 dy dx =2 Integrating, we get y+3 x+4 or log (y + 3) = 2 log (x + 4) + log c or log (y + 3) – log (x + 4)2 = log c =

ò

ò

9

y +3 y+3 =c 2 = log c \ (x + 4) (x + 4) 2 The curve passes through (–2, 1) 1+ 3 4 = c = =1 2 (-2 + 4) 4 y +3 Equation of the curve is =1 (x + 4)2 or y + 3 = (x + 4)2 or y = (x + 4)2 – 3. The volume of a spherical balloon being inflated. Changes at a constant rate. If initially its radius is 3 units and offer 3 seconds it is 6 units. Find the radius of balloon after t seconds. Let v be volume of the balloon. dv d æ4 3ö = k or we have pr ÷ = k dt dt çè 3 ø 4 2 dr 2 dr p× 3r = k 4 p r or =k or 3 dt dt Þ 4 p r2 dr = k dt Integrating, we get r3 4p ò r2 dr = k ò dt or 4p × = kt + c ...(i) 3 when t = 0, r = 3 4 p (3)3 = k . 0 + c Þ 36 p = c 3 4 3 when t = 3, r = 6 (6) = 3k + c = 3k + 36 p 3 216 ´ 4 p = 3k + 36p 3 3k = 72 × 4p – 36p = 288p – 36p = 252p 252 p = 84 p \ k= 3 Putting the values of k and c in (i) 4 3 84 ´ 3 3 t + 36 ´ pr = 84p + 36p r3 = 3 4 4 r3 = 63t + 27 or r 3 = 9 (7t + 3) r = [9 (7t + 3)]1/3 In a bank principal increases at the rate of r% per year. Find the value of r if `100 double itself in 10 years (loge 2 = 0.6931). Let P be the principal at any time t. According to the problem dP r = .P dt 100 dP r ò P = ò 100 dt r t + C1 log P = 100 i.e., log

19.

Sol.

20.

Sol.

P =e

rt /100 C1

.e

( where e

C1

=C

)

Now P = 100, when t = 0 \ P = ert /100 ´ 100 When P = 200, t = 10 r Þ log 2 = 10 r = 6.931 Hence r = 6.931% per annum. 21. In a bank interest increases at the rate of 5% per year. An amount of ` 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1×648) Sol. Let p be the principal, Rate of interest is 5% dp dp 5 = p \ p = 0×05 dt dt 100 Integrating, we get log p = 0×05t + log c p Þ log = 0×05t c p 0×05t =e \ p = ce 0×05t ...(i) c Initially, p = ` 1000, t = 0 1000 = c, e0 = c \ c = 1000 Putting this value in (i) p = 1000e0×05t when t = 10 Þ p = 1000 e0×05 × 10 = 1000e0×5 p = 1000 × 1×648 Q e0×5 = 1×648 p = 1648 After 10 year ` 1000 will amount to ` 1648. 22. In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many house will the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present. Sol. Let y denote the number of bacteria at any instant t × then according to the question dy dy ay Þ = k dt ...(i) dt y k is the constant of proportionality, taken to be + ve on integrating (i), we get log y = kt + c ... (ii) c is a parameter. let y0 be the initial number of bacteria i.e., at t = 0 using this in (ii), c = log y0 Þ log y = kt + log y0 y Þ log = kt ...(iii) y0 10 ö 11y0 when t = 2 y = æç y0 + y0 ÷ = , 100 10 è ø 11y 0

So, from (iii), we get log 10 = k (2) y0

10 Þ k=

1 11 log 2 10

y 1 æ 11 ö Using (iv) in (iii) log = ç log t ... (v) y0 2è 10 ÷ø let the number of bacteria become 1, 00, 000 to 2,00,000 in t1 hours. i.e., y = 2y0 when t = t1 hours. from (v)

23.

2y0

Sol.

dv 1 + v 2 1 +v dx = -vÞ dv = dx 1+ v 1- v x 1+ v dx Integrating, 1 - v dv = x

Þ x

=

dy = e x + y is : dx

(b) e x + e y = C

(c) e – x + e y = C

(d) e – x + e – y = C

–y

dy = ò e x dx

Þ – e– y = e x + K Þ ex + e– y = C

where C = – K Thus option (a) is correct.

dy x 2 + y 2 = = f(x, y) dx x 2 + xy x 2 + y2 f (x, y) = 2 x + xy Replacing x by lx and y by ly

dx

Put v =

y x

æ yö g ç ÷ . So it is a homogeneous function of è xø degree 1. Let y = vx Þ v + x

l 2 x 2 + l 2 y2

Þ

l2 æ x 2 + y2 ö

æ x 2 + y2 ö 0 çè x 2 + xy ÷ø = l f (x, y) Hence f (x, y) is a homogeneous function of degree zero to solve it put y = vx

ò dv = ò

dv = 1+ v dx

dx Þ v = log x + C x

y = log x + C x Hence y = x log x + Cx . (x – y) dy – (x + y) dx = 0 Þ

...(i)

f (lx, ly) = l 2 x 2 + l 2 xy = l 2 ç x 2 + xy ÷ è ø = l0

2 ö

ò çè - 1 + 1- v ÷ø dv = ò x

– v – 2 log (1 – v) = log | x | – log c

Show that the given differential equation is homogeneous and solve each of them in Exercises 1 to 10 1. (x2 + xy)dy = (x2 + y2) dx Sol.

ò

y y - 2 log 1 - = log | x | - log c x x y x-y or + 2 log + log | x| - log c = 0 x x (x - y)2 x y (x - y)2 = e- y / x ´ = log or cx x2 c x \ solution is (x – y)2 = cx e– y/x x+ y y' = 2. x dy x + y y = = 1+ Sol. Given dx x x R.H.S of this differential equation is of the form

dy = e x .e y dx

òe

æ

or

ò

–

(a) e x + e – y = C

Þ

Equ (i)

dv x 2 + v2 x 2 x 2 (1 + v 2 ) = 2 = 2 dx x + (xvx) x (1 + v)

v+ x

2 log 2 1æ 11 ö ç log ÷ t1 Þ t1 = log 1110 y0 2è 10 ø 2 log 2 Hence, the reqd. no. of hours = log 1110 The general solution of a differential equation log

dy dy =v+x dx dx

or

... (iv)

3.

y x y 1– x R.H.S of the differential equation is of the form

dy x + y = = Sol. dx x – y

1+

æ yö g ç ÷ , so it is a homogeneous function of è xø degree zero.

11 Þ v+ x

dv 1 + v = dx 1 – v

2

Sol.

dv æ 1 + v 2 ö =ç x ÷ Þ dx è 1 – v ø

v –1

Now R.H.S. of the differential equation is of the

æ yö form g çè ÷ø , hence the given equation is a x homogeneous function. Let y = vx

dx x

Þ

ò v2 + 1 dv = – ò

Þ

dv 1 2v dv – ò 2 = – log x 2 ò v2 + 1 v +1

Þ x

æ y2 ö 1 –1 y log = – log x + C ç 2 + 1÷ – tan Þ 2 x èx ø Þ

dy y æ yö = 1– 2ç ÷ + è xø dx x

Þ

1 y log( y 2 + x 2 ) – tan –1 = C 2 x

Þ

y 1 = log( x 2 + y 2 ) – C. x 2 (x2 – y2)dx + 2xy dy = 0 –1 Þ tan

4.

Thus f ( x, y ) = R.H.S is a homogeneous function of zero degree. Put y = vx Þ

dy dv =v+x dx dx

Þ x Þ

dx 2v dv dv v 2 –1 =– = –v Þ 2 x 1+ v dx 2v 2v dv

ò 1 + v2 = – ò

1 2ò

dx x

dv æ 1 ö v –ç è 2 ÷ø

2

dv

ò 2v 2 –1 = – ò

= – log x

2

æ 2v – 1ö C log ç ÷ = log x + 2 2 v 2 1 è ø

æ yö 2 ç ÷ –1 2 è xø æCö =ç ÷ y è xø 2 +1 x

2

æ Cö =ç ÷ 2y +1 è x ø

2y – x

Hence 6.

x dy – y dx = x 2 + y 2 dx

Sol.

dy = dx

2 2

.

x2 + y 2 + y x x2 + y 2

Þ

dy = dx

Þ

dy y æ yö = 1+ ç ÷ + è xø dx x

Þ log(v + 1) = – log x + log C

dx x

1

Þ

dy y 2 – x 2 = Sol. dx 2 xy

dv = 1 – 2v 2 Þ dx

x

2

+

y x

2

R.H.S. of the differential equation is of the form

x2 + y 2 =C x Þ x2 + y2 = Cx.

æ yö f ç ÷ , thus the differential equation is è xø homogeneous.

Þ

5.

2

æ x2 + y 2 ö ÷ + log x = log C Þ log ç è x2 ø

x2

dy = x 2 – 2 y 2 + xy dx

Now let y = vx Þ x

dv = 1 + v2 dx

12 Þ

ò

dv 1 + v2

=ò

vx 2 [cos v + v sin v] v (cos v + v sin v) = = x2 [v sin v - cos v] v sin v - cos v Transposing v to R.H.S.

dx x

dv v (cos v + vsin v) = -v dx v sin v - cos v v cos v + v 2 sin v - v 2 sin v + v cos v = v sin v - cos v dv 2v cos c Þ x dx = vsin v - cos v

Þ log[v + 1 + v 2 ] = log x + log C

x

Þ v + 1 + v 2 = Cx Þ y + x 2 + y 2 = Cx 2 is the required solution.

7.

ì æ yö æ yöü í x cos ç ÷ + y sin ç ÷ ý y dx è xø è xøþ î

Sol.

dy = dx

é æ yö æ y öù y ê x cos ç ÷ + y sin ç ÷ ú x è ø è x øû ë = f (x, y) é y æ ö æ y öù x ê y sin ç ÷ - x cos ç ÷ ú èxø è x øû ë

ò

Replacing x by lx and y by ly × in f (x, y)

( ) ( )

( ) ( )

ù ly éê lx cos ly + ly sin ly lx lx úû ë f (lx , ly) = ù lx éê lx sin ly - ly cos ly lx lx úû ë

( ) ( )

( ) ( )

é ù l 2 × y ê x cos y + y sin y ú x x û ë = l 2 × y éê y sin y + y cos y ùú x x û ë

( ) ( )

8.

( ) ( )

é y y ù y êx cos x + y sin x ú ë û = l0 f (x, y) = l0 é ù y y - y cos x êx sin x x úû ë

\ f (x, y) is homogeneous function of degree zero.

dy Now, = dx

( ) ( )

( ) ( )

y éê x cos y + y sin y ùú x x û ë é ù y y + x sin x ê y cos x x úû ë

...(i)

dy dv =v+x dx dx Putting this value in (i)

Put y = vx Þ

vx vx ù é vx êx cos + vx sin ú dv x xû ë v+ x = dx x évx sin v - x cos vx ù x x ûú ëê

( )

( )

vsin v - cos v 2dx dv = v cos v x 1ö 2dx æ or ç tan v - ÷ dv = vø x è 1ö dx æ or ç tan v - ÷ dv = 2 v x è ø Integrating, log sec v – log v = 2 logx + log c sec v – log x2 = log c or log v sec v sec v Þ log 2 = log c i.e., vx 2 = c vx y y sec y x Putting v = , = c or sec = cxy x x y ´ x2 x dy æ yö x – y + x sin ç ÷ = 0 è xø dx

Þ

ì æ yö æ yö ü = í y sin ç ÷ - x cos ç ÷ ý xdy è ø è xøþ x î

ò

dy æ yö dy y y – y + x sin ç ÷ = 0 Þ = – sin è xø dx dx x x R.H.S. of the differential equation is of the type æ yö f ç ÷ , so the differential equation is a è xø homogeneous. dv Now let y = vx Þ x = – sin v dx dv dx = –ò Þ ò Þ ò cos ec v dv = – log x sin v x Þ log(co sec v – cot v) = – log x + log C C Þ cosec v – cot v = x y y C Þ cosec – cot = x x x y yù é x êcosec – cot ú = C x x ë û is the required solution.

Sol. x

13

æ yö y dx + x log ç ÷ dy – 2 x dy = 0 è xø y dy y x = = Sol. dx y y 2 x – x log 2 – log x x R.H.S. of the differential equation is of the form æ yö f ç ÷ , thus the given differential equation is è xø homogeneous. Put y = vx dy dv =v+ x ......(ii) dx dx From (i) and (ii), we get dv – v + v log v = x 2 – log v dx

9.

2 – log v dx dv = – v + v log v x Integrating both sides, we get dv dv dx ò v(log v –1) – ò v = ò x Þ

For

\

.....(iii)

1

ò v(log v –1) dv é ù êë Put log v –1 = t so that v = dt úû dt = ò = log t = log(log v – 1) t From (iii), we have

æ xö æx ö e x / y ç 1 - ÷ ç -1 ÷ e x / y yø èy ø dx è == = f (x, y) dy 1 + ex / y 1 + ex / y

æx ö x/y ç - 1÷ e y è ø \ f (x, y) = 1 + ex / y

æ lx ö æx ö x/y - 1÷ elx / ly ç y - 1ø÷ e èç ly ø è f (lx, ly) = = l0 1 - el x / l y 1 + ex / y

Hence, f (x, y) is a homogeneous function of degree zero. æx ö x/y ç - 1÷ e y ø Now, dx = è dy 1 + ex / y dx dv =v+y Put x = vy is dy dy

æ vy ö - 1÷ evy / y ç ø (v -1)e v \ v + y dv = è y = vy / y dy 1+ e 1 + ev or y

y ö æ log ç log –1÷ – log y + log x = log x + log C è x ø y ö æ Þ log ç log –1÷ = log y + log C = log Cy è x ø y Þ log – 1 = Cy x which is the required solution. æ xö 1- ÷ 10. (1+ ex/y) dx + ex/y çè y ø dy = 0

Sol.

Replace x by lx and y by ly

dv (v - 1) e v ve v - e v - v - ve v = -v= v dy 1+ e 1 + ev - (v + e v ) = 1 + ev

æ 1 + ev ö dy \ ç v + e v ÷ dv = - y è ø 1 + ev dy dv = Integrating , v + ev y Put v + ev = t, (1+ ev) dv = dt dt \ = – log y + log c t or log t = – log y + log c or log t y = log c x \ ty = c Putting t = v + ev = + ex/y y æx ö x/y \ ç y + e ÷ y=c è ø \ Reqd solu. is x + y. ex/y = cy. For each of the following differential equation in Q 11 ot 15 find the particular solution satisfying the given condition : 11. (x + y) dy+ (x – y) dx = 0, y = 1 when x = 1

ò

ò

ò

y –1 dy x Sol. = dx y +1 x

R.H.S. of the differential equation is of the type

æ yö f ç ÷ , hence is a homogeneous function. è xø

14 æ v 2 + 1ö y dv =v Þ x = –ç ÷ x dx è v +1 ø v +1 dx dv = – ò Þ ò 2 x v +1 1 2 log(v + 1) + tan –1 v = – log x + C Þ 2 1 y log( y 2 + x 2 ) + tan –1 = C Þ 2 x Now x = 1, y = 1

Now let

1 p 1 log 2 + tan –1 (1) = C Þ log 2 + = C 2 2 4 \ Particular solution is Þ

π 1 æ yö 1 log(y 2 + x 2 ) + tan –1 ç ÷ = log 2 + . è xø 2 2 4 2 2 12. x dy + (xy + y ) dx = 0, y = 1 when x = 1 dy xy + y 2 =Sol. = f (x, y) dx x2 f (x, y) is homogeneous \ Put y = vx dy dv =v+x or dx dx \ v+x

dv x (vx) + (vx)2 x 2 (v + v 2 ) ==2 dx x x2

1 dx \ v 2 + 2v dv = - x

Integrating,

ò

1 dv = 2 v + 2v

ò

dx + log c x

ò

1 dv = – log x + log c 2 (v + 2v + 1) - 1 1 or dv = – log x + log c (v + 1)2 -1 1 v + 1 -1 log = – log x + log c or 2 v + 1+ 1

ò

i.e., log Þ log \

v + log x = log c v+2

x v = log c v+2 x v y = c, Putting v = v+2 x

\ x2y = c2 (y + 2x) Putting x = 1, y = 1

... (ii)

1 1 = c2 (1+ 2) \ c2 = 3

1 1 in (i) x2 y = (y + 2x) 3 3 Particular solution is 3x2 y = y + 2x. æ ö p 2 y - y ÷ dx + xdy = 0, y = , when x = 1 13. ç x sin x 4 è ø y é ù 2 - y ú dx + xdy = 0 Sol. ê x sin x ë û dy y y = - sin 2 which is homogeneous ...(i) dx x x dv Put y = vx , \ v + x = v – sin2 v from (i) dx dv dx or =2 sin v x dv dx Integrating sin 2 v = - x dx òcosec2 vdx = – x – cot v = – log x + c log x – cot v = c y So general solution is Putting v = x p y log x – cot = c Putting x = 1, y = 4 x p log 1 = cot = c or 0 – 1 = c Þ c = – 1 4 y Particular solution is cot – log x = 1 x

Putting c2 =

ò

ò

ò

14. Sol.

dy y æyö - + cosec ç ÷ = 0, y = 0 when x = 1 dx x èxø

dy y y - + cosec = 0 ... (i) dx x x which is a homogeneous differential equation dy dv =v+x Put y = vx so that ...(ii) dx dx from (i) & (ii), we get dv ö æ ç v + x dx ÷ – v + cosec v = 0 è ø dv x + cosec v = 0 dx dv = – cosec v Þ x dx dx dv dx Þ – sin v dv = Þ = x cos ec v x Integrating ò – sin v dv =

ò

dx x

Þ cos v = log | x | + c y cos = log | x | + c Now, y (1) = 0, i.e., x

15 when x = 1, y = 0 cos 0 = log | 1 | + c Þ 1 = 0 + c, c = 1 y \ cos = log | x | + 1 x y Þ log | x | = cos – 1, (x ¹ 0) x which is reqd. solution. dy 15. 2xy – y2 – 2x2 = 0 , y = 2, when x = 1 dx Sol.

dy y 1 æ y ö = + ç ÷ dx x 2 è x ø

2

...(i)

Which is a homogeneous differential equation of the form

dy æ yö = fç ÷ è xø dx

(i) becomes, v + x

dv 1 = v + v2 dx 2

dx x v Integrating both sides, we get

Þ

2

2

dv =

2x 2 = log | x | + C = log | x | + C Þ y v It is given that y (1) = 2 i.e., When x = 1, y = 2 -

2x , ( x ¹ 0, ± e) 1 - log | x | which is the required solution. 16. A homogeneous equation of the form Þ

y=

æ xö dx = h ç ÷ can be solved by making the dy è yø

substitution. (a) y = vx (c) x = vy

e 2x (2 sinx – cos x ) + c 5 Putting the value of I1 in (i), the general solution e 2x (2 sin x – cos x) + c ye2x = 5 or 5y = 2 sin x – cos x + 5 ce–2x dy + 3 y = e -2 x 2. dx dy + 3 y = e -2 x Sol. dx

I1 =

Put y = vx, \

Find the general solution of the following differential equations in Q.1 to 12. dy + 2y = sin x 1. dx Sol. Here, P = 2, Q = sin x I.F. = eò2dx = e2x Solution to the diff. equation is ye2x = ò(sin x) e2x dx = I1 ...(i) I1 = òe2x sin x dx = e2x (– cos x) – ò2e2x (– cos x ) dx = – e2x cos x + 2 [e2x sin x – ò2e2x sin x dx] = – e2x cos x + 2 sin x e2x – 4 òe2x sin x dx I1 = e2x (2sin x – cos x) – 4I1 \ 5I1 = e2x (2 sin x – cos x )

(b) v = yx (d) x = v

x is correct. y 17. Which of the following is a homogeneous differential equation? (a) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0 (b) xydx – (x3 – y3) dy = 0 (c) (x3 + 2y2) dx + 2xydy = 0 (d) y2dx + (x2 – xy – y2) dy = 0 Sol. (d) y2dx + (x2 – xy – y2) dy = 0

Sol. (c) Option x = vy or v =

P = 3, integrating factor = e3x

y . e3 x = ò e x dx

Þ y = e -2 x + C e -3 x which is the required solution. dy y + = x2 3. dx x dy y + = x2 Sol. dx x 1 Þ Intergrating factor = ò x dx e = e log x = x dy 3 \ x + y = x Þ y.x = ò x3dx dx x4 x3 C Þ y.x = +C Þ y = + 4 4 x (General solution) pö æ dy + sec (x . y) = tan x ç 0 £ x < ÷ 4. 2ø dx è dy Sol. Linear equation of the form + Py = Q. dx Here, P = sec x, Q = tan x I.F = sec x + tan x i.e., The solu. is y × I.F. = òQ × I.F. dx + c or y × (sec x + tan x) = òtan x (sec x + tan x) dx + c Reqd. solu. is \ y (sec x + tan x) = (sec x + tan x) – x + c

16

dy pö æ + y = tan x ç 0 £ x £ ÷ è dx 2ø dy 2 Sol. cos x + y = tan x dx dy + y sec2 x = sec2 x tan x Þ dx 2 Þ Integrating factor = e ò sec x dx = e tan x 5.

tan x \e

Þ 8.

dy + y sec 2 x. e tan x = e tan x sec 2 x tan x dx

dy 2x cot x + y= dx 1 + x 2 1 + x2

Þ

2x

dx Þ Integrating factor = e ò 1+ x2 =1 + x 2

y . e tan x = tet - t + C y .e

tan x

= tan x e

tan x

-e

tan x

2 Þ y (1 + x ) = ò cot x dx + C

+C

Þ y (1 + x 2 ) = log sin x + C

y = tan x – 1 + Ce–tan x dy + 2y = x2 log x dx dy 2 + y = x log x Linear equation of the form Sol. dx x dy 2 + Py = Q Here, P = and Q = x log x dx x

6.

x

2 dx x

= e 2log x = e = x2 \ I.F. = e ò Pdx = e ò The solu. of the given equ. is y × I.F. = òQ × I.F. dx + c or y × x2 = ò(x log x) x2 dx + c = ò(x3 log x) dx + c

I.F = e \ Þ Þ

1 ò x log x dx

= e log(log x ) = log x

dy 1 2 (log x) + y = 2 log x dx x x y. log x = 2ò (log x) ( x -2 )dx é log x ù y.log x = 2 ê + ò x -2 dx ú x êë úû

y=

Þ

9.

x

log sin x 1+ x

2

+

C 1 + x2

dy + y – x + xy cot x = 0 (x ¹ 0) dx

Sol. x

log x 2

x4 1 x4 × = (log x), dx + c 4 x 4 1 1 = x 4 log x - ò x 3 dx + c 4 4 x2 x2 + c. x 2 Reqd. Solu. is y = log x 4 16 or 16y = x2 (4 log x – 1) + 16 c. x2 dy 2 x log x + y = log x 7. dx x dy 1 2 y= 2 + Sol. dx x log x x

ò

(1 + x 2 )dy + 2 xy dx = cot x dx ( x ¹ 0)

Sol. (1 + x 2 )dy + 2 xy dx = cot x dx

Þ y . e tan x = ò e tan x sec 2 x tan x dx Let tan x = t Þ sec2 x dx = dt Þ y . e tan x = ò te t dt Þ

é log x 1 ù y.log x = -2 ê + ú+C xû ë x é1 1 ù C + y = -2 ê + ú ë x x log x û log x

Þ

cos 2 x

dy + y - x + xy cot x = 0 dx

Þ x Þ

dy + (1 + x cot x ) y = x dx

dy æ 1 + x cot x ö +ç ÷ø y = 1 dx è x

1 + x cot x 1 = + cot x x x Þ Integrating factor P=

= eò

P dx

æ1

òç = e èx

ö + cot x ÷ dx ø

= elog x + log sin x = elog (x sin x) = x sin x Þ y . x sin x =

ò

x sin x dx + C

Þ y . x sin x = - x cos x +

ò 1 . cos x dx + C

Þ y . x sin x = – x cos x + sin x + C x cos x sin x C + + Þ y= x sin x x sin x x sin x Here y = - cot x +

1 C + x x sin x

17 dy =1 dx dy =1 Sol. ( x + y ) dx

10.

( x + y)

Þ

d ( xy ) = y 2 dy

Þ

xy = ò y 2 dy + C

Þ

1 dx =1 ( x + y ) dy

Þ

xy =

Þ

dx = x+ y dy

Þ

x=

dx -x= y dy Now P = – 1 Þ

\e

-y

dx - xe - y = ye - y dy

d [ xe - y ] = ye - y dy

y2 C + y 3

dy = y (y > 0) dx dx dx x - = 3y Sol. y dy = x + 3y2 or dy y dx linear equation of the form + Px = Q dy 1 Where P = – , Q = 3y,, y 1

12.

Þ Integrating factor = eò P dy = e ò ( -1)dy = e - y

y3 +C 3

(x + 3y2)

I. F.= e ò

Pdy

= eò

-

1

log = -1 1 y y – log y= e log y = e y dy = e

The solution is x × I.F.= òQ × I.F. dy + c 1 1 or x × = 3y ´ dy + c = 3y + c y y Hence, the reqd. solu. is x = 3y2 + cy.

-y -y Þ xe = ò ye dy + C

ò

Þ xe - y = - ye - y - e - y + C Þ x = - y - 1 + Ce y

For each of the following Exercises 13 to 15 find a particular solution, satisfying the given condition : dy p + 2 y tan x = sin x, y = 0 when x = 13. dx 3 dy + (2 tan x) y = sin x , P = 2 tan x Sol. dx Þ Integrating factor

Þ x + y + 1 = Ce y

11.

y dx + ( x - y 2 )dy = 0

Sol. y.dx + ( x - y 2 )dy = 0

dx + x - y2 = 0 dy

Þ

y.

Þ

dx y. + x = y 2 dy

Þ

dx 1 + x=y dy y

2 tan x dx = e 2log sec x = sec2 x = eò

\ sec 2 x

2 2 Þ y.sec x = ò sec x sin x dx 2 Þ y.sec x = ò sec x tan x dx

1 P= y Þ Integrating factor = e

y

dx + x = y2 dy

dy + sec 2 x(2 tan x) y = sec2 x sin x dx

1 ò y dy

= elog y = y

Þ y.sec 2 x = sec x + C p When x = and y = 0 Þ C = – 2 3

y = cos x - 2cos 2 x is the required solution.

18

dy 1 , y = 0 when x = 1 + 2xy = dx 1 + x2 dy 2x 1 + y= Sol. 2 dx 1 + x (1 + x2 )2 dy Linear equation of the form + Py = Q dx 2x 1 and Q = 2 where P= 2 x +1 (x + 1) 2 Now.

14.

(1 + x2)

2x

I.F. =

e ò Pdx

=e

ò x 2 +1 dx

= elog (x

2 + 1)

= x 2 +1

y × (I.F.) = òQ × I.F. dx + c, y (x2 + 1) =

ò (x

2

1 (x2 + 1) dx + c + 1)2

ò

tan -1 x c + 2 , (x Î R) x 2 +1 x +1

y = 0 when x = 1 \ c = – tan–1. 1 = –

p 4

Reqd. particular solution is tan -1 x p y = 2 x +1 4 (x 2 + 1) or y (x2 + 1) = tan–1 x –

Sol.

dy dy =x+ yÞ -y=x dx dx P = – 1 Þ Integrating factor = e ò - dx = e - x - x dy - ye - x = xe - x Þ e dx -x -x Þ y.e = ò xe dx Þ y.e - x = - xe - x - e - x + C

Þ y = - x - 1 + Ce x When x = 0, y = 0 Þ C = 1

1 Þ (x2 + 1) y = x 2 + 1 dx + c Þ (x2 + 1) y = tan–1 x + c Þ y=

16. Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

p 4

dy p – 3y cot x = sin 2x, y = 2, when x = dx 2 dy Sol. Linear equation of the form + Py = Q dx where P = – 3 cot x, Q = sin 2x \ Integrating function = e ò Pdx = e log cosec3x = cos ec3 x Þ Solution is y × I.F. = òQ. × I.F. dx + c or y × cosec3 x = òsin 2x cosec3 x dx + c 2 sin x cos x dx + c sin 3 x

15.

ò

= 2 òcosec x cot x dx + c = – 2 cosec x + c or y = – 2sin 2 x + c sin3 x p Now, y = 2, x = , 2 = – 2 + c \ c = 4 2 \ Reqd particular solution is y = – 2 sin2 x + 4 sin3 x y = – 2 sin 2 x (1 – 2 sin x)

\ y = - x - 1 + e x (Particular solution). 17. Find the equation of the curve passing through the point (0,2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5 dy Sol. By the given condition x + y – =5 dx dy dy = x + y– 5 = (x + y – 5) or \ dx dx dy (i) Taking + ve – y= x – 5 dx I.F. = eò(–1) dx = e–x is y × e–x = òe–x × (x – 5) dx + c Integrating by parts taking (x – 5) as first function ye–x = (x – 5) (–e–x) – ò 1 . (–e–x) dx + c = – (x – 5)e–x – e–x + c y = – x + 5 – 1 + cex = 4 –x + c . ex The curve passes through (0, 2) \ x = 0, y = 2 Þ 2 = 4 – 0 + c. e0 \ c = – 2 Regd. equ. of the curve is y = 4 – x – 2ex dy (ii) Taking – ve = – (x + y – 5) = – x – y + 5 dx dy or + y= – x + 5 dx ò1dx \ I.F. = e = ex Solution is yex = ò(5 – x)ex dx + c = (5 – x)ex – ò (–1). exdx + c = (5 – x) ex + ex + c or y = 5 – x + 1 + cex = 6 – x + cex The curve passes through (0, 2) \ x = 0, y = 2 2 = 6 – 0 + c. e0 or c = – 4 \ Equation of the curve y = 6 – x – 4e–x

19 18. The integrating factor of the differential dy 2 equation x - y = 2 x is dx (a) e –x (b) e–y (c) 1/x (d) x -1 Sol. P = x 1 - ò dx 1 \ Integrating factor = e x = e - log x = x Thus option (c) is true . 19. The integrating factor of the differential equation dx (1 - y 2 ) + yx = ay (-1 < y < 1) is : dy 1 1 (b) (a) y2 -1 y2 - 1 1 1 (c) (d) 2 1- y 1 - y2 y , Integrating factor Sol. P = 1 - y2

ò =e

y 1- y 2

-

dy

=e

1 (-2) y dy ò 2 1- y 2

1 - log(1- y 2 ) =e 2

Hence option (d) is correct.

1.

For each of the differential equations given below, indicate its order and degree (if defined). 2

d y æ dy ö + 5x ç ÷ – 6y = log x 2 dx è dx ø 2

(i)

3

2

æ dy ö æ dy ö (ii) ç ÷ - 4 ç ÷ + 7y = sin x dx è ø è dx ø æ d3 y ö d4 y - sin ç 3 ÷ = 0 4 dx è dx ø Sol. (i) Order = 2, Degree = 1 (ii) Order = 1, Degree = 3 (iii) Order = 4, Degree not defined 2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. (i) y = a ex + b e–x + x2 : d2 y dy +2 – xy + x2 – 2 = 0 x dx 2 dx (ii) y = ex (a cos x + b sin x) : d2 y dy -2 + 2y = 0 dx 2 dx

(iii)

(iii) y = x sin 3x : d2 y + 9y – 6 cos 3x = 0 dx 2 (iv) x2 = 2y2 log y :

dy – xy = 0 dx Sol. (i) y = aex + be–x + x2

(x2 + y2)

dy = aex – be–x + 2x dx d2 y = aex + be–x + 2 dx 2 d2 y dy +2 – xy + x2 – 2 2 dx dx = 2x + 2 (aex – be–x) + 4x – x3 + x2 – 2 = 2 (aex – be–x) – x3 + x2 + 6x – 2 ¹ 0 Hence, y = aex + bex + x2 is not the solution of the differential equation d2 y dy x 2 +2 – xy + x2 – 2 = 0 dx dx

Now, x

(ii) y = ex (a cos x + b sin x)

dy = ex [– a sinx + b cos x] + ex [a cosx + b dx sin x] = ex [(a + b) cos x – (a – b) sin x] d2 y = ex [–2a sin x + 2b cos x] dx 2

d2 y dy –2 + 2y = ex [(2a + 2a – 2b + dx 2 dx 2b) sin x + (2b – 2a – 2b + 2a) cos x] = 0 Hence, y = ex (a cos x + b sin x) is the solution of the differential equation

Now,

d2 y dy +2 + 2y = 0 2 dx dx

(iii) y = x sin 3x,

dy = sin 3x + 3x cos 3x dx

d2 y = 6 cos 3x – 9y dx 2 d2 y + 9y – 6 cos 3x = 0 dx 2 Hence, y = x sin 3x is the solution of

Þ

d2 y + 9y – 6 cos 3x = 0 dx 2

20 (iv) x2 = 2y2 log y ...(i) Differentiating w.r.t. x é 1 ù dy 2x = 2 ê 2y log y + y 2 ´ ú y û dx ë dy = 2 [2y log y + y] dx dy x x = = Þ dx 2y log y + y y (2 log y + 1) x2 from (i) 2 log y = 2 y dy x xy = = 2 \ 2 dx éx ù x + y2 y ê 2 + 1ú ëy û dy Þ (x2 + y2) – xy = 0 dx Hence, x2 = 2y2 log y is the solution of differential

dy – xy = 0 dx 3. Form the differential equation representing the family of curves given by (x – a)2 + 2y2 = a2, where a is an arbitrary constant. Sol. The equation of the curve is (x– a)2 + 2y2 = a2 i.e., x2 + 2y2 – 2ax = 0 ... (ii) dy Differentiating w.r.t. x 2x + 4y – 2a = 0 dx dy – 2ax = 0 ...(ii) Multiply it with x, 2x2 + 4xy dx dy =0 subtracting (i) from (ii) x2 – 2y2 + 4xy dx

equation (x2 + y2)

dy 2y2 - x 2 = dx 4xy 2 2 2 2 2 4. Prove that x – y = c (x + y ) is the general solution of differential equation (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter. Sol. The differential equation is (x3 – 3xy2) dx = (y3 – 3x2y) dy

\ Reqd. diff. equation is

dy x3 - 3xy 2 = dx y3 - 3x 2 y which is homogeneous equation dy dv = v+ x Put y = vx or dx dx

\ v+x

3 dv 1 - 3v 2 v - 3v dv = dx = 3 , 4 x dx v - 3v 1 - v

ò

Integrating, we get

v 3 - 3v dv = 1- v4

òx

dx

v3 v dv - 3 dv 1- v4 1- v4 = I1 – 3I2 ...(i)

\ log x =

ò

ò

1 -4 v3 I1 = 4 1 - v 4 dv Put 1 – v4 = t, – 4v3 dv = dt

ò

ò

\ I1 1 dt = - 1 log t = - 1 4 t 4 4 1 log t = – log (1 – v4) 4 1 2v dv I2 = 2 1 - v4 Put v2 = t, 2 vdv = dt

ò

1 dt 1 1 + t 1 log 1 + v 2 = log = 2 1 - v2 2 1- t 4 1- t 4 log c¢ + log x = I1 – 3 I2 [from (ii)] \ I2 =

ò

log c¢x = – 1 =4

1 1+ v2 1 log (1 – v4) – 3 × 4 log 1 - v 2 4

2 é 4 ) + 3log (1+ v ) ù log(1 v ê ú (1- v2 ) û ë

-1 = 4

= log

é (1+ v 2 )4 ù log ê ú (1 - v2 )3 û ë 1- v2 = log 1+ v 2

x2 - y x2 - y2 ×x = log x x2 + y2 x2 + y2 x2

x 2 - y4 ×x x 2 + y2 Þ x2 – y2 = c¢ (x2 + y2)2 Putting c¢ = c Þ x2 – y2 = c (x2 + y2)2 Hence x2 – y2 = c (x2 + y2) is the solution of the differential equation (x3 – 3xy2) dx = (y3 – 3x2y) dy 5. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes. Sol. Family of circles which touch the coordinates axes in first quadrant is

\ c¢ . x =

21 y

Sol. The differential equation is dy dx + 2 =0 2 y + y +1 x + x +1 dy

a

(a, a)

2

1ö 3 æ ç y+ ÷ + 2ø 4 è Integrating, dy

a x

0

òæ

(x – a)2 + (y – a)2 = a2 ...(i) a is radius of the circle. Differentiating w.r.t. x dy =0 2 (x – a) + 2 (y – a) dx dy =p Putting dx \ (x – a) + (y – a) p = 0 or x – a + py – pa = 0 x + py x + py – a (1 + p) = 0 \ a= 1+ p Putting value of a in (i) 2

2

é (x - y) p ù é (y - x) ù é x + py ù ê 1 + p ú + ê 1+ p ú = ê 1 + p ú ë û ë û ë û 2 Multiplying both side by (1 + p) \ (x – y)2 P2 + (x – y)2 = (x + py)2 or (x – y)2 (1 + p2) = (x + py)2

1ö 3 çy+ 2 ÷ + 4 è ø

or

2

equation

dy + dx

1 - y2 1- x2

=0

Sol. The differential equation is

Integrating 7.

ò

dy 1- y

2

=

ò

dy 1 - y2

=

dx 1- x 2

dx 1- x 2

or sin–1 y – sin–1 x = c This is the reqd. equ. Show that the general solution of the differential dy y2 + y + 1 + 2 = 0 is given by dx x + x + 1 (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.

equation

2

1ö 3 æ çx+ ÷ + 2ø 4 è

+

òæ

=0

dx 2

1ö 3 çx+ 2 ÷ + 4 è ø

=c

1 ö æ ÷ ç x+ 2 2 -1 tan ç ÷+ 3 ÷÷ çç 3 2 ø è

ö ÷ ÷ =c ÷÷ ø

æ 2y + 1 ö æ 2x + 1 ö 2 2 tan -1 ç tan -1 ç ÷+ ÷=c 3 3 3 è ø è 3 ø

é 3 (2y + 1+ 2x + 1) ù ê ú=c êë 3 - (2y + 1) (2x + 1) úû

or

2 tan -1 3

or

é 3 (2x + 2y + 2) ù 3 tan -1 ê c = tan -1 3 A ú= 2 2x 2y 4xy 2 ëê ûú

Þ c=

2 2 é dy ù æ dy ö ù é (x – y)2 ê1 + ç ÷ ú = ê x + y dx ûú è dx ø úû ë êë This is the reqd. equ. Find the general solution of the differential

dx

1 æ ç y+ 2 2 -1 tan ç or 3 çç 3 2 è

Þ

6.

2

+

2 tan -1 3

(

3A

)

2 3 (x + y + 1) = 3A 2(1- x - y - 2xy) Reqd. general solution is x + y + 1 = A (1 – x – y – 2xy) Find the equation of the curve passing through Þ

8.

æ pö the point ç 0, ÷ whose differential equation is è 4ø sin x cos y dx + cos x sin y dy = 0. Sol. The given differential equation is sin x cos y dx + cos x sin y dy = 0 sin x sin y dx + dividing by cos x cos y Þ dy = 0 cos x cos y Integrating, òtan x dx + òtan y dy = log c or log sec x sec y = log c or sec x sec y = c æ pö curve passes through the point ç 0, ÷ è 4ø p sec 0 sec = c = 2 4 Hence, the reqd. equ. of the curve is sec x sec y = 2

22 9.

Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0. Sol. The differential equation is (1 + e2x) dy + (1 + y2) ex dx = 0 Dividing by (1 + e2x) (1 + y2) dy e x dx + =0 2 1+ y 1 + e 2x dy ex + Integrating, dx = c 1 + y2 1 + e 2x

ò

ò

Put ex = t, ex dx = dt or tan–1 y +

ò 1+ t dt

2

=c

tan–1 y + tan–1 t = c i.e., tan–1 y + tan–1 ex = c \ tan–1 1 + tan–1 1 =c

10.

put y = 1, x = 0

æ x ö = çè xe y + y 2 ÷ø dy (y ¹ 0) Sol. The differential equation is x ye y dx

æ dx ö dx = xex/y + y2 or ex/y ç y dy - x ÷ = y2 è ø dy æ ö dx -x÷ ex / y ç y è dy ø =1 ...(i) or y2 let ex/y = z Differentiating w.r.t. y

y ex/y

ex / y

Þ

æ dx ö ç dy × y - x ×1 ÷ dz çç ÷÷ = y2 è ø dy

from (i)

t +1 dt = dx 2t t +1 dt = dx + c Integrating, 2t 1 æ 1ö 1 + ÷ dt = x + c 2 çè tø 1 or (t + log 1 + 1) = x + c put, t = x – y 2 1 [(x – y) + log | x – y |] = x + c \ 2 1 when x = 0, y = –1, (1 + 0) = 0 + c 2 1 \ c= 2 1 1 solution is [(x – y) + log |x – y|] = x + 2 2 log |x – y| = x + y + 1. 12. Solve the differential equation

ò dz = ò dy + c

ò

ò

ò

æ e-2 x y ö dx = 1 (x ¹ 0). çç ÷ x ÷ø dy è x Sol. The differential equation æ e -2 x dy 1 e -2 x y ö dx y= + = 1 or çç ÷÷ dx x x x ø dy è x

æ dx ö ç y dy - x ÷ çç y2 ÷÷ = 1 è ø dz =1 or dz = dy dy

Integrating,

dt t +1 + t -1 2t dt t -1 = =1+ Þ dx = t +1 t +1 dx t +1 or

p p or 2 tan–1 1 = c or 2 × = c \ c = 4 2 The particular solution of the diff. equ. p tan–1 y + tan–1 ex = 2 Solve the differential equation

ex / y

dy + x – y –1 = 0 dx dy x - y -1 =put dx x - y +1 dy dt dy dt =1= put x – y = t, 1 – or dx dx dx dx dt t -1 =\ 1dx t +1

or (x – y – 1)

dy + Py = Q dx

linear equation of the form where P = Þ

z = y+ c

Required solution is ex/y = y + c. 11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = – 1, when x = 0. Sol. The differential equation is (x – y) (dx + dy) = dx – dy

1 e2 x ,Q= x x

I.F. = e ò Pdx = e2

x

The solution is y × e 2 =

ò

1 dx + c = 2 x

Reqd. sol. is y e 2

x

x

=

ò

x

e -2 x

x +c

=2 x +c

´ e2

x

dx + c

23 13. Find a particular solution of the differential dy equation + y cot x = 4x cosec x (x ¹ 0), given dx p that y = 0 when x = . 2 dy Sol. The differential equation + y cot x = 4x cosec x dx dy + Py = Q linear equation of the form dx Þ P = cot x , q = 4x cosec x I.F. = eòpdx = elog sin x = sin x \ The solution is y sin x = ò4x cosec x sin x dx + c = ò4x dx + c = 2x2 + c 2 i.e., y sin x = 2x + c p y = 0, x = 2 p2 2 Reqd. particular solution is

\c= –

p2 (sin x ¹ 0) 2 14. Find a particular solution of the differential dy equation (x + 1) = 2e–y – 1, given that y = 0 dx when x = 0. Sol. The differential equation is dy (x + 1) = 2e–y – 1 variables are separable dx dy dx = -y 2e - 1 x +1

y sinx = 2x2 –

Integrating Þ

ey

ò 2 - ey

ò 2e

dy -y

-1

=

ò

dx x +1

dx x +1 \ – ey dy = dt

dy = ò

Put 2 – ey = t dt =– = – log t = – log (2 – ey) t

ò

=ò =ò

e 2 e

dy = ò y

dx xe1

ey 2-e

y

(

dy = ò

Þ - log 2 - e y

dx 1

Þ - log 2 - e

Þ

1 2 - ey

(

= log ( x + 1) + log c = log ( x + 1) c

)

1 = A ( say ) c \ (x + 1) (2 – ey) = A Put x = 0, y = 0, 2 – 1 = A = 1 Reqd. particular solution is (x + 1) (2 – ey) = 1 Þ ( x + 1) 2 - e y =

or

2 – ey =

1 1 2x + 1 = or ey = 2 – x +1 x +1 x +1

2x + 1 , x ¹ –1 x +1

y = log 15.

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009? Sol. Let y be the population at an instant t. Now population increase at a rate a No. of inhabitants \

dy dy a y or = ky dt dt

ò

dy dy \ y = kdt Integrating y = òkdt + c or log y = kt + c ...(i) In 1999, t = 0 population = 20,000 \ log 20,000 = c Put the value of c in (i) log y = kt + log 20,000 or log y – log 20000 = kt

y = kt 20000 In 2004, t = 5, y = 25000

or log

log

...(ii)

25000 1 5 = k × 5 Þ k = log 20000 5 4

Equ (ii) as log

y 5ö æ1 = ç log ÷ t 20000 è 5 4ø

In 2009, t = 10 Þ log

xe

)

= ( x + 1) c

y 5ö 5 æ1 = log ÷ ´ 10 = 2 log 20000 çè 5 4ø 4 2

25 æ5ö Þ log ç ÷ = log 4 16 è ø Þ y=

Þ

y 25 = 20000 16

25 × 20000 = 25 × 1250 = 31250 16

24 16.

The general solution of the differential y dx - x dy equation = 0 is y (a) xy = c (b) x = cy2 (c) y = cx (d) y = cx2

Sol. The differential equation is

ydx - xdy =0 y

ò x -ò dx

log x – log y = c¢

17.

dy = c¢ (say) y x or y = c¢

1 x 1 Put c¢ = , = y = cx is the reqd. sol. c y c Option (c) is correct The general solution of a differential equation of the type

(b) y × eò P1 dx (c) x e ò P1 dy ò P dx (d) x e 1

1

1

1

1

ò P1 dy

ò P1 dx

ò P1 dy

ò P1 dx

Sol. The linear differential equation is

dx dy or =0 x y

Integrate,

ò ( Q e ) dy + C = ò (Q e ) dx + C = ò (Q e ) dy + C = (Q e ) dx + C ò

ò P dy (a) y e 1 =

dx + P1 x = Q 1 is dy

P1 and Q1 are the function of y Hence the solu. is x.e ò P1.dy =

ò (Q ´ e 1

ò P1.dy

dx P x = Q1 dy 1

I.F. = eò P1.dy

) dy + C

option (c) is correct 18. The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is (a) x ey + x2 = C (b) x ey + y2 = C x 2 (c) y e + x = C (d) y ey + x2 = C Sol. The differential equation is dy ex dy + (yex + 2x) dx = 0 or e x + exy = – 2x dx dy = 1.y = – 2x e–x I.F. = eòpdx = ex or dx \ Solution is yex = ò (–2x)e–x × ex dx + c = – ò2x dx + c = – x2 + c Þ yex + x2 = c Option (c) is correct