Class 12th PCMB

Q53.

The degree of the differntial equation 5 +

dy

dx

2

5

3 = x5 d2y

dx2 is:

1. 4

2. 2

3. 5

4. 10

Ans: Solution:

We have,

5 +

dy

dx

2

5

3 = x5 d2y

dx2

Taking cube power on both sides, we get

5 +

dy

dx

2

5

= x15 d2y

dx2

3

The highest order derivative is

d2y

dx2 and its power is 3.

Hence, the degree is 3.

Disclaimer: The correct potion is not given in the quation.

Q54. The differention equationdy/dx + Py = Qyn, n > 2 can be reduced to linear from by substituting:

1. z = yn − 1

2. z = yn

3. z = yn + 1

4. z = y1 − n

Ans: 4. z = y1 − n

Solution:

We have,

dy

dx + Py = Qyn

⇒ y − n dy

dx + Py1 − n = Q . . . (i)

Put z = y1 − n

Integrating both sides with respect to x, we get

dz

dx = (1 − n)y − n dy

dx

⇒ y − n dy

dx =

1

( 1 − n )

dz

dx

Now, (i),

1

( 1 − n )

dz

dx + Pz = Q

⇒

dz

dx + P(1 − n) = Q(1 − n)

Which is linear from of differential equation.

Therefore the given differential equation can be to linear by the z = y1 − n.

Q55. The order of the differential equation

2×2 d2y

dx2 − 3

dy

dx + y = 0 is

1. 2

2. 1

3. 0

4. not defined.

Ans: 1. 2

The given differential equation is 2×2 d2y

dx2 − 3

dy

dx + y = 0

{ ( ) } ( )

{ ( ) } ( )

{ ( ) } ( )

The highest order derivative present in the differential equation is

d2y

dx2 .

∴ its order is 2

Q56. The order of the differential whose general solution is given by

y = C1cos(2x + C2) + (C3 + C4)ax + C5 + C6sin(x − C7).

1. 3

2. 4

3. 5

4. 2

Ans: 3. 5

Solution:

The given equation can be reduced to :

y = C1cos(2x + C2) + (C3 + C4)ax + C5 + C6sin(x − C7)

Where C = C + C be a constant

There are 5 constant (C , C , C , C , C ) in the given differential equation.

Hence, the order of the dfifferential equation is 5.

Q57. Choose the correct answer from the given four option.

If y = e − x(Acosx + Bsinx), then y is a solution of:

1.

d2y

dx2 + 2

dy

dx = 0

2.

d2y

dx2 − 2

dy

dx + 2y = 0

3.

d2y

dx2 + 2

dy

dx + 2y = 0

4.

d2y

dx2 + 2y = 0

Ans: 3.

d2y

dx2 + 2

dy

dx + 2y = 0

Solution:

Given that, y = e − x(Acosx + Bsinx)

Differentiating both sides w.r.t.x, we get

dy

dx = e − x(Acosx + Bsinx) + e − x( − Asinx + Bcosx)

dy

dx = − y + e − x( − Asinx + Bcosx)

Again, Differentiating both sides W.r.t.x, we get

d2y

dx2 = −

dy

dx + e − x( − Acosx − Bsinx) − e − x( − Asinx + Bcosx)

⇒

d2y

dx2 = −

dy

dx − y −

dy

dx + y

⇒

d2y

dx2 = − 2

dy

dx − 2y

⇒

d2y

dx2 + 2

dy

dx + 2y = 0

Q58. The differential equation of the ellipse

x2

a2 +

y2

b2 = C is:

1.

y ″

y ′ +

y ′

y −

1

x = 0

2.

y ″

y ′ +

y ′

y +

1

x = 0

3.

y ″

y ′ −

y ′

y −

1

x = 0

4. None of these.

Ans: 1.

y ″

y ′ +

y ′

y −

1

x = 0

Solution:

We have,

3 4

1 2 3 6 7

[ ]

x2

a2 +

y2

b2 = C . . . (i)

Differentiating with respect to x, we get

2×2

a2 +

2y2

b2 y ′ = 0

x2

a2 +

y2

b2 y ′ = 0 . . . (ii)

Again differentiating with respect to x, we get

⇒

1

a2 +

1

b2 (y’2) +

xy

b2 y ″ = 0 . . . (iii)

Multiplying throughout by x, we get

⇒

x

a2 +

x

b2 (y’2) +

xy

b2 y ″ = 0 . . . (iv)

Subtracting (ii) from (iv),

1

b2 [x(y ′ )2 + xyy ″ − yy ″ ] = 0

⇒ x(y ′ )2 + xyy ″ − yy ″ = 0

Diving both sides by,

y ″

y ′ +

y ′

y −

1

x = 0

⇒

y ″

y ′ +

y ′

y −

1

x = 0

Q59. Integrating factor of the differential equation cosx

dy

dx + ysinx = 1 is:

1. cosx

2. tanx

3. secx

4. sinx

Ans: 3. secx

Solution:

We have,

cosx

dy

dx + ysinx = 1

Dividing both sides by, we get

dy

dx +

sin x

cos x y =

1

cos x

⇒

dy

dx + (tanx)y =

1

cos x

Comparing with we get,

P = tanx, Q =

2

cos x

Now,

I.F = e ∫ tan xdx

= elog ( sec x )

= secx

Q60. The solution of the differential equartion y y = y is:

1. x = C1eC2y + C3

2. y = C1eC2x + C3

3. 2x = C1eC2y + C3

4. None of these.

Ans: 2. y = C1eC2x + C3

Solution:

y1y3 = y22

y3

y2

=

y2

y1

⇒

d3y

dx3

d2y

dx2

=

d2y

dx2

dy

dx

1 3 2

( )

( )

( )

( )

⇒ ∫

d

dx

d3y

dx3

d2y

dx2

= ∫

d

dx

dy

dx

dy

dx

⇒ log

d2y

dx2 = log

dy

dx + logC4

⇒

d2y

dx

2

= C4

dy

dx

⇒ ∫

d

dx

d3y

dx3

d2y

dx2

= ∫C4 dx

⇒ log

dy

dx = C4x + C5

⇒ ∫dy = ∫eC4x+C5 dx

⇒ y =

eC4x + C3

C4

+ C6

⇒ y = C1eC2x + C3

Where,

C1 =

eC5

C4

C4 = C2

C6 = C3

Q61. Choose the correct answer from the given four options.

The order and degree of the differential equation

d3y

dx3

2 − 3

d2y

dx2 + 2

dy

dx

4 = y4 are:

1. 1, 4.

2. 3, 4.

3. 2, 4.

4. 3, 2.

Ans: 4. 3, 2.

Solution:

Given that

d3y

dx3

2 − 3

d2y

dx2 + 2

dy

dx

4 = y4

∴ Order = 3 and degree = 2

Q62. The general solution of the differential equation

y dx − x dx

y = 0 is

1. xy = C

2. x = Cy2

3. y = Cx

4. y = Cx2

Ans: 3. y = Cx

The given differential equation is

y dx − x dy

y = 0

or

y dx − x dy

x2 = 0 or d

x

y = 0

∴

x

y = constant.

∴

x

y = C or y = Cx

∴ (C) is correct answer.

The given differential equation is

y dx − x dy

y = 0

or

y dx − x dy

x2 = 0 or d

x

y = 0

∴

x

y = constant.

∴

y

x = C or y = Cx

( )

( )

( )

( )

( ) ( )

( )

( )

( )

( ) ( )

( ) ( )

( )

( )

Q63. Choose the correct answer from the given four option.

Which of the following is a second order differential equation?

1. (y ′ )2 + x = y2

2. y ′ y ″ + y = sinx

3. y ‴ + (y ″ )2 + y = 0

4. y ′ = y2

Ans: 2. y ′ y ″ + y = sinx

Solution:

The second order differential equation is y ′ y ″ + y = sinx.

Q64. Choose the correct answer from the given four option.

The degree of the differential equation

d2y

dx2 +

dy

dx

2 − sin

dy

dx is:

1. 1

2. 2

3. 3

4. Not defined

Ans: 4. Not difined

Solution:

The degree of above differential equation is not defined because on solving sin

dy

dx we will get an infinite series in the increasing

powers of

dy

dx . Therefore its degree is not defined.

Q65. The solution of the differention equation

dy

dx =

x2 + xy + y2

x2 is:

1. tan − 1 (x

y ) − logy + C

2. tan − 1 (y

x ) − logx + C

3. tan − 1 (x

y ) = logx + C

4. tan − 1 (y

x ) = logy + C

Ans: 2. tan − 1 (y

x ) = logy + C

Solution:

We have,

dy

dx =

x2 + xy + y2

x2 . . . (i)

This is homogenous differential equation.

Let y = ux

⇒

dy

dx = u + x

du

dx

Now, putting equation (i),

u+x

du

dx =

x2 + x2u + x2u2

x2

⇒ u + x

du

dx = 1 + u + u2

⇒ x

du

dx = 1 + u2

⇒ ( 1

1 + u2 )du =

1

x dx

Intergreting both sides, we get

∫ ( 1

1 + u2 )du = ∫

1

x dx

⇒ tan − 1u = logx + C

⇒ tan − 1 (y

2 ) = logx + C

Q66. Choose the correct answer from the given four options.

( ) ( ) ( )

( )

Which of the following is the general solution of

d2y

dx2 − 2

dy

dx + y = 0?

1. y = (Ax + B)ex

2. y = (Ax + B)e − x

3. y = Axex + Bex

4. y = Acosx + Bsinx

Ans: 1. y = (Ax + B)ex

Solution:

Given that,

d2y

dx2 − 2

dy

dx + y = 0

D2y − 2Dy + y = 0,

Where, D =

d

dx

D2y − 2Dy + y = 0

The auxiliary equation is m2 − 2m + 1 = 0

(m − 1)2 = 0

⇒ m = 1, 1

Since, the roots are real and equal.

∴ CF = (Ax + B)ex

⇒ y = (Ax + B)ex

[ Since, if roots of Auxiliary equation are real and equal say (m), then CF = (C1x + C2)emx ]

Q67. The Integrating Factor of the differential equation x

dy

dx − y = 2×2 is

1. e − x

2. e − y

3.

1

x

4. x

Ans: 3.

1

x

The given differential equation is:

x

dy

dx − y = 2×2

⇒

dy

dx −

y

x = 2x

This is a linear differential equation of the form:

dy

dx + py = Q (where p = −

1

x and Q = 2x)

The integrating factor (I.F) is given by the relation,

e ∫ pdx

∴ I.F = e ∫ −

1

x dx = e − log x = elog ( x − 1 ) = x − 1 =

1

x

Q68. Choose the correct answer from the given four option.

Integrating factor of the differential equation cos

dy

dx + ysinx = 1 is:

1. cosx

2. tanx

3. secx

4. sinx

Ans: 3. secx

Solution:

we have, cos

dy

dx + ysinx = 1

⇒

dy

dx + ytanx = secx

This is a linear differential equation.

On comparing it with

dy

dx + Py = Q, we get

P = tanx and Q = secx

I.F. = e ∫ Pdx = e ∫ tan xdx

= elog sec x = secx

Q69. Which of the following is the integrating factor of (xlogx)

dy

dx + y = 2logx?

1. x

2. ex

3. logx

4. log(logx)

Ans: 3. logx

Solution:

We have,

(xlogx)

dy

dx + y = 2logx

Dividing both sides by, we get

dy

dx +

y

xlog x = 2

log x

xlog x

⇒

dy

dx +

y

xlog x =

2

x

⇒

dy

dx + ( 1

xlog x )y =

2

x

Comparing with we get,

P =

1

xlog x , Q =

2

x

I.F = e ∫

1

xlog x dx

= elog ( log x )

= logx

Q70. The solution of the differential equation

dy

dx = 1 + x + y2 + xy2, y = (0) is:

1. y2 = exp(x +

x2

2 − 1)

2. y2 = 1 + C exp x +

x2

2

3. y = tan(C + x + x2)

4. y = tan x +

x2

2

Ans: 4. y = tan(x +

x2

2 )

Solution:

We have,

dy

dx = 1 + x + y2 + xy2

⇒

dy

dx = (x + 1)y2(x + 1)

⇒

dy

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

⇒

dy

( 1 + y2 )

= (x + 1)dx

Integrating both sides, we get

∫

dy

( 1 + y2 )

= ∫(x + 1)dx

⇒ tan − 1 =

x2

2 + x + C . . . (i)

Now, y(0) = 0

∴ tan − 1(0) =

0

2 + 0 + C

⇒ C = 0

Putting the value of C in (i),

⇒ tan − 1y =

x2

2 + x

⇒ y = tan(x2

2 + x)

Q71.

( )

( )

The general solution of a differential equation of the type

dx

dy + P1x = Q1 is

1. y e ∫ P1dy = ∫ Q1e ∫ P1dy dy + C

2. y.e ∫ P1dx = ∫ Q1e ∫ P1dx dx + C

3. x e ∫ P1dy = ∫ Q1e ∫ P1dy dy + C

4. x e ∫ P1dx = ∫ Q1e ∫ P1dx dx + C

Ans: 3. x e ∫ P1dy = ∫ Q1e ∫ P1dy dy + C

The given differential equation is

dx

dy + P1 x = Q1

Multiplying through by e ∫ P1dy, we get,

dx

dy e ∫ P1dy + P1 x e ∫ P1dy = Q1 e ∫ P1dy

or

d

dy (x e ∫ P1dy) = Q1 e ∫ P1dy

∵

d

dy (x e ∫ P1dy) = x e ∫ P1dy d

dy (∫P1dy) + e ∫ P1dy dx

dy

= x e ∫ P1dy P1 + e ∫ P1dy

dx

dy

=

dx

dy e ∫ P1dy + P1 x e ∫ P1dy

Integrating both sides w.r.t.y, we get,

xe ∫ P1dy = ∫Q1e ∫ P1dydy + C which is required solution.

∴ (C) is correct answer.

The given differential equation is

dx

dy + P1x = Q1

Multiplying through by e ∫ P1dy, we get,

dx

dy e ∫ P1dy + P1x e ∫ P1dy = Q1 e ∫ P1dy

or

d

dy (x e ∫ P1dy) = Q1 e ∫ P1dy

∵

d

dy (x e ∫ P1dy) = x e ∫ P1dy. d

dy (∫P1dy) + e ∫ P1dy dx

dy

= x e ∫ P1dy P1 + e ∫ P1dy

dx

dy

=

dx

dy e ∫ P1dy + P1 xe ∫ P1dy

Integrating both sides w.r.t.y, we get,

x e ∫ P1dy = ∫Q1 e ∫ P1dydy + C which is required solution.

Q72. The famliy of curve in which the sub tangent at any point of a curve is double is the abscissae, is

1. x = Cy

2. y = Cx

3. x = Cy

4. Y = Cx

Ans: 1. x = Cy

Solution:

It is given that subtangent at any point of a curve is doble of the abscissa.

∴

y

dy

dx

= 2x

y = 2x

dy

dx

∫

dx

x = 2∫

dy

y

( )

( )

( )

( )

( )

[ ]

[ ]

2

2

2 2

2

logx = 2logy + a

logx = logy2 + logC

logx = logCy2

x = Cy2

Q73. Choose the correct answer from the given four option.

Family y = Ax + A of curves is represented by the differential equation of degree:

1. 1.

2. 2.

3. 3.

4. 4.

Ans: 1. 1.

Solution:

Given is, y = Ax + A

Differentiating both sides w.r.t. x, we get

dy

dx = A

This equation can be differentiated only once because it has only one arbitrary constant.

∴ Degree = 1

Q74. The general solution of the differential equation

dy

dx + y g(x) = g(x) g ′ (x) is a given function of x, is:

1. g(x) + log(1 + y + g(x)) = C

2. g(x) + log(1 + y − g(x)) = C

3. g(x) − log(1 + y − g(x)) = C

4. None of these.

Ans: 2. g(x) + log(1 + y − g(x)) = C

Solution:

We have,

dy

dx + y g ′ (x) = g(x) g ′ (x) . . . (i)

Clearly, it is a linear differential equation of the form

dy

dx + Py = Q

Where P = g ′ (x), Q = g(x) g ′ (x)

I.F = e ∫ Pdx

= e ∫ g ′ ( x ) dx

= eg ( x )

Multiplying both sides, we get

eg ( x ) dy

dx + yg ′ (x) = eg ( x ) g(x) g ′ (x)

eg ( x ) dy

dx + eg ( x ) yg ′ (x) = eg ( x ) g(x) g ′ (x)

Integrating both sides with respect to x, we get

y eg ( x ) = eg ( x ) g(x) g ′ (x) dx + K

y eg ( x ) = I + K

Where, I = eg ( x ) g(x) g ′ (x) dx

Now,

I = eg ( x ) g(x) g ′ (x) dx

Putting g ′ (dx) = dt

I = ∫t et dt

= t∫ et dt − ∫

d

dx (t)∫et dt dt

= tet − et

= g(x)eg ( x ) − eg ( x )

⇒ y eg ( x ) = g(x)eg ( x ) − eg ( x ) + K

⇒ y eg ( x ) = g(x)eg ( x ) − eg ( x ) = K

Taking log on both sides, we get

log y+1 − g(x) = − g(x) + logK

3

3

( )

[ ]

[ ]

Q75. Choose the correct answer from the given four options.

General solution of

dy

dx + ytanx = secx is:

1. ysecx = tanx + c

2. ytanx = secx + c

3. tanx = secx + c

4. xsecx = tany + c

Ans: 1. ysecx = tanx + c

Solution:

Given differential equation is

dy

dx + ytanx = secx

This is a linear differential equation

Here, P = tanx, Q = secx,

∴ I.F. = e ∫ tan xdx

= elog | sec x | = secx

Thus, the general solution is

y. secx = ∫secx. secx + C

⇒ y. secx = ∫sec2xdx + C

⇒ y. secx = tanx + C

Q76. Choose the correct answer from the given four options.

The solution of the equation (2y − 1)dx − (2x + 3)dy = 0 is:

1.

2x − 1

2y + 3 = k

2.

2y + 1

2x − 3 = k

3.

2x + 3

2y − 1 = k

4.

2x − 1

2y − 1 = k

Ans: 3.

2x + 3

2y − 1 = k

Solution:

Given is, (2y − 1)dx − (2x + 3)dy = 0

⇒ (2y − 1)dx = (2x + 3)dy

⇒

dx

2x + 3 =

dy

2y − 1

On integrating both sides, we get

1

2 log(2x + 3) =

1

2 log(2y − 1) + logC

⇒

1

2 log(2x + 3) − log(2y − 1) = logC

⇒

1

2 log

2x + 3

2y − 1 = C2

⇒

2x + 3

2y − 1

1

2 = C

⇒

2x + 3

2y − 1 = C2

⇒

2x + 3

2y − 1 = C2

⇒

2x + 3

2y − 1 = k, Where k = c2

Q77. Choose the correct answer from the given four option.

The number of solutions of

dy

dx =

y+1

x − 1 when y(1) = 2:

1. None.

2. One.

3. Two.

4. Infinity.

Ans: 2. One.

( )

( )

Solution:

Given that,

dy

dx =

y+1

x − 1

⇒

dy

y+1 =

dx

x − 1

On integrating both sides, we get

∫

dy

y+1 = ∫

dx

x − 1

log(y + 1) = log(x − 1) − logC

C(y + 1) = (x − 1)

⇒ C =

x − 1

y + 1

When x = 1 and y = 2, then C = 0

So, the required solution is x – 1 = 0

Hence, only one solution exists.

Q78. Choose the correct answer from the given four option.

tan − 1 + tan − 1y = C is the general solution of the differential equation:

1.

dy

dx =

1 + y2

1 + x2

2.

dy

dx =

1 + x2

1 + y2

3. (1 + x2)dy + (1 + y2)dx = 0

4. (1 + x2)dx + (1 + y2)dy = 0

Ans: 3. (1 + x2)dy + (1 + y2)dx = 0

Solution:

Given is, tan − 1 + tan − 1y = C

On differentiating above eqaution w.r. t. x, we get

1

1 + x2 +

1

1 + y2 .

dy

dx = 0

⇒

1

1 + y2 .

dy

dx = −

1

1 + x2

⇒ (1 + x2)dy + (1 + y2)dx = 0

Q79. Choose the correct answer from the given four option.

y = aemx + be − mx satisfies which of the following differential equation?

1.

dy

dx + my = 0

2.

dy

dx − my = 0

3.

d2y

dx2 − m2y = 0

4.

d2y

dx2 + m2y = 0

Ans: 3.

dy

dx − m2y = 0

Solution:

Given that, y = aemx + be − mx

On differentiating both sides w.r.t.x, we get

dy

dx = maemx + bme − mx

Again, differentiating both sides w.r.t.x, we get

d2y

dx2 = m2aemx + bm2e − mx

⇒

d2y

dx2 = m2(aemx + be − mx)

⇒

d2y

dx2 = m2y

⇒

d2y

dx2 − m2y = 0

Q80.

The equation of the curve whose slope is given by

dy

dx =

2y

x ; x > 0, y > 0 and which passes through the point (1,

1) is:

1. x2 = y

2. y2 = x

3. x2 = 2y

4. y2 = 2x

Ans: 1. x2 = y

Solution:

We have,

dy

dx =

2y

x

⇒

1

2 ×

1

y dy =

1

x dx

Interating both sides, we get

⇒

1

2 ∫

1

y dy = ∫

1

x dx

⇒

1

2 logy = logx + logC

⇒ logy

1

2 − logx = logC

⇒ log(√y

2 ) = logC

⇒

√y

2 = C

⇒ √y = Cx . . . (i)

As (i) passes through (1, 1), we get

1 = C

Putting the value of C in (1), we get

⇒ √y = x

⇒ y = x2

Q81. The differential equation which respresents the famliy of curves y = eCx is:

1. y1 = C2y

2. xy1 − logy = 0

3. xlogy = yy1

4. ylogy = xy1

Ans: 2. ylogy = xy1

Solution:

We have,

y = eCx

Taking in both sides, we get

⇒ logy = Cx . . . (1)

Differentiating both sides of (i) with respect to x, we get

1

y1

= C

Substituting the value of C in in (i). we get

logy =

y1

y x

⇒ y logy = y1x

Q82. Choose the correct answer from the given four options.

Family y = Ax + A3 of curves will correspond to a differential equation of order:

1. 3

2. 2

3. 1

4. Not defined

Ans: 3. 1

Solution:

Given family of curves is y = Ax + A3 . . . . . (i)

⇒

dy

dx = A

Replacting A by

dy

dx in Eq. (i), we get

y = x

dy

dx +

dy

dx

3

∴ Order = 1

Q83. Choose the correct answer from the given four options.

The solution of

dy

dx + y = e − x, y(0) = 0 is:

1. y = e − x(x − 1)

2. y = xex

3. y = xe − x + 1

4. y = xe − x

Ans: 4. y = xe − x

Solution:

We have,

dy

dx + y = e − x

Which is a linear differential equation.

Here, P = 1 and Q = e − x

I.F. = e ∫ dx = ex

The general solution is

y. e − x = ∫e − x. ex dx + C

⇒ yex = ∫dx + C

⇒ ye − x = x + C . . . . (i)

Given, when x = 0 and y = 0

⇒ 0 = 0 + C

⇒ C = 0

Eq, (i) resuces to y. e − x = x or y = xe − x.

Q84. Integrating factor of the differntial equation cosx

dy

dx + ysinx = 1is:

1. sinx

2. secx

3. tanx

4. cosx

Ans: 2. secx

Solution:

We have,

cosx

dy

dx + ysinx = 1

Dividing both sides by we get

dy

dx +

sin x

cos x y =

1

cos x

⇒

dy

dx + (tanx)y =

1

cos x

Comparing with

dy

dx + Py = Q

P = tanx

Q =

1

cos x

Now,

I.F = e ∫ tan xdx

= elog ( sec x )

= secx

Q85. The solution of the differention

dy

dx + 1 = ex + y is:

1. (x + y)ex + y = 0

( )

2. (x + C)ex + y = 0

3. (x − C)ex + y = 1

4. (x − C)ex + y + 1 = 0

Ans: 4. (x − C)ex + y + 1 = 0

Solution:

We have,

dy

dx + 1 = ex + y

Let x+y = u

⇒ 1 +

dy

dx =

du

dx

⇒

dy

dx + 1 =

du

dx

⇒

dy

dx = eu

⇒ e − udu = dx

Intergrating both sides, we get

⇒ e − u = x − C

⇒ − 1 = e − u(x − C)

⇒ (x − C)ex + y + 1 = 0

Q86. Choose the correct answer from the given four options.

The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the

point is:

1. An ellipse.

2. parabola.

3. circle.

4. rectangular hyperbola.

Ans: 4. rectangular hyperbola.

Solution:

Slope of langent to the curve =

dy

dx

According to the question,

dy

dx =

x

y

⇒ ydy = xdx

On integrating both sides, we get

y2

2 =

x2

2 + C

⇒ y2 − x2 = 2C which is an equation of rectangular hyperbola.

Q87. Choose the correct answer from the given four options.

The solution of the differential equation

dy

dx = ex − y + x2e − y is:

1. y = ex − y − x2e − y + c

2. ey − ex =

x3

3 + c

3. ex + ey =

x3

3 + c

4. ex − ey =

x3

3 + c

Ans: 2. ey − ex =

x3

3 + c

Solution:

We have,

dy

dx = ex − y + x2e − y

⇒ eydy = (ex + x2)dx

⇒ ∫eydy = ∫(ex + x2)dx

⇒ ey = ex +

x3

3 + C

⇒ ey − ex =

x3

3 + C

Q88. The integrating factor of the differential equation (xlogx)

dy

dx + y = 2 logx is given by:

1. log(logx)

2. ex

3. logx

4. x

Ans: 3. logx

Solution:

We have,

(xlogx)

dy

dx + y = 2 logx

Dividing both sides by,

dy

dx +

y

x + log x =

2

x

dy

dx +

1

x + log x y =

2

x

Comparing with

dy

dx + Py = Q

P =

1

xlog x

Q =

2

x

Now,

I.F = e ∫ Pdx

I.F = e ∫

1

xlog x dx

= elog ( log x )

= logx

Q89. The degree of the differential equation 2×2 d2y

dx2 + 3

dy

dx + y = 0 is:

1. 2

2. 1

3. 0

4. Not defined.

Ans: 1. 2

Solution:

We have,

2×2 d2y

dx2 + 3

dy

dx + y = 0

Here, the highest order is

d2y

d2x

.

Hence, the order is 2.

Q90. Choose the correct answer from the given four option.

The solution of the differential equation cosx siny dx + sinx cosy dy = 0 is:

1.

sin x

sin y = C

2. sinx siny = C

3. sinx + siny = C

4. cosx cosy = C

Ans: 2. sinx siny = C

Solution:

Given differential equation is

cosx siny dx + sinx cosy dy = 0

⇒ cosx siny dx = − sinx cosy dy = 0

⇒

cos x

sin x dx = −

cos y

sin y dy

⇒ cotx dx = − coty dy

On integrating both sides, we get

( )

log sinx = − log siny + log C

⇒ log sinx siny = log C

Carrying the exponent on both sides, we get

⇒ sinx siny = C

Q91. The differential equation satisfied by ax2 + by2 = 1 is:

1. xyy2 + y21

+ yy1 = 0

2. xyy2 + xy21

− yy1 = 0

3. xyy2 + xy21

+ yy1 = 0

4. None of these.

Ans: 2. xyy2 + xy21

− yy1 = 0

Solution:

We have,

ax2 + by2 = 1 . . . (i)

Differential both sides of (i) with x, we get

2ax + 2by

dy

dy = 0 . . . (ii)

Differential both sides of (ii) with x, we get

2ax + 2b

dy

dy

2 + 2by

d2y

dx2 = 0

y

d2y

dx2

dy

dx2 = −

2a

2b

x y

d2y

dx2

dy

dx2 = − −

y

x

dy

dx

xy

d2y

dx2 + x

dy

dx )2 − y

dy

dx = 0

xyy2 + x(y21

) − yy1 = 0

Q92. Which of the following is a homogeneous differential equation?

1. (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0

2. (xy) dx – (x + y ) dy = 0

3. (x + 2y ) dx + 2xy dy = 0

4. y dx + (x – xy – y ) dy = 0

Ans: 4. y dx + (x – xy – y ) dy = 0

Out of the given four options, option (D) is the only option in which all coefficients of dx and dy are of same degree i.e., 2. It may be

noted that xy is a term of second degree. Hence differential equation in option (D) is Homogeneous differential equation.

Q93. Choose the correct answer from the given four options.

The general solution of the differential equation (ex + 1)ydy = (y + 1)ex is:

1. (y + 1) = k(ex + 1)

2. y+1 = ex + 1 + k

3. y = log k(y + 1)(ex + 1)

4. y = log

ex + 1

y + 1 + k

Ans: 3. y = log k(y + 1)(ex + 1)

Solution:

Given differential equation

(ex + 1)ydy = (y + 1)exdx

⇒

dy

dx =

ex ( 1 + y )

( ex + 1 ) y

⇒

dy

dx =

( ex + 1 ) y

ex ( 1 + y )

⇒

dy

dx =

exy

ex ( 1 + y )

y

ex ( 1 + y )

( )

[ ( )]

[ ( )] ( )

(

3 3

3 2

2 2 2

2 2 2

{ }

{ }

{ }

⇒

dy

dx =

y

1 + y +

y

( 1 + y ) ex

⇒

dy

dx =

y

1 + y 1 +

1

ex dx

⇒

dy

dx =

y

1 + y 1 +

1

ex

⇒

dy

dx =

y

1 + y

ex + 1

ex

⇒

y

1 + y dy =

ex

ex + 1

dx

On integrating both sides, we get

∫

y

1 + y dy = ∫

ex

1 + ex dx

⇒ ∫

1 + y − 1

1 + y dy = ∫

ex

1 + ex dx

⇒ ∫1dy − ∫

y

1 + y dy = ∫

ex

1 + ex dx

⇒ y − log | (1 + y) + log(1 + ex) | + log(k)

⇒ y = log k(1 + y)(1 + ex)

Q94. A homogeneous differential equation of the from

dx

dy = h

x

y can be solved by making the substitution.

1. y = vx

2. v = yx

3. x = vy

4. x = v

Ans: 3. x = vy

We know that a homogeneous differential equation of the form

dx

dy = h

x

y can be solved by the substitution

x

y = v i.e., x = vy.

Q95. The solution of the differential equation

dy

dx − Ky = 0, y(0) = 1 approaches to zero when x →∝ if,

1. K = 0

2. K > 0

3. K < 0

4. None of these.

Ans: 3. K < 0

Solution:

We have,

⇒

dy

dx − Ky = 0

⇒

dy

dx = Ky

⇒

1

y dy = K dx

Integrating both sides, we get

∫

1

y dy = K∫dx

⇒ log | y | = Kx + C . . . (i)

Now,

y(0) = 1

C = 0

Putting C = 0 in (i),

log | y | = Kx

⇒ eKx = y

According to the quation,

eK ∝ = 0

Q96. The general solution of differention eqution of the e dy + (ye + 2x)dx = 0 is:

1. xe + x = C

2. xe + y = C

( )

( )

( )

( ) ( )

{ }

( )

( )

x x

y 2

y 2

3. ye + y = C

4. ye + x = C

Ans: 3. ye + x = C

Solution:

We have,

e dy + (ye + 2x) dx = 0

Diving both sides by we get,

dy

dx + (y +

2x

ex ) = 0

⇒

dy

dx + y = −

2x

ex

Comping with

dy

dx = Q we get,

P = 1, Q = −

2x

ex

Now,

I.F = e ∫ dx

= ex

Solution is given by,

y × I.F = ∫(Q × I.F)dx + C

⇒ yex = − ∫ex ×

2x

ex dx + C

⇒ yex = − 2∫x dx + C

⇒ yex = − x2 + C

⇒ yex + x2 = C

Q97. Choose the correct answer from the given four option.

Solution of differential equation xdy – ydx = 0 represents:

1. A rectangular hyperbola.

2. Parabola whose vertex is at origin.

3. Straight line passing through origin.

4. A circle whose centre is at origin.

Ans: 3. Straight line passing through origin.

Solution:

Given that, xdy−ydx = 0

⇒ xdy = ydx

⇒

dy

y =

dx

x

On integrating both sides, we get

⇒ ∫

dy

y = ∫

dx

x

⇒ logy = logx + logC

⇒ logy = logCx

⇒ y = Cx

Which is a straight line passing through origin.

Q98. The solution of the differential equation 2x

dy

dx − y = 3 resresents:

1. circles

2. straight lines

3. ellipses

4. parabolas

Ans: 4. parabolas

Solution:

We have,

2x

dy

dx − y = 3

⇒ 2x

dy

dx = 3 + y

x 2

y 2

x 2

x x

⇒

1

3 + y dy =

1

2x dx

Interating both sides, we get

⇒ ∫

1

3 + y dy =

1

2 ∫

1

x dx

⇒ log | 3 + y | =

1

2 log | x | + log | C |

⇒ log |

3 + y

√x

| = logC

⇒

3 + y

√x

= C

⇒ 3 + y = C√x

Squaring both sides, we get

(3 + y)2 = Cx . . . (i)

Thus, (i) the equation of parabolas.

Q99. A homogeneous dofferential equation of the from

dx

dy = h(

x

y ) can be solved by making the substitution:

1. y = vx

2. v = yx

3. x = vy

4. x = v

Ans: 3. x = vy

Solution:

A homogeneous differential of the from

dx

dy = h(

x

y ) can be solved by sunstituting x = vy.

Q100. Choose the correct answer from the given four option.

The degree of the differential equation

d2y

dx2 +

dy

dx

3 + 6y5 = 0 is:

1. 1.

2. 2.

3. 3.

4. 5.

Ans: 1. 1.

Solution:

d2y

dx2 +

dy

dx

3 + 6y5 = 0

We know that, the degree of a differential equation is highest exponent of order dervivatibve.

∴ Degree = 1

Q101. The solution of the differential equartion

dy

dx −

y ( x + 1 )

x = 0 is given by:

1. y = xex + C

2. x = yex

3. y = x + c

4. xy = ex + C

Ans: 1. y = xex + C

Solution:

We have,

dy

dx −

y ( x + 1 )

x = 0

⇒

dy

dx −

y ( x + 1 )

x

⇒

dy

dx =

( x + 1 )

x dx

Integrating both sides, we get

∫

dy

y = ∫

( x + 1 )

x dx

⇒ ∫

dy

y = ∫dx + ∫

1

x dx

( )

( )

⇒ logy = x + logx + C

⇒ logy − logx = x + C

⇒ log

y

x = x + C

⇒

y

x = ex + C

⇒ y = xex + C

Q102. Which of the following differentials equation has y = x as one of its particular solution?

1.

d2y

dx2 − x2 dy

dx + xy = x

2.

d2y

dx2 + x2 dy

dx + xy = x

3.

d2y

dx2 − x2 dy

dx + xy = 0

4.

d2y

dx2 + x2 dy

dx + xy = 0

Ans: 3.

d2y

dx2 − x2 dy

dx + xy = 0

Solution:

We have,

y = x . . . (i)

Differentiating both sides of (i) with respect to x, we get

dy

dx = 1 . . . (ii)

Differentiating both sides of (ii) with respect to x, we get

⇒

d2y

dx2 = 0

⇒

d2y

dx2 + x2 = x2

⇒

d2y

dx2 + x × x = x2 × 1

⇒

d2y

dx2 + xy = x2 × 1

⇒

d2y

dx2 + xy = x2 dy

dx

⇒

d2y

dx2 − x2 dy

dx + xy = 0

Q103. The degree of the differntial equation

d2y

dx2

2 =

dy

dx = y3 is:

1.

1

2

2. 2

3. 3

4. 4

Ans: 2. 2

Solution:

We have,

d2y

dx2

2 =

dy

dx = y3

The highest order derivative is

d2y

dx2 and its power is 2.

Hence, the degree is 2.

Q104. The solution of the differential equation x

dy

dx = y + x tan

y

x is:

1. sin

x

y = x + C

2. sin

y

x = Cx

3. sin

x

y = Cy

( ) ( )

( ) ( )

4. sin

y

x = Cy

Ans: 1. sin

y

x = Cx

Solution:

We have,

x

dy

dx = y + x tan

y

x

⇒

dy

dx =

y

x + tan

y

x . . . (i)

Let y = υx

⇒

dy

dx = υ + x

dυ

dx

Putting both value in (i)

υ + x

dυ

dx = υ + tanυ

⇒

dυ

tan υ =

dx

x

Integrating both sides, we get

logsinυ = logx + logC

⇒ log

sin υ

x = logC

⇒

sin υ

x = C

⇒ sinυ = Cx

⇒ sin(

y

x ) = Cx