Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 9 Differential Equations. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Maths Chapter 9 Important Extra Questions Differential Equations

### Differential Equations Important Extra Questions Very Short Answer Type

Question 1.

Find the order and the degree of the differential equation: \(x^{2} \frac{d^{2} y}{d x^{2}}=\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{4}\)

(Delhi 2019)

Solution:

Here, order = 2 and degree = 1.

Question 2

Determine the order and the degree of the differential equation:\(\left(\frac{d y}{d x}\right)^{3}+2 y \frac{d^{2} y}{d x^{2}}=0\) (C.B.S.E. 2019 C)

Solution:

Order = 2 and Degree = 1.

Question 3.

Form the differential equation representing the family of curves: y = b (x + a), where « and b are arbitrary constants. (C.B.S.E. 2019 C)

Solution:

Wehave:y= b(x + a) …(1)

Diff. w.r.t. x, b.

Again diff. w.r.t. x, \(\frac{d^{2} y}{d x^{2}}\) = 0,

which is the reqd. differential equation.

Question 4.

Write the general solution of differential equation:

\(\frac{d y}{d x}\) = e^{x+y} (C.B.S.E. Sample Paper 2019-20)

Solution:

We have: \(\frac{d y}{d x}\) = e^{x+y}

⇒ e^{-y} dy = e^{x} dx [Variables Separable

Integrating, \(\int e^{-y} d y+c=\int e^{x} d x\)

⇒ – e^{-y} + c = e^{x}

⇒ e^{x} + e^{-y} = c.

Question 5.

Find the integrating factor of the differential equation:

y\(\frac{d y}{d x}\) – 2x = y^{3}e^{-y}

Solution:

The given equation can be written as.

Question 6.

Form the differential equation representing the family of curves y = a sin (3x – b), where a and b are arbitrary constants. (C.B.S.E. 2019C)

Solution:

We have: y – a sin (3x – b) …(1)

Diff. W.r.t y \(\frac{d y}{d x}\) = a cos (3x – b) .3

= 3a cos (3x – b)

\(\frac{d^{2} y}{d x^{2}}\) = -3a sin (3x – b) 3

= -9a sin (3x – b)

= -9y [Using (1)]

\(\frac{d^{2} y}{d x^{2}}\) + 9y = 0,m

which in the reqd. differential equation.

### Differential Equations Important Extra Questions Short Answer Type

Question 1.

Determine the order and the degree of the differential equation:

\(\left(\frac{d y}{d x}\right)^{3}+2 y \frac{d^{2} y}{d x^{2}}=0\) (Outside Delhi 2019C)

Solution:

Order = 2 and Degree = 1.

Question 2.

Form the differential equation representing the family of curves: y = e^{2x} (a + bx), where ‘a’ and ‘h’ are arbitray constants. (Delhi 2019)

Solution:

We have : y = e^{2x} (a + bx) …(1)

Diff. w.r.t. x, \(\frac{d y}{d x}\) = e^{2x} (b) + 2e2x (a + bx)

⇒ \(\frac{d y}{d x}\) = be^{2x }+ 2y ………….. (2)

Again diff. w.r.t. x,

\(\frac{d^{2} y}{d x^{2}}\) = 2be^{2x} + 2^{2x}

\(\frac{d^{2} y}{d x^{2}}\) = 2(\(\frac{d y}{d x}\) – 2y) + \(\frac{d y}{d x}\)

[Using (2)]

Hence, \(\frac{d^{2} y}{d x^{2}}\) -4 \(\frac{d y}{d x}\) + 4y = 0, which is the reqd. differential equation.

Question 3.

Solve the following differentia equation:

\(\frac{d y}{d x}\) + y = cos x – sin x (Outside Delhi 2019)

Solution:

The given differential equation is :

\(\frac{d y}{d x}\) + y = cos x – sin x dx Linear Equation

∴ I.F. = e^{∫1dx} = e^{x}

The solution is :

y.e^{x} = ∫ (cos x — sin x) e^{x} dx + C

⇒ y.e^{x} = e^{x} cos x + C

or y = cos x + C e^{-x}

Question 4.

Solve the following differential equation :

\(\frac{d x}{d y}\) + x = (tan y + sec^{2}y). (Outside Delhi 2019 C)

Solution:

The given differential equation is :

\(\frac{d x}{d y}\) + x = (tany + sec^{2}y).

Linear Equation

∵ I.F. = Jldy = ey

∴ The solution is :

x. ey = ∫ ey (tan y + sec^{2} y)dy + c

⇒ x. ey = ey tan y + c

= x = tan y + c e-y, which is the reqd. solution.

### Differential Equations Important Extra Questions Long Answer Type 1

Question 1.

Solve the differential equation

(x^{2} – y^{2})dx + 2xydy = 0 (C.B.S.E. 2018)

Solution:

log x = -log (1 + v^{2}) + log C

x(1 + v^{2}) = C

x(1 + \(\frac{y^{2}}{x^{2}}\)) = C

x^{2} + y^{2} = C.

Question 2.

Find the particular solution of the differential equation (1 + x^{2})\(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^{2}}\), given that y = 0 when x = 1(C.B.S.E. 2018 C)

Solution:

Solution is y( 1 + x^{2}) = \(\int \frac{1}{1+x^{2}} d x\)

= tan^{-1} x + C

When y = 0,x = 1,

then 0 = \(\frac{\pi}{4}\) + C

C = \(\frac{\pi}{4}\)

∴ y(1 + x^{2}) = tan ^{-1} x – \(\frac{\pi}{4}\)

i.e, y = \(\frac{\tan ^{-1} x}{1+x^{2}}-\frac{\pi}{4\left(1+x^{2}\right)}\)

Question 3.

Find the differential equation representing the family of curves y = ae^{bx + 5}, where ‘a’and ‘A’are arbitrary constants. {C.B.S.E. 2018)

Solution:

We have: y = ae^{bx + 5} + 5 …(1)

Diff. w.r.t. x, \(\frac{d y}{d x}\) = ae^{bx + 5}. (b)

\(\frac{d y}{d x}\) = dy ……(2) [Using (1)]]

Again diff. w.r.t x.,

\(\frac{d^{2} y}{d x^{2}}=b \frac{d y}{d x}\) ………(3)

Dividing (3) by (2),

which is the required differential equation.

Question 4.

Find the particular solution of the differential equation x dx – ye^{y} \(\sqrt{1+x^{2}}\) dy = 0, given that y = 1 when x = 0. (C.B.S.E. 2019 C)

Solution:

The given differential equation is:

When x = 0,y = 1, ∴ 1 = c + c(0) ⇒ c = 1.

Putting in (2), \(\sqrt{1+x^{2}}\) = 1 + e^{y}(y -1),

which is the reqd. particular solution.

Question 5.

Obtain the differential equation of the family of circles, which touch the x-axis at the origin. (N.C.E.R.T.; C.B.S.E. Sample Paper 2018)

Solution:

Let (0, α) be the centre of any member of the family of circles.

Then the equation of the family of circles is : x^{2} + (y-α)^{2} = α^{2}

⇒ x^{2} + y^{2} – 2αy = 0 …(1)

Diff. w.r.t. x, 2x + 2y \(\frac{d y}{d x}\) 2α \(\frac{d y}{d x}\) = 0

which is the required differential equation.

Question 6.

Obtain the differential equation representing the family of parabolas having vertex at the origin and axis along the positive direction of x-axis. (N.C.E.R.T.)

Solution:

Let S (a, 0) be the focus of any member of the family of parabolas.

Then the equation of the family of curves is y^{2} = 4 ax …………. (1)

Diff. w.r.t. x, 2y\(\frac{d y}{d x}\) = 4a ……………. (2)

Using (2) in (1), we get:

y^{2} = (2y\(\frac{d y}{d x}\))x

y^{2} – 2xy\(\frac{d y}{d x}\) = 0

which is the required differential equation.

Question 7.

Find the general solution of the differential equation:

(tan^{2}x + 2 tanx + 5) \(\frac{d y}{d x}\) = 2 (1 + tanx) sec^{2} x

Solution:

We have : (tan^{2} x + 2 tan x + 5)\(\frac{d y}{d x}\)

= 2 (1 + tanx) sec^{2}x

Integrating

Put t^{2} + 2t + 5 = z

so that (2t+ 2) dt = dz.

∴ From (2),

1 = \(\int \frac{d z}{z}\) = log| z | = log |t^{2} + 2t + 5|

= log |tan^{2} x| + 2 tan x + 5|

From (1), y = log |tan^{2} x| + 2 tan x + 5| + c,

which is the required general solution.

Question 8.

Solve the differential equation:

(x + 1) \(\frac{d y}{d x}\) = 2e^{-y} – 1 ; y(0) = 0. (C.B.S.E. 2019)

Solution:

The given equation can be written as :

\(\frac{d y}{2 e^{-y}-1}=\frac{d x}{x+1}\)

| Variables Separable

Integrating, \(\int \frac{e^{y} d y}{2-e^{y}}=\int \frac{d x}{x+1}\)

⇒ – log |2 – e^{y}| + log |C| = log |x + 1|

⇒ (2 – e^{y}) (x + 1) = C.

When x = 0, y = 0, then C = 1.

Hence, the solution is (2 – e^{y}) (x + 1) = 1.

Question 9.

Solve: (a) (i) \(\frac{d y}{d x}\) = 1 + x + y + xy

(ii) xyy’ = 1 + x + y + xy. (N.C.E.R.T.)

(b) Find the particular solution of the differential dy

equation \(\frac{d y}{d x}\) = 1 + x + y + xy, given that y = 0 when x = 1. (A.I.C.B.S.E. 2014)

Solution:

(a), (i) The given equation is :

\(\frac{d y}{d x}\) = 1 + x + y + xy

\(\frac{d y}{d x}\) = (1 +x) (1 +y)

\(\frac{d}{1+y}\) = (1 + x)dx.

|Variables Separable

Integrating,

\(\int \frac{d y}{1+y}\) = ∫(1 + x)dx + c

⇒ log |1 + y | = x + \(\frac { 1 }{ 2 }\) x^{2} + c,

which is the required solution.

(ii) The given equation is xyy’ = 1 + x + y + xy

⇒ xy\(\int \frac{d y}{1+y}\) = (1 + x)(1 + y)

= y – log|1 + y| = log|x|+ x + c

= x + log |x (1 + y)|+ c,

which is the required solution.

(b) From part (a) (i),

Iog|1+y| = x + \(\frac { 1 }{ 2 }\)x^{2} + c …(1)

When x = 1, y = 0, then:

log |1+0|= 1 + \(\frac { 1 }{ 2 }\)(1)^{2} + c

log 1 = 1 + \(\frac { 1 }{ 2 }\) + c

0 = \(\frac { 3 }{ 2 }\) + c

c = \(-\frac { 3 }{ 2 }\)

Putting in (1), log |1 + y| = x + \(\frac { 1 }{ 2 }\)x^{2} – \(\frac { 3 }{ 2 }\)

which is the required particular solution.

Question 10.

Solve the differential equation:

\(\frac{d y}{d x}\) = 1 + x^{2} + y^{2} + x^{2}y^{2}

given thaty = 1 when x = 0. (Outside Delhi 2019)

Solution:

The given equation is \(\frac{d y}{d x}\) = 1 + x^{2} + y^{2} + x^{2}y^{2}

⇒ \(\frac{d y}{d x}\) = (1+x^{2})(1 +y^{2})

⇒ \(\frac{d y}{1+y^{2}}\) (x^{2} + 1) dx

|Variables Separable

Integrating, ∫\(\frac{d y}{1+y^{2}}\) = ∫(x^{2} + 1) dx + C

tan^{-1}y = \(\frac{x^{3}}{3}\) + x + C ………. (1)

Where x = 0,y = 1,

∴ tan^{-1} (1) = C

C = \(\frac{\pi}{4}\)

Putting in (1),

tan^{-1} y = \(\frac{x^{3}}{3}+x+\frac{\pi}{4}\)

which is the reqd. particular solution.

Question 11.

Find the particular solution of the differential equation: e^{x} tan y dx + (2 – e^{x}) sec^{2} y dy = 0, given thaty = \(\frac{\pi}{4}\) where x = 0.

Solution:

The given differential equation is:

∴ From (1),

log C + log |e^{x} – 2| = log |tan y|

⇒ tany = C(e^{x} – 2) …(2)

When x = 0, y = y = \(\frac{\pi}{4}\)

∴ 1 = C(1 – 2)

C = -1

Putting in (2),

tan y = 2 – e^{x}

which is the required particular solution.

Question 12.

Find the particular solution of the following differential equation:

cosy dx + (1 + 2e^{-x}) siny dy = 0; y(0) = \(\frac{\pi}{4}\) .

(CBSE Sample Paper 2018 – 19)

Solution:

The given differential equation is:

cos y dx + (1 + 2e^{-x}) sin ydy = 0

log |e^{x}| – log |cos y| = log|C|

e^{x} + 2 = C cos y ….. (1)

Substituting, x = 0, y = \(=\frac{\pi}{4}\) in(1),we get:

1 + 2 = C cos \(=\frac{\pi}{4}\)

C = \(\frac{3}{1 / \sqrt{2}}\)

C = 3√2

Putting in (1),

e^{x} + 2 = 3√2 cos y

which is the required particular solution.

Question 13.

Find the particular solution of the following differential equation:

(x + 1) \(\frac{d y}{d x}\) = 2e^{-y} – 1 ; y = 0 when x = 0. (C.B.S.E. 2012)

Solution:

The given equation is (x + 1) \(\frac{d y}{d x}\) = 2e^{-y} – 1

-log|2 – e^{y}| = log|x + 1| + log|c|,

where c’ = log |c|

⇒ log \(\left|\frac{1}{2-e^{y}}\right|\) = log |c(x + 1)|

⇒ \(\frac{1}{2-e^{y}}\) = c(x + 1)

When x = 0, y = 0;

\(\frac{1}{2-e^{0}}\) = c(0 + 1)

⇒ c = \(\frac{1}{2-1}=\frac{1}{1}\) = 1

Putting (1)

\(\frac{1}{2-e^{y}}\) = 1.(x + 1)

⇒ (x + 1)(2 – e^{y}) = 1

which is the required particular solution.

Question 14.

Solve the differential equation:

(1 – y^{2}) (1 + log x)dx + 2xy dy = 0, given that when x = 1 ,y = 0.

(C.B.S.E. 2016)

Solution:

The given equation is :

(1 – y^{2}) (1 + log x)dx + 2xy dy = 0

Putting in (1),

\(\frac{(1+\log x)^{2}}{2}\) = log|1 – y^{2}| + 1/2

(1 + log x)^{2} = 2 log|1 – y^{2}| + 1

which is the required solution.

Question 15.

Solve the differential equation:

x cos x (\(\frac { y }{ x }\)) \(\frac{d y}{d x}\) = y cos(\(\frac { y }{ x }\)) = x (C.B.S.E. 2019 C)

Solution:

The given equation is:

Put \(\frac { y }{ x }\) = v i.e, y = vx so that \(\frac { dy }{ dx }\) = v + x \(\frac { dv }{ dx }\)

∴ (1) becomes:

cos v (v + x\(\frac { dv }{ dx }\)) = vcosv + 1

= v + x \(\frac { dv }{ dx }\) = v + \(\frac { 1 }{ cos v }\)

x \(\frac { dv }{ dx }\) = \(\frac { 1 }{ cos v }\)

cos v dv = \(\frac { dx }{ x }\)

[Variables Seperable]

Integrating, ∫ cos v dv = ∫ \(\frac{d x}{ x}\) + c

sin = log|x| + c

sin \(\frac{y}{ x}\) = log|x| + c,

which is the reqd. solution.

Question 16.

Find the general solution of the differential equation: \(\frac{d x}{d y}=\frac{y \tan y-x \tan y-x y}{y \tan y}\) (C.B.S.E. Sample Paper 2018-19)

Solution:

The given differential equation is :

|Linear Equation

Comparing with \(\frac{d x}{d y}\) + Py = Q, we have :

‘P’ = \(\frac{1}{y}+\frac{1}{\tan y}\) and ‘Q’ = 1

∴ I.F = \(e^{\int\left(\frac{1}{y}+\frac{1}{\tan y}\right) d y}=e^{\log y+\log \sin y}\)

e^{log(ysiny)} = ysiny.

∴ The solution is

x x I.F. = ∫ (Q x I.F.)dy + C

⇒ x(y sin y) = ∫(1 x y sin y)dy + C

⇒ xy sin y = ∫ y sin y dy + C

⇒ xy sin y = y (- cos y) – ∫(1) -cos y)dy + C

(Integrating by Parts)

⇒ xy sin y = – y cos y + ∫ cos ydy + C

⇒ xy sin y = – y cos y + sin y + C

⇒ x = \(\frac{\sin y-y \cos y+C}{y \sin y}\)

which is the required general solution.

Question 17.

Solve the differential equation :

xdy – ydx = \(\sqrt{x^{2}+y^{2}}\) dx , given that y = 0 when x = 1. (C.B.S.E. 2019)

Solution:

The given equation can be written as :

where x = 1, y = 0

∴ 0 + \(\sqrt{1+0} \) = c(1)

⇒ c = 1

Putting in (2), y + \(\sqrt{x^{2}+y^{2}}\) = x^{2},

which is the reqd. solutions.

Question 18.

Solve the following differential equation:

\(x y \log \left(\frac{y}{x}\right) d x+\left(y^{2}-x^{2} \log \left(\frac{y}{x}\right)\right) d y=0\) (C.B.S.E. 2010 C)

Solution:

The given equation can be written as:

Question 19.

Solve the differential equation:

xdy-y dx= \(\sqrt{x^{2}+y^{2}}\) dx.

(iC.B.S.E. Sample Paper 2019)

Solution:

The given differential equation is:

Put y = vx, so that \(\frac{d y}{d x}\) = dv + x \(\frac{d v}{d x}\)

(1) becomes:

⇒ y+ \(\sqrt{x^{2}+y^{2}}\) = cx^{2},

which is the reqd. solution.

Question 20.

Prove that x^{2} – y^{2} = c(x^{2} + y^{2})^{2} is the general solution of the differential equation: (x^{3} – 3xy^{2})dx = (y^{3} – 3x^{2}y)dy, where c is a parameter. (C.B.S.E. 2017)

Solution:

We have :(x^{3} – 3xy^{2}) dx = (y^{3 }– 3x^{2}y)dy

∴ v^{3} – 3v = A(1 + v)(1 + v^{2}) + B(1 – v) (1 + v^{2}) + (Cv + D) (1 – v^{2}).

Putting v= 1,

1-3 = A(2) (2)

⇒ A = -1/2

Putiing v = -1, -1 + 3 = B(2)(1 + 1)

⇒ B = 1/2

Putting v = 0

⇒ 0 = A(1)(1) + B(1)(1) + D(1)

⇒ 0 = \(-\frac{1}{2}+\frac{1}{2}\) + D

D = 0

Comparing coeff. of v^{3}, 1 = A – B – C

⇒ C = A – B – 1 = \(-\frac{1}{2}-\frac{1}{2}\) – 1 = -2

∴ From (2),

Squaring , x^{2} – y^{2} = c’^{2}(x^{2}+ y^{2})^{2}

⇒ x^{2} – y^{2} = c’^{2} (x^{2} + y^{2} )^{2}

where c’^{2} = c,

which is the required solution.

Question 21.

Show that the differential equation:

2ye^{x/y} dx + (y – 2x e^{x/y}) dy = 0 is homogeneous and find the particular solution, given that x = 0 when y = 1. (C.B.S.E. 2013)

Solution:

(i) The given equation is :

2ye^{x/y} dx + (y – 2x e^{x/y}) dy = 0

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

⇒ 2e^{v} = – log |y| + c

⇒ e^{x/y} = -log|y| + c …………….(2)

Now x = 0 when y = 1,

∴ 2(1) = -log |1| + c

⇒ 2 = – log 1 + c

⇒ 2 = —0 + c

⇒ c = 2.

Putting in (2),

2e^{x/y} = – log |y|+ 2,

which is the required solution.

Question 22.

Solve the differential equation:

(1 + x^{2})\(\frac{d y}{d x}\) + 2xy – 4x^{2} =0, subject to the initial condition y (0) = 0. (C.B.S.E. 2019)

Solution:

The given equation can be written as :

Multiplying (1) by (1 + x^{2}), we get:

(1 + x^{2})\(\frac{d y}{d x}\) + 2xy = 4x^{2} dx

⇒ \(\frac{d}{d x}\) y.(1 + x^{2}) = 4x^{2}

Integrating, y.(1 + x^{2}) = ∫ 4x^{2} dx = \(\frac{4 x^{3}}{3}\) + C

⇒ y(1 + x^{2}) = \(\frac{4}{3}\) x^{2} + C …(2)

When x = 0, y = 0,

∴ C = 0 + C

⇒ C = 0.

ry 4 T

Putting in (2), y(1 + x^{2}) = \(\frac{4}{3}\) x^{3} , which is the required solution.

Question 23.

Solve the differential equation :

\(\frac{d y}{d x}-\frac{2 x}{1+x^{2}} y\) = x^{2} + 2 (C.B.S.E. 2019)

Solution:

The given equation is :

Hence, y = (1 + x^{2}) (x + tan^{-1} x + C).

Question 24.

Find the general solution of the following differential equation:

(1 + y^{2}) + (x – e^{tan-1y})\(\frac{d y}{d x}\) (C.B.S.E. 2016)

Solution:

The given equation is:

Putting xee^{tan-1y} = \(\frac{1}{2}\)e^{2tan-1y} + c

x = \(\frac{1}{2}\)e^{tan-1y} + c

which is the required solution.

Question 25.

(i) Solve the differential equation:

(tan^{-1} y – x)dy = (1 + y^{2}) dx. (N.C.E.R.T.; C.B.S.E. 2015)

(ii) Find the particular solution when x = 0, y = 0. (A.I.C.B.S.E. 2013)

Solution:

(i) The given equation can be written as :

\(\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}}\) …… (1)

Comparing with \(\frac{d x}{d y}\)+ Px = Q , we have:

which is the required solution.

(ii) When

x – 0, y = 0.

0 = (tan^{-1}0 – 1) + ce-tan^{-1}0,

⇒ 0 – (0 – 1) + ce^{-0}

⇒ 0 = – 1 + c

⇒ c = 1.

Putting in (3),

x = (tan^{-1} y – 1) + e^{-tan-1y},

which is the required particular solution.