Class 12th PCMB

Q1. The integrating factor of the differential equation(1 − y2)

dx

dy + yx = ay( − 1 < y < 1) is:

1.

1

y2 − 1

2.

1

√y2 + 1

3.

1

1 − y2

4.

1

√1 − y3

Ans: 4.

1

√1 − y2

Solution:

We have,

(1 − y2)

dx

dy + yx = ay

dx

dy +

y

1 − y2 x =

ay

1 − y2

Comparing with we get,

P =

y

1 − y2 , Q =

ay

1 − y2

Now,

I.F = e ∫

y

1 − y2 dy

= e −

1

2 ∫

− 2y

1 − y2 dy

= e −

1

2 log | 1 − y2 |

= elog

1

√1 − y2

=

1

√1 − y2

Q2. The general solution of the differential equation

dy

dx = ex + yis

1. ex + e − y = C

2. ex + ey = C

3. e − x + ey = C

4. e − x + e − y = C

Ans: 1. ex + e − y = C

The given differential equation is

dy

dx = ex+y or

dy

dx = ex. ey

Separating the variables, we get,

1

ey dy = ex dx

Integrating, ∫e − ydy = ∫ex dx

∴

e − y

− 1 = ex + c ′

∴ − e − y = ex + c ′ or ex + e − y = − c ′

∴ ex + e − y = C, which is required solution.

Q3. Choose the correct answer from the given four option.

Ans: 3.

1

x

Solution:

Given that

xdy

dx − y = x4 − 3x

Dividing both sides by x, we get

⇒

dy

dx −

y

x = x3 − 3

Here, P = −

1

x , Q = x3 − 3

∴ I.F. = e ∫ Pdx = e − ∫

1

x dx

e − log x =

1

x

Q4. The number of arbitrary constants in the particular solution of differential equation of fourth order is:

1. 3

2. 2

3. 1

4. 0

Ans: 4. 0

Solution:

The number of arbitray constant in the particular solution of a differential equation is always zero.

Q5. Choose the correct answer from the given four option.

Integrating factor of the differential equation (1 − x2)

dy

dx − xy = 1is:

1. − x

2.

x

1 + x2

3. √1 − x2

4.

1

2 log(1 − x2)

Ans: 3. √1 − x2

Solution:

Given is, (1 − x2)

dy

dx − xy = 1

⇒

dy

dx −

x

1 − x2 y =

1

1 − x

2

Which is a linear differential equation.

∴ I.F. = e − ∫

x

1 − x2 dx

Put 1 − x2 = t

⇒ − 2xdx = dt

⇒ xdx = −

dt

2

Now, I.F. = e

1

2 ∫

dt

t

e

1

2 log t = e

1

2 log ( 1 − x2 )

⇒ √1 − x2

Q6. Choose the correct answer from the given four option.

The degree of the differential equation 1 +

dy

dx

2

3

2 =

d2y

dx2 is:

1. 4

Integrating factor of

xdy

dx − y = x4 − 3x is:

1. x

2. logx

3.

1

x

4. − x

[ ( ) ]

2.

3

2

3. Not defined

4. 2

Ans: 4. 2

Solution:

Given is, 1 +

dy

dx

2

3

2 =

d2y

dx2

On squaring both sides, we get

1 +

dy

dx

2 3 =

d2y

dx2

2

So, the degree of differential equation is 2.

Q7. The solution of the differential equation

dy

dx +

2y

x = 0 with y(1) = 1 is given by.

1. y =

1

x2

2. x =

1

y2

3. x =

1

y

4. y =

1

x

Ans: 1. y =

1

x2

Solution:

We have,

dy

dx +

2y

x = 0

⇒

dy

dx =

− 2y

x

⇒

1

2 ×

1

y dy =

− 1

x dx

Integrating both sides, we get

⇒

1

2 ∫

1

y dy = − ∫

1

x dx

⇒

1

2 logy = − logx + logC

⇒ logy

1

2 + logx = logC

⇒ log(√yx) = logC

⇒ √yx = logC . . . (i)

As (i) y(1) = 1, we get

1 = C

Putting the valur of C in (i)

⇒ √yx = 1

⇒ y =

1

x2

Q8. Choose the correct answer from the given four option.

The integrating factor of the differential equation

dy

dx + y =

1 + y

x is:

1.

x

ex

2.

ex

x

3. xex

4. ex

Ans: 2.

ex

x

Solution:

We have,

dy

dx + y =

1 + y

x

[ ( ) ]

[ ( ) ] ( )

⇒

dy

dx =

1 + y

x − y

⇒

dy

dx =

1 + y − xy

x

⇒

dy

dx =

1

x +

y ( 1 − x )

x

⇒

dy

dx −

1 − x

x

⇒ y =

1

x

Here, P =

− ( 1 − x )

x , Q =

1

x

I.F. = e ∫ Pdx

= e − ∫

1 + x

x dx = e

x − 1

x

= e ∫ 1 −

1

x dx = e ∫ x − log x

= ex. elog

1

x = ex.

1

x

Q9. The number of arbitrary constants in the particular solution of a differential equation of second order is (are):

1. 0

2. 1

3. 2

4. 3

Ans: 1. 0

Solution:

In the particular solution of a differential equation of third order, there is no arbitrary constant because in the particular solution of

any differential equation, we remove all the arbitrary constant by substituting some particular values.

Q10. The general solution of the differntial equation

dy

dx =

y

x is:

1. logy = kx

2. y = kx

3. xy = k

4. y = klogx

Ans: 2. y = kx

Solution:

We have,

dy

dx =

y

x

⇒

1

y dy =

1

x dx

Integrating both sides, we get

∫

1

y dy = ∫

1

x dx

logy = logx + logk

logy − logx = logk

log

y

x = logk

⇒

y

x = k

⇒ y = kx

Q11. Choose the correct answer from the given four option.

The differential equation for y = Acosαx + Bsinαx, where A and B are arbitrary constants is:

1.

d2y

dx2 − α2y = 0

2.

d2y

dx2 + α2y = 0

3.

d2y

dx2 + αy = 0

4.

d2y

dx2 − αy = 0

( )

( )

( )

Ans: 2.

d2y

dx2 + α2y = 0

Solution:

Given, y = Acosαx + Bsinαx

⇒

dy

dx = − αAsinαx + αBcosαx

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

d2y

dx2 = − Aα2cosαx − α2Bsinαx

⇒

d2y

dx2 = − α2(Acosαx − Bsinαx)

⇒

d2y

dx2 = − α2y

⇒

d2y

dx2 + α2y = 0

Q12. Choose the correct answer from the given four options.

The general solution of the differential equation

dy

dx e

x2

2 + xy is:

1. y = ce

− x2

2

2. y = ce

x2

2

3. y = (x + c)e

x2

2

4. y = (c − x)e

x2

2

Ans: 3. y = (x + c)e

x2

2

Solution:

Given that,

dy

dx e

x2

2 + xy

⇒

dy

dx − xy = e

x2

2

It is a linear differential equation.

Here, P = − x, Q = e

x2

2

∴ I.F. = e ∫ − xdx = e

− x2

2

The general solution is

ye

− x2

2 = ∫e

− x2

2 . e

− x2

2 dx + c

⇒ ye

− x2

2 = ∫1dx + c

⇒ ye

− x2

2 = x + c

⇒ y = xe

x2

2 + ce

x2

2

⇒ y = (x + c)e

x2

2

Q13. What is integrating factor of

dy

dx + ysecx = tanx?

1. secx + tanx

2. log(secx + tanx)

3. esec x

4. secx

Ans: 1. secx + tanx

Solution:

We have,

dy

dx + ysecx = tanx

Comparing with We get,

P = secx, Q = tanx

Now,

I.F = e ∫ sec xdx

= elog ( sec x + tan x )

= secx + tanx

Q14. Which of the following differential equations has y = x as one of its particular solution?

1.

d2y

dx2 − x2 dy

dx + xy = x

2.

d2y

dx2 + x

dy

dx + xy = x

3.

d2y

dx2 − x2 dy

dx + xy = 0

4.

d2y

dx2 + x

dy

dx + xy = 0

Ans: The given equation of curve is y = x.

Differentiting with respect to x, we get:

dy

dx = 1 . . . (1)

Again, differentiating with respect to x, we get:

d2y

dx2 = 0 . . . (2)

Now, on substituting the values of y,

d2y

dx2 , and

dy

dx from equation (1) and (2) in each of the given alternatives, we find that only the

differential equation given in alternative C is correct.

d2y

dx2 − x2 dy

dx + xy = 0 − x2 ⋅ 1 + x ⋅ x

= − x2 + x2

= 0

Hence, the correct answer is C.

Q15. The differential equation x

dy

dx − y = x2 has the general solution:

1. y – x = 2Cx

2. 2y – x = Cx

3. 2y + x = 2Cx

4. y + x = 2Cx

Ans: 2. 2y − x3 = Cx

Solution:

We have,

x

dy

dx − y = x2

⇒

dy

dx −

1

x y = x2

Comparing with we get,

P = −

1

x

Q = x2

Now,

I.F = e − ∫

1

x dx

= e − log | x |

= elog |

1

x |

=

1

x

y × I.F = ∫x2 × I.Fdx + C

⇒ y

1

x = ∫x2 ×

1

x dx + C

⇒ y

1

x = ∫x2dx + C

⇒ y

1

x =

x2

2 + C

⇒ 2y − x3 = Cx

Q16. The general solution of differention eqution

y dx − x dy

y = 0 is:

3

3

2

2

1. xy = C

2. x = Cy

3. y = Cx

4. y = Cx

Ans: 3. y =Cx

Solution:

We have,

y dx − x dy

y = 0

⇒ y dx = x dy

⇒

1

y dy =

1

x dx

Integrating both sides, we get,

∫

1

y dy = ∫

1

x dx

⇒ logy = logx + D

⇒ logy − logx = C

⇒ log(y

2 ) = logC

⇒

y

2 = C

⇒ y = Cx

Q17. Choose the correct answer from the given four option.

Solution of the differential equation tanysec2xdx + tanx sec2ydy = 0is:

1. tanx + tany = k

2. tanx − tany = k

3.

tan x

tan y = k

4. tanx. tany = k

Ans: 4. tanx. tany = k

Solution:

We have, tanysec2xdx + tanx sec2ydy = 0

⇒ tanysec2xdx = − tanx sec2ydy

⇒

sec2 x

tan x dx = −

sec2 y

tan y dy

⇒ ∫

sec2 x

tan x dx = − ∫

sec2 y

tan y dy

⇒ logtanx = − logtany + logk

⇒ logtanx + logtany = logk

⇒ log(tanxtany) = logk

⇒ tanxtany = k

Q18. Choose the correct answer from the given four options.

The general solution of

dy

dx = 2x ex2 − y is:

1. ex2 − y = C

2. e − y + ex2 = C

3. ey = ex2 + C

4. ex2 + y = C

Ans: 3. ey = ex2 + C

Solution:

We have

dy

dx = 2x ex2 − y

⇒ ey =

dy

dx = 2x ex2

⇒ ∫eydy = 2∫xex2dx

Put x2 = t in R.H.S. integral we get

2xdx = dt

2

2

⇒ ∫eydy = ∫etdt

⇒ ey = et + C

⇒ ey = ex2 + C

Q19. Which of the following transformation reduce the differential quation into the form

du

dx + P(x)u = Q(x) into the

from

dz

dx +

z

x logz =

z

x2 (logz)2

1. u = logx

2. u = ez

3. u = (logz) − 1

4. u = (logz)2

Ans: 3. u = (logz) − 1

Solution:

We have,

dz

dx +

z

x logz =

z

x2 (logz)2 . . . (i)

Let u = (logz) − 1

du

dx = −

1

( log )2 ×

1

z ×

dz

dx

du

dx = − z(logz)2 du

dx

Substituting the value of the equation (i),

−z(logz)2 du

dx +

z

x logz =

z

x2 (logz2)

du

dx −

1

x

1

log z = −

1

x2

du

dx −

1

x (logz) − 1 = −

1

x2

du

dx −

1

x (u) = −

1

x2

It can be written as,

du

dx + P(x)u = Q(x)

Where, p(x) = −

1

x

q(x) = −

1

x2

Q20. The solution of x2 + y2 dy

dx = 4 is:

1. x + y = 12x + C

2. x + y = 3x + C

3. x + y = 3x + C

4. x + y = 12x + C

Ans: 4. x + y = 12x + C

Solution:

We have,

x2 + y2 dy

dx = 4

⇒ y2 dy

dx = 4 − x2

⇒ y2 dy

dx = (4 − x2)dx

Integrating both sides, we get

∫y2 dy

dx = ∫(4 − x2)dx

⇒

y3

3 = 4x −

x3

3 + D

⇒ y3 = 12x − x3 + 3D

⇒ x3 + y3 = 12x + C

Q21. Choose the correct answer from the given four option.

2 2

2 2

3 3

3 3

3 3

The differential equation y

dy

dx + x = C represents:

1. Family of hype.

2. Family of parabolas.

3. Family of ellipses.

4. Family of circles.

Ans: 4. Family of circles.

Solution:

Given that, y

dy

dx + x = C

⇒ y

dy

dx = C − x

⇒ ydy = (C − x)dx

On integrating both sides, we get

∫ydy = ∫(C − x)dx

⇒

y2

2 = Cx −

x2

2 + k

⇒

x2

2 +

y2

2 = Cx + k

⇒

x2

2 +

y2

2 − Cx = k

which represent family of circles.

Q22. The equation of the curve aatisfying the differential y(x + y3)dx = x(y3 − x) dy and passing through the point (1,

1) is:

1. y3 − 2x + 3x2y = 0

2. y3 + 2x + 3x2y = 0

3. y3 + 2x − 3x2y = 0

4. None of these.

Ans: 3. y3 + 2x − 3x2y = 0

Solution:

We have,

y(x + y3)dx = x(y3 − x)dy

⇒ (xy + y4)dx = (xy3 − x2)dy = 0

⇒ xy dx + y4dx − xy3dy + x2dy = 0

⇒ x(ydx + xdy) + y3(ydx − xdy) = 0

⇒ xd(xy) + x2y3 ( ydx − xdy )

x2 = 0

⇒

d ( xy )

x2y2 −

y

x d

y

x = 0

⇒

d ( xy )

x2y2 −

y

x d

y

x

Integrating both sides we get,

⇒ ∫

d ( xy )

x2y2 = ∫

y

x d

y

x

⇒ −

1

xy =

y

x

2

2 − C

⇒ −

1

xy −

1

2

y2

x2 + C = 0

⇒ y3 + 2x + 2Cx2y = 0

It is given that the curve passes through (1, 1).

Hence,

y3 + 2x + 2Cx2y = 0

(1)3 + 2(1) + 2C(1)(1) = 0

1 + 2 + 2C = 0

2C = − 3

C = −

3

2

The required curve is,

y3 + 2x − 3x2y = 0

( )

( )

( )

( )

( )

Q23. Solution of the differential equation

dy

dx +

y

x = sinx is:

1. x(y + cosx) = sinx + C

2. x(y − cosx) = sinx + C

3. x(y + cosx) = cosx + C

4. None of these.

Ans: 1. x(y + cosx) = sinx + C

Solution:

We have,

dy

dx +

y

x = sinx

⇒

dy

dx +

1

x y = sinx . . . (i)

Comparing with we get,

P =

1

x

Q = sinx

Now,

I.F = e ∫

1

x dx

= elog | x |

= x

Therefore, intergration of (i) is given by,

y × I.F = ∫x2 × I.F. dx + C

⇒ yx = ∫x sinx dx + C

⇒ yx = x∫sinx dx − ∫

d

dx (x)∫sinx dx dx + C

⇒ yx = − xcosx + ∫cosx dx + C

⇒ yx + xcosx = sinx + C

⇒ x(y + cosx) = sinx + C

Q24. Choose the correct answer from the given four option.

Integrating factor of the differential equation

dy

dx + ytanx − secx = 0 is:

1. cosx

2. secx

3. ecos x

4. esec x

Ans: 2. secx

Solution:

Given that,

dy

dx + ytanx − secx = 0

⇒

dy

dx + ytanx = secx

It is a linear differential equation of form

dy

dx + Py = Q

Here, P = tanx, Q = secx

= I.F. = e ∫ Pdx

= e ∫ tan xdx

= e ( log sec x )

= secx

Q25. Choose the correct answer from the given four option.

The differential equation of the family of curves x2 + y2 − 2ay = 0, where a is arbitrary constant, is:

1. (x2 − y2)

dy

dx = 2xy

2. 2(x2 + y2)

dy

dx = xy

3. 2(x2 − y2)

dy

dx = xy

4. (x2 + y2)

dy

dx = 2xy

[ ]

( )

Ans: 1. (x2 − y2)

dy

dx = 2xy

Solution:

Given equation is, x2 + y2 − 2ay = 0

⇒

x2 + y2

y = 2a

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

y 2x + 2y

dy

dx − ( x2 + y2 )

dy

dx

y2 = 0

⇒ 2xy + 2y2 dy

dx − (x2 + y2)

dy

dx = 0

⇒ (2y2 − x2 − y2)

dy

dx = − 2xy

⇒ (y2 − x2)

dy

dx = − 2xy

⇒ (x2 − y2)

dy

dx = 2xy

Q26. Choose the correct answer from the given four option.

The general solution of excosydx − exsinydy = 0 is:

1. excosy = k

2. exsiny = k

3. ex = kcosy

4. ex = ksiny

Ans: 1. excosy = k

Solution:

Given is, excosydx − exsinydy = 0

⇒ excosydx = exsinydy

⇒

dy

dx = tanydy

⇒ dx = tanydy

On integrating both sides, we get

x = logsecy + C

⇒ x − C = logsecy

⇒ secy = ex − c

⇒

1

cos y =

ex

ec

⇒ excosy = k [where, K = ec]

Q27. The solution of the differential equation

dy

dx =

ax + g

by + f represents a circle when,

1. a = b

2. a = − b

3. a = − 2b

4. a = 2b

Ans: 2. a = − b

Solution:

We have,

dy

dx =

ax + g

by + f

⇒ (by + f)dy = (ax + g)dx

Intergrating both sides, we get

⇒ ∫(by + f)dy = ∫(ax + g)dx

⇒ b

y2

2 + fy = a

x2

2 + gx + C

⇒ b

y2

2 + fy − a

x2

2 − gx = C

⇒ by2 + 2fy − ax2 − 2gx − 2C = 0

The above equation resprasents a circle.

Therefore, the coffrcients of x and y must be equal.

−a = b

( )

2 2

⇒ a = − b

Q28. Choose the correct answer from the given four options.

The differential equation of the family of curves y2 = 4a(x + a) is:

1. y2 = 4

dy

dx x +

dy

dx

2. 2y

dy

dx = 4a

3. y

d2y

dx2 +

dy

dx

2 = 0

4. 2x

dy

dx + y

dy

dx

2 − y = 0

Ans: 4. 2x

dy

dx + y

dy

dx

2 − y = 0

Solution:

We have equation of the curve as

y2 = 4a(x + a) . . . . . (i)

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

2y

dy

dx = 4a

⇒

1

2 y

dy

dx = a

On putting the value of a in Eq. (i), we get

y2 = 2y

dy

dx x +

1

2 y

dy

dx

⇒ y2 = 2xy

dy

dx + y2 dy

dx

2

⇒ 2x

dy

dx + y

dy

dx

2 − y = 0

Q29. The general solution of the differential equation ex dy + (y ex + 2x)dx = 0 is

1. x ey + x2 = C

2. x ey + y2 = C

3. y ex + x2 = C

4. y ey + x2 = C

Ans: 3. y ex + x2 = C

The given differential equation is

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

or ex dy

dx + y ex + 2x = 0 or

dy

dx + y +

2x

ex = 0 or

dy

dx + y = − 2x e − x

Comparing it with

dy

dx + Py = Q, we get, P = 1, Q = − 2x e − x

∴ ∫P dx = ∫1 dx = ‘x, e ∫ P dx = ex

Solution of given differential equation is

y e ∫ P dx = ∫Q e ∫ P dx + C or y ex = ∫( − 2x ex). e − x dx + C

or y ex = − 2∫x dx + C or y ex = − x2 + C

or y ex + x2 = C

∴ (C) is correct answer.

Q30. The solution of the differential equation xdy + ydy = x y dy – y x dx, is:

1. x – 1 = C(1 + y )

2. x + 1 = C(1 + y )

3. x – 1 = C(1 + y )

4. x + 1 = C(1 – y )

Ans: 1. x – 1 = C(1 + y )

Solution:

We have,

x dx+y dy = x2 dy − y2x dx

⇒ (x + xy2)dx = (x2y − y)dy

⇒

x

( x2 − 1 )

dx =

y

( 1 + x )2 dy

( )

( )

( )

( )

( )

( )

( )

2 2

2 2

2 2

3 3

3 3

2 2

⇒

2x

2 ( x2 − 1 )

dx =

2y

2 ( 1 + y )2 dy

Integrating both sides, we get

1

2 ∫

2y

( 1 + y )2 dy =

1

2 ∫

2x

( 1 + x )2 dx

⇒ log | (1 + y2) | =

1

2 log | (x2 − 1) | −

1

2 log | C |

⇒ log | (1 + y2) | = log | (x2 − 1) | − log | C |

⇒ log | (1 + y2) | = log |

x2 − 1

C |

⇒ 1 + y2 =

x2 − 1

C

⇒ C(1 + y2) = x2 − 1

Q31. The general solution of the differntial equation

dy

dx + ycotx = coses x is:

1. x+ysinx = C

2. x+ycosx = C

3. y+x(sinx + cosx) = C

4. ysinx = x + C

Ans: 4. ysinx = x + C

Solution:

dy

dx + ycotx = coses x

Comparting with

dy

dx + Py = Q we get

P = cotx

Q = coses x

Now,

I.F = e ∫ cot xdx

= elog ( sin x )

= sinx

So, the solution is given by

⇒ ysinx = ∫sinx × cosec xdx + C

⇒ ysinx = x + C

Q32. Choose the correct answer from the given four option.

Solution of

dy

dx − y = 1, y(0) = 1is given by:

1. xy = − ex

2. xy = − e − x

3. xy = − 1

4. y = 2ex − 1

Ans: 4. y = 2ex − 1

Solution:

Given is,

dy

dx − y = 1

⇒

dy

dx = y + 1

⇒

dy

1 + y = dx

On integrating both sides, we get

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

When x = 0 and y = 1, then

log2 = 0 + C

⇒ C = log2

The required solution is

log(1 + y) = x + log2

⇒ log

1 + y

2 = x

⇒

1 + y

2 = ex

( )

⇒ 1 + y = 2ex

⇒ y = 2ex − 1

Q33. The solution of the differential equation (x2 + 1)

dy

dx + (y2 + 1) = 0 is:

1. y = 2 + x2

2. y =

1 + x

1 − x

3. y = x(x − 1)

4. y =

1 + y

1 − y

Ans: 4. y =

1 − x

1 + x

Solution:

We have,

(x2 + 1)

dy

dx = − (y2 + 1) = 0

⇒ (x2 + 1)

dy

dx = − (y2 + 1)

⇒

1

( y2 + 1 )

dy = −

1

( x2 + 1 )

dx

Intergrating both sides, we get

⇒ ∫

1

( y2 + 1 )

dy = − ∫

1

( x2 + 1 )

dx

⇒ tan − 1y = − tan − 1x + tan − 1C

⇒ tan − 1y + tan − 1x = tan − 1C

⇒ tan − 1 x + y

1 − xy = tan − 1C

⇒

x + y

1 + xy = C

Disclaimer : The initial value given, So the find will be C = 1, So

⇒ x + y = 1 − xy

⇒ y + xy = 1 − x

⇒ y(1 + x) = 1 − x

⇒ y =

1 − x

1 + x

Q34. The solution of the differention equation (1 + x2)

dy

dx + 1 + y2 = 0 is:

1. tan − 1x − tan − 1y = tan − 1C

2. tan − 1y − tan − 1x = tan − 1C

3. tan − 1y ± tan − 1x = tan − 1C

4. tan − 1y + tan − 1x = tan − 1C

Ans: 4. tan − 1y + tan − 1x = tan − 1C

Solution:

We have,

(1 + x2)

dy

dx + 1 + y2 = 0

⇒ (1 + x2)

dy

dx = − (1 + y2)

⇒

1

( 1 + y2 )

dy = −

1

( 1 + x2 )

dx

Integrating both sides, we get

∫

1

( 1 + y2 )

dy = − ∫

1

( 1 + x2 )

dx

⇒ tan − 1y = − tan − 1x + tan − 1C

⇒ tan − 1y + tan − 1x = tan − 1C

Q35. Which of the following differentials equation has y = C1ex + C2e − x as the general solution?

1.

d2y

dx2 + y = 0

2.

d2y

dx2 − y = 0

( )

3.

d2y

dx2 + 1 = 0

4.

d2y

dx2 − 1 = 0

Ans: 2.

d2y

dx2 − 1 = 0

Solution:

We have,

y = C1ex + C2e − x . . . (i)

Differentiating both sides of (i) with we get,

dy

dx = C1ex + C2e − x . . . (ii)

Differentiating both sides of (ii) with we get,

d2y

dx2 = C1ex + C2e − x

⇒

d2y

dx2 = y

⇒

d2y

dx2 − y = 0

Q36. The general solution of the dofferential equation

dy

dx = ex + y is:

1. ex + e − y = C

2. ex + ey = C

3. e − x + ey = C

4. e − x + e − y = C

Ans: 1. ex + e − y = C

Solution:

We have,

dy

dx = ex + y

⇒

dy

dx = ex × ey

⇒ e − ydy = exdx

Integrating both sides, we get

∫e − ydy = ∫exdx

⇒ e − y = ex + D

⇒ ex + e − y = − D

⇒ ex + e − y = − C

Q37. Choose the correct answer from the given four option.

The solution of

dy

dx + y = e-x, y(0)is:

1. y = ex(x − 1)

2. y = xe − x

3. y = xex + 1

4. y = (x + 1)e − x

Ans: 2. y = xe − x

Solution:

We have,

dy

dx + y = e-x

This is a linear differential equation.

On comparing it with

dy

dx + Py = Q, we get

P = 1, Q = e − x

I.F. = e ∫ Pdx = e ∫ dx

So, the general solution is,

y. ex = ∫e − xexdx + C

⇒ y. ex = ∫dx + C

⇒ y. ex = x + C

Given that when x = 0 and y = 0

⇒ 0 = 0 + C

⇒ C = 0

Eq. (i) becomes y. ex = x

⇒ y = xe − x

Q38. Choose the correct answer from the given four option.

The order and degree of the differential equation

d2y

dx2 +

dy

dx

1

4 + x

1

5 respectively, are:

1. 2 and 4

2. 2 and 2

3. 2 and 3

4. 3 and 3

Ans: 1. 2 and 4

Solution:

We have,

d2y

dx2 +

dy

dx

1

4 + x

1

5

⇒

dy

dx

1

4 = − x

1

5 +

d2y

dx2

⇒

dy

dx = x

1

5 +

d2y

dx2

4

∴ order = -2, Degree = 4

Q39. Choose the correct answer from the given four options.

Solution of the differential equation

dy

dx +

y

x = sinx is:

1. x(y + cosx) = sinx + c

2. x(y − cosx) = sinx + c

3. xycosx = sinx + c

4. x(y + cosx) = cosx + c

Ans: 1. x(y + cosx) = sinx + c

Solution:

Given differential equation is

dy

dx + y

1

x = sinx

Which is liner differential equations.

Here, P =

1

x and Q = sinx

∴ I.F. = e ∫

1

x dx

= elog x = x

The general solution is

yx = ∫xsinxdx + c . . . . . (i)

Take I = ∫xsinxdx

−xcosx − ∫ − cosxdx

= − xcosx + sinx

Put the value of 1 in Eq. (i), we get

xy = − xcosx + sinx + c

⇒ x(y + cosx) = sinx + c

Q40.

If m and n are the order and degree of the differential equation (y2)5 +

4 ( y2 )3

y3 + y3 = y3 = x2 − 1, then

1. m = 3, n = 3

2. m = 3, n = 2

3. m = 3, n = 5

4. m = 3, n = 1

Ans: 2. m = 3, n = 2

Solution:

( )

( )

( ) ( )

( )

We have,

(y2)5 +

4 ( y2 )3

y3 + y3 = x2 − 1

y3(y2)5 + 4(y2)3 + (y3)2 = y3(x2 − 1)

The highest order is y and its highest in this equation is 2.

Hence, m = 3, n = 2.

Q41. The order of the differential equartion √1 − x4 + √1 − y4 = a(x2 − y2) is:

1. 1

2. 2

3. 3

4. 4

Ans: 1. 1

Solution:

The order of a differention depends on the number of constent in it.

Since √1 − x4 + √1 − y4 = a(x2 − y2) constant only 1 constant, the order of the differential equation is 1.

Q42. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

1. 0

2. 2

3. 3

4. 4

Ans: 4. 4

The number of arbitrary constants (c , c , c , etc.) in the general solution of a differential equation of n order is n.

Q43. Choose the correct answer from the given four option.

The solution of the differential equation

dy

dx =

1 + y2

1 + x2 is:

1. y = tan − 1x

2. y−x = k(1 + xy)

3. x = tan − 1y

4. tan(xy) = k

Ans: 2. y−x = k(1 + xy)

Solution:

Given that,

dy

dx =

1 + y2

1 + x2

⇒

dy

1 + y2 =

dx

1 + x2

On integrating both sides, we get

tan − 1y = tan − 1x + C

⇒ tan − 1y − tan − 1x = C

⇒ tan − 1 y − x

1 + xy = C

⇒

y − x

1 + xy = tanC

⇒ y − x = tanC(1 + xy)

⇒ y − x = K(1 + xy)

Where, k = tanC

Q44. The differential equation obtained on eliminating A and B from y = Acosωt + Bsinωt is:

1. y ″ + y ′ = 0

2. y ″ − ω2y = 0

3. y ″ = − ω2y = 0

4. y ″ + y = 0

Ans: 3. y ″ = − ω2y

3

1 2 3

th

( )

Solution:

We have,

y = Acosωt + Bsinωt . . . (i)

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

dy

dt = − Aωsinωt + Bωcosωt . . . (ii)

Differentiating both sides of (ii)

d2y

dt2 = − Aω2cosωt + Bω2sinωt

⇒

d2y

dt2 = − ω2(Acosωt + Bsinωt)

⇒

d2y

dt2 = − ω2y

y ″ = − ω2y

Q45. Choose the correct answer from the given four options.

The solution of the differential equation

dy

dx +

2xy

1 + x2 =

1

( 1 + x2 )2 is:

1. y(1 + x2) = C + tan − 1x

2.

y

1 + x2 = C + tan − 1x

3. ylog(1 + x2) = C + tan − 1x

4. y(1 + x2) = C + sin − 1x

Ans: 1. y(1 + x2) = C + tan − 1x

Solution:

Given is,

dy

dx +

2xy

1 + x2 =

1

( 1 + x2 )2

Here, P =

2x

1 + x2 and Q =

1

( 1 + x2 )2

This ia a linerar differential equation.

∴ I.F. = e ∫

2x

1 + x2 dx

Put 1 + x2 = t ⇒ 2x dx = dt

∴ I.F. = e ∫

dt

t = elog t

= elog ( 1 + x2 ) = 1 + x2

Thus, the general solution is

y. (1 + x2) = ∫(1 + x2)

1

( 1 + x2 )

+ C

⇒ y(1 + x2) = ∫

1

( 1 + x2 )

dx + C

⇒ y(1 + x2) = tan − 1x + C

Q46. If P and q are the order and degree of the differention y

dy

dx + x3 d2y

dx3 + xy = cosx then:

1. p < q

2. p = q

3. p > q

4. None of these.

Ans: 3. p > q

Solution:

We have,

y

dy

dx + x3 d2y

dx3 + xy = cosx

The highest order is

d2y

dz2 and it’s degree is 1.

So, the order is 2 and the degree is 1.

p = 2, q = 1

Clearly, p > q

Q47. The general solution of differention eqution of the type

dx

dy + P1x = Q1 is:

1. ye ∫ P1dy = ∫ Q1e ∫ P1dy dy + C

2. ye ∫ P1dy = ∫ Q1e ∫ P1dx dx + C

3. xe ∫ P1dy = ∫ Q1e ∫ P1dx dx + C

4. xe ∫ P1dx = ∫ Q1e ∫ P1dx dx + C

Ans: 3. xe ∫ P1dy = ∫ Q1e ∫ P1dy dy + C

Solution:

We have,

dx

dy + P1x = Q1

Comparing with the equation

dx

dy + Px = Q we get,

P = P1, Q = Q1

The solution of the equation

dx

dy + Px = Q is given by

xe ∫ P1dy = ∫ Q1e ∫ P1dy dy + C . . . (i)

Putting the value of P and Q in (i),

xe ∫ P1dy = ∫ Q1e ∫ P1dy dy + C

Q48. Which of the following is a homogeneous differnetial 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 ) = 0

Ans: 4. y dx + (x – xy – y )dy

Solution:

A differential equation is said to be homogenous if all the in the terms in the equation have equal degree and it can be written in the

from

dy

dx =

f ( x,y )

g ( x,y ) .

In (a), (b) and (c), the degree of all the terms is not equal.

But in the equation y dx + (x – xy – y )dy = 0, the degree of all the terms is 2.

Thus, (d) constant a homogeneous differential equation.

Q49. The number of arbitrary constants in the particular solution of a differential equation of third order are:

1. 3

2. 2

3. 1

4. 0

Ans: 4. 0

The number of arbitrary constants in a particular solution of a differential equation of any order is zero (0) as a particular solution is a

solution which contains no arbitrary constant. Therefore, option (d) is correct.

Q50. Choose the correct answer from the given four options.

The differential equation for which y = acosx + bsinx is a solution, is:

1.

d2y

dx2 + y = 0

2.

d2y

dx2 − y = 0

3.

d2y

dx2 + (a + b)y = 0

4.

d2y

dx2 + (a − b)y = 0

Ans:

{ }

{ }

{ }

{ }

{ }

{ }

{ }

3 3

3 2

2 2 2

2 2 2

2 2 2

1.

d2y

dx2 + y = 0

Solutuion:

Given equation is, y = acosx + bsinx

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

dy

dx = − asinx + bcosdx

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

d2y

dx2 = − asinx + bcosdx

⇒

d2y

dx2 = − y

⇒

d2y

dx2 + y = 0

Q51. Choose the correct answer from the given four options.

The order and degree of the differential equation 1 +

dy

dx

2 =

d2y

dx2 are:

1. 2,

3

2

2. 2, 3

3. 2, 1

4. 3, 4

Ans: 3. 2, 1

Solution:

Given that, 1 +

dy

dx

2 =

d2y

dx2

∴ Order = 2 and degree = 1

Q52.

The solution of the differential equation

dy

dx =

y

x +

ϕ (

y

x )

ϕ ′ (

y

x )

is:

1. ϕ(

y

x ) = Kx

2. xϕ(

y

x ) = K

3. ϕ(

y

x ) = Ky

4. yϕ(

y

x ) = K

Ans: 1. ϕ(

y

x ) = Kx

Solution:

We have,

dy

dx =

y

x +

ϕ (

y

x )

ϕ ′ (

y

x )

Let y = ux

⇒

dy

dx = u + x

du

dx

∴ u + x

du

dx = u +

ϕ ( u )

ϕ ′ ( u )

⇒ x

du

dx =

ϕ ( u )

ϕ ′ ( u )

⇒

ϕ ( u )

ϕ ′ ( u )

du =

1

x dx

Integrating both sides, we get

∫

ϕ ( u )

ϕ ′ ( u )

du = ∫

1

x dx

⇒ log | ϕ(v) | = log | x | + log | K |

⇒ log | ϕ(

y

2 ) | − log | x | = logK

⇒ log | ϕ(

y

2 ) | = logK

[ ( ) ]

[ ( ) ]

⇒ ϕ(

y

2 ) | = Kx