Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type

Model I – Problems on Variables Separable Method

Question 1.

Solve \(\frac{\mathbf{d y}}{\mathbf{d x}}\) = e^{x-y} + x^{2} (e^{-y}). [(AP) May ’18, (TS) ’17]

Solution:

Question 2.

Solve (e^{x} + 1) y dy + (y + 1) dx = 0. [May ’10, mar. ’04]

Solution:

Question 3.

Solve tan y sec^{2}x dx + tan x sec^{2}y dy = 0.

Solution:

Given D.E is tan y sec^{2}x dx + tan x sec^{2}y dy = 0

⇒ \(\frac{\sec ^2 x d x}{\tan x}+\frac{\sec ^2 y d y}{\tan y}=0\)

⇒ \(\int \frac{\sec ^2 x}{\tan x} d x+\int \frac{\sec ^2 y}{\tan y} d y=c\)

⇒ log|tan x| + log|tan y| = c

⇒ log|tan x| + log|tan y| = log c

⇒ log|tan x tan y| = log c

⇒ tan x tan y = c

Question 4.

Solve (xy^{2} + x) dx + (yx^{2} + y) dy = 0. [(TS) Mar. ’20, (AP) ’15; May ’13]

Solution:

Question 5.

Solve \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\). [Mar. ’05]

Solution:

Question 6.

Solve \(\frac{d y}{d x}=\frac{-\left(y^2+y+1\right)}{\left(x^2+x+1\right)}\). [May ’08, Mar. ’03]

Solution:

which is the required general solution.

(where k = √3c = constant)

Question 7.

Solve \(\sin ^{-1}\left(\frac{d y}{d x}\right)\) = x + y. [(TS) Mar. ’20; May ’07]

Solution:

⇒ ∫sec^{2}t dt – ∫tan t sec t dt = x + c

⇒ tan t – sec t = x + c

⇒ tan(x + y) – sec(x + y) = x + c

which is the required general solution.

Question 8.

Solve \(\frac{d y}{d x}\) = tan^{2}(x + y).

Solution:

Question 9.

Solve \(\frac{d y}{d x}\) – x tan(y – x) = 1. [(TS) May ’15]

Solution:

Put y – x = z, then

\(\frac{d y}{d x}-1=\frac{d z}{d x}\)

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

∴ The solution is sin(y – x) = \(c \mathrm{e}^{\mathrm{x}^2 / 2}\), where c is an arbitrary constant.

Question 10.

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

Solution:

⇒ x + y + 1 – log(1 + x + y + 1) = x + c

⇒ y + 1 – log(x + y + 2) = c

⇒ y – log(x + y + 2) = c

Which is the required solution.

Question 11.

Solve \(\sqrt{1+x^2} d x+\sqrt{1+y^2} d y=0\). [Mar. ’09]

Solution:

Question 12.

Solve y(1 + x) dx + x(1 + y) dy = 0. [(AP) Mar. ’20]

Solution:

Given D.E. is y(1 + x) dx + x(1 + y) dy = 0

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

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

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

Now, Integrating on both sides, we get

\(\int\left(\frac{1}{x}+1\right) d x=\int\left(\frac{-1}{y}-1\right) d y\)

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

log x + x = -log y – y + c

log x + x + log y + y = c

x + y + log xy = c

Which is the required solution.

Question 13.

Solve \(\frac{d y}{d x}=\frac{x y+y}{x y+x}\). [(TS) May ’16]

Solution:

Question 14.

Solve \(\frac{d y}{d x}\) = sin(x + y) + cos(x + y)

Solution:

Given D.E. is \(\frac{d y}{d x}\) = sin(x + y) + cos(x + y)

Put x + y = z

differential w.r.t ‘x’

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

⇒ \(\frac{d y}{d x}=\frac{d z}{d x}-1\)

(1) ⇒ \(\frac{\mathrm{dz}}{\mathrm{d} \mathrm{x}}\) – 1 = sin z + cos z

Question 15.

Solve \(\frac{d y}{d x}\) = (3x + y + 4)^{2}

Solution:

Given D.E. is \(\frac{d y}{d x}\) = (3x + y + 4)^{2} …….(1)

Put 3x + y + 4 = z

differential w.r.t ‘x’

\(3+\frac{d y}{d x}=\frac{d z}{d x}\)

⇒ \(\frac{d y}{d x}=\frac{d z}{d x}-3\)

Question 16.

Find the equation of the curve whose slope, at any point (x, y) is \(\frac{y}{x^2}\) and which satisfies the condition y = 1, when x = 3.

Solution:

We know that the slope at any point (x, y) on the curve is m = \(\frac{\mathrm{dy}}{\mathrm{dx}}\)

Given that, slope of at any point (x, y) on the curve is m = \(\frac{y}{x^2}\)

Question 17.

Solve (y^{2} – 2xy) dx + (2xy – x^{2}) dy = 0. [May ’01]

Solution:

Question 18.

Solve \(\frac{d y}{d x}=\frac{y^2-2 x y}{x^2-x y}\). [Mar. ’19 (AP)]

Solution:

Given D.E is \(\frac{d y}{d x}=\frac{y^2-2 x y}{x^2-x y}\) ……..(1)

The given equation is a homogeneous equation since both the numerator and denominator are homogeneous functions of degree 2.

Now, put y = vx

Diff. w.r. to x, we get

This is the required general solution of the given equation.

Question 19.

Find the equation of a curve whose gradient is \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\), where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)). [(AP) May ’19, ’16, (TS) ’16]

Solution:

Given D.E. is \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\) ………(1)

The given equation is a homogeneous equation since both the numerator and denominator are homogeneous functions of degree 1.

Put y = vx

diff. w.r.t ‘x’ on both sides

Question 20.

Give the solution of \(x \sin ^2\left(\frac{y}{x}\right) d x=y d x-x d y\) which passes through the point (1, \(\frac{\pi}{4}\)). [Mar. ’14]

Solution:

Given differential equation is

The given equation is a homogeneous equation since both the numerator and denominator are homogeneous functions of degree 1.

Put y = vx

Diff. w.r. to ‘x’ on both sides

which is the required particular solution.

Model II – Problems on Homogeneous D.E.

Question 21.

Solve (x^{2} – y^{2}) dx – xy dy = 0. [(AP) May ’17; Mar. ’09]

Solution:

Question 22.

Solve \(\frac{d y}{d x}=\frac{(x+y)^2}{2 x^2}\). [Mar. ’05]

Solution:

Question 23.

Solve (x^{2} – y^{2}) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy

Solution:

Given D.E. is (x – y ) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy

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

This is a homogeneous equation since both the nr and dr are homogeneous functions of degree 2.

Let y = vx

⇒ -x^{2} = 2y^{2} log(cy)

⇒ x^{2} + 2y^{2} log(cy) = 0

Which is the required general solution.

Question 24.

Solve (x^{2} + y^{2}) dx = 2xy dy. [(AP) Mar. ’20, Mar. ’17, ’16]

Solution:

Given diff. equation is \(\frac{d y}{d x}=\left(\frac{2 x y}{x^2+y^2}\right)^{-1}\) …….(1)

Clearly, it is a homogeneous diff. equation of degree zero.

which is the required general solution.

Question 25.

Solve y^{2} dx + (x^{2} – xy + y^{2}) dy = 0.

Solution:

Given D.E is y^{2} dx + (x^{2} – xy + y^{2}) dy = 0

(x^{2} – xy + y^{2}) dy = -y^{2} dx

Which is the required general solution.

Question 26.

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

Solution:

Question 27.

Solve x dy = \(\left(y+x \cos ^2 \frac{y}{x}\right) d x\). [Mar. ’13, ’11]

Solution:

Given diff. equation is

x dy = \(\left(y+x \cos ^2 \frac{y}{x}\right) d x\)

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

Clearly, it is a homogeneous differential equation of degree zero.

Let y = vx then v + x \(\frac{d v}{d x}=\frac{d y}{d x}\)

From (1), x \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = v + cos^{2}v – v = cos^{2}v

sec^{2}v dv = \(\frac{\mathrm{dx}}{\mathrm{x}}\)

Integrating on both sides

∫sec^{2}v dv = ∫\(\frac{\mathrm{dx}}{\mathrm{x}}\)

tan v = log x + c

⇒ tan(\(\frac{y}{x}\)) = log x + c

Which is the required general solution.

Question 28.

Solve (2x – y) dy = (2y – x) dx. [Mar. ’12]

Solution:

Given diff. equation is \(\frac{d y}{d x}=\frac{2 y-x}{2 x-y}\) …….(1)

Clearly, it is a homogeneous diff. equation of degree zero.

Which is the required general solution.

Question 29.

Solve xy^{2} dy – (x^{3} + y^{3}) dx = 0. [(TS) May ’18]

Solution:

Given D.E. is xy^{2} dy – (x^{3} + y^{3}) dx = 0

xy^{2} dy = (x^{3} + y^{3}) dx

\(\frac{d y}{d x}=\frac{x^3+y^3}{x y^2}\) ……(1)

The given equation is a homogeneous equation since both the numerator and denominator are homogeneous functions of degree 3.

Put y = vx

diff. w.r.t ‘x’ on both sides

Which is the general solution of the given equations.

Question 30.

Solve (x^{3} – 3xy^{2}) dx + (3x^{2}y – y^{3}) dy = 0. [(AP) May ’18; May ’14]

Solution:

Given D.E is (x^{3} – 3xy^{2}) dx + (3x^{2}y – y^{3}) dy = 0

(3x^{2}y – y^{3}) dy = -(x^{3} – 3xy^{2}) dx

\(\frac{d y}{d x}=\frac{-\left(x^3-3 x y^2\right)}{3 x^2 y-y^3}\)

\(\frac{d y}{d x}=\frac{3 x y^2-x^3}{3 x^2 y-y^3}\) ……..(1)

It is a homogeneous equation since both nr and dr are homogeneous functions of degree 3.

Put y = vx

Then diff. w.r. t. x

Which is the required solution of the given equation.

Question 31.

Solve (x^{2}y – 2xy^{2}) dx = (x^{3} – 3x^{2}y) dy. [Mar. ’18 (AP)]

Solution:

Question 32.

Solve \(\frac{d y}{d x}=\frac{2 x-y+1}{x+2 y-3}\)

Solution:

Given D.E. is \(\frac{d y}{d x}=\frac{2 x-y+1}{x+2 y-3}\) …….(1)

Comparing (1) with \(\frac{d y}{d x}=\frac{a x+b y+c}{a^1 x+b^1 y+c^1}\)

We get a = 2, b = -1, c = 1

a’ = 1, b’ = 2, c’ = -3

Now b = -1 = -(1) = -a’

∴ b = -a’

Now (x + 2y – 3) dy = (2x – y + 1) dx

⇒ x dy + 2y dy – 3dy = 2x dy – y dx + dx

⇒ x dy + 2y dy – 3 dy – 2x dx + y dx – dx = 0

⇒ (x dy + y dx) + 2y dy – 3 dy + 2x dx – dx = 0

⇒ d(xy) + 2y dy – 3 dy + 2x dx – dx = 0

By integrating, we get

∫d(xy) + 2∫y dy – 3 ∫1 dy + 2 ∫x dx – ∫dx = c

xy + 2 . \(\frac{y^2}{2}\) – 3y + 2 . \(\frac{x^2}{2}\) – x = c

xy + y^{2} – 3y + x^{2} – x = c

Which is the required solution.

Model III – Problems on non-homogeneous D.E.

Question 33.

Solve \(\frac{d y}{d x}=\frac{-3 x-2 y+5}{2 x+3 y+5}\)

Solution:

Given diff. equation is \(\frac{d y}{d x}=\frac{-3 x-2 y+5}{2 x+3 y+5}\) …….(1)

Comparing it with \(\frac{d y}{d x}=\frac{a x+b y+c}{a^{\prime} x+b^{\prime} y+c^{\prime}}\)

We get a = -3, b = -2, c = 5, a’ = 2, b’ = 3, c’ = 5

Then, we can solve equation (1) by the case (i)

∴ b = -a’

\(\frac{d y}{d x}=\frac{-3 x-2 y+5}{2 x+3 y+5}\)

⇒ dy (2x + 3y + 5) = dx (-3x – 2y + 5)

⇒ 2x dy + 3y dy + 5 dy + 3x dx + 2y dx – 5 dx = 0

⇒ 2(x dy + y dy) + 3y dy + 5 dy + 3x dx – 5 dx = 0

⇒ 2d(xy) + 3y dy + 5 dy + 3x dx – 5 dx = 0

Integrating on both sides, we get

2∫d(xy) + 3∫y dy + 5∫dy + 3∫x dx – 5∫dx = c

⇒ 2xy + 3 \(\frac{y^2}{2}\) + 5y + 3 \(\frac{x^2}{2}\) – 5x = c

⇒ 4xy + 3y^{2} + 10y + 3x^{2} – 10x = 2c = k

(k = constant)

Which is the required general solution.

Question 34.

Solve 2(x – 3y + 1) \(\frac{d y}{d x}\) = 4x – 2y + 1

Solution:

Given D.E is 2(x – 3y + 1) \(\frac{d y}{d x}\) = 4x – 2y + 1

\(\frac{d y}{d x}=\frac{4 x-2 y+1}{2 x-6 y+2}\) …….(1)

Comparing (1) with \(\frac{d y}{d x}=\frac{a x+b y+c}{a^{\prime} x+b^{\prime} y+c^{\prime}}\)

We get a = 4, b = -2, c = 1

a’ = 2, b’ = -6, c’ = 2

Now, b = -2 = -(2) = -a’

∴ b = -a’

Hence we can solve by the case (1).

(2x – 6y + 2) dy = (4x – 2y + 1) dx

⇒ 2x dy – 6y dy + 2 dy = 4x dx – 2y dx + 1 dx

⇒ 2x dy – 6y dy + 2 dy – 4x dx + 2y dx – 1 dx = 0

⇒ 2(x dy + y dx) – 6y dy – 4x dx + 2dy – 1 dx = 0

⇒ 2d(xy) – 6y dy – 4x dx + 2 dy – dx = 0

By integrating we get,

2∫d(xy) – 6∫y dy – 4∫x dx + 2∫dy – ∫dx = c

⇒ 2(xy) – 6(\(\frac{y^2}{2}\)) – 4(\(\frac{x^2}{2}\)) + 2y – x = 0

⇒ 2xy – 3y^{2} – 2x^{2} + 2y – x = c

Which is the required solution.

Question 35.

Solve x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2y = log x

Solution:

Given D.E. is x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2y = log x

dividing on both sides by ‘x’

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

Which is in the form \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + Py = Q

It is linear in y.

Model IV(a) – Problems on L.D.E in y

Question 36.

Solve x log x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = 2 log x. [(AP) Mar. ’20; May ’14]

Solution:

Question 37.

Solve (1 + x^{2}) \(\frac{d y}{d x}\) + y = \(e^{\tan ^{-1} x}\). [(AP) Mar. ’18, ’16; May ’16, (TS) Mar. ’15 (TS); May ’13]

Solution:

Question 38.

Solve \(\frac{d y}{d x}\) = y tan x + e^{x} sec x. [(TS) May ’19; Mar. ’09]

Solution:

Given diff. equation is

\(\frac{d y}{d x}\) – y tan x = e^{x} sec x ……..(1)

It is a linear diff. equation in y.

Comparing it with \(\frac{d y}{d x}\) + Py = Q

We get P = -tan x, Q = e^{x} sec x

I.F. = e^{2∫Pdx}

= e^{-∫tan x dx}

= e^{log|cos x|}

= cos x

The solution of (1) is,

y (I.F) = ∫Q(I.F) dx + c

y (cos x) = ∫e^{x} sec x (cos x) dx + c = ∫e^{x} + c

y (cos x) = e^{x} + c

Which is the required solution.

Question 39.

Solve \(\frac{d y}{d x}\) + y sec x = tan x. [May ’10]

Solution:

Given diff. equation is

\(\frac{d y}{d x}\) + y sec x = tan x

It is a linear diff. equation in y.

It compared with \(\frac{d y}{d x}\) + Py = Q

We get P = sec x, Q = tan x

I.F. = e^{∫Pdx}

= e^{∫sec x dx}

= e^{log|sec x + tan x|}

= sec x + tan x

The solution of (1) is,

y (I.F) = ∫Q(I.F) dx + c

y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c

= ∫sec x tan x dx + ∫tan^{2}x dx + c

= ∫sec x tan x dx + ∫(sec^{2}x – 1) dx + c

= sec x + tan x – x + c

Which is the required solution.

Question 40.

Solve \(\frac{d y}{d x}\) + y tan x = sin x. [(TS) Mar. ’16, ’12]

Solution:

Given diff. equation is

\(\frac{d y}{d x}\) + y tan x = sin x ……..(1)

It is a linear diff. equation in y.

Comparing it with \(\frac{d y}{d x}\) + Py = Q

Where P = tan x, Q = sin x

I.F = e^{∫Pdx}

= e^{∫tan x dx}

= sec x

The solution of (1) is,

y(I.F) = ∫Q(I.F) dx + c

y sec x = ∫sin x . sec x dx + c

y sec x = ∫tan x dx + c

y sec x = log|sec x| + c

Which is the required solution.

Question 41.

Solve \(\frac{d y}{d x}\) + y tan x = cos^{3}x. [(AP) May ’19, ’18; Mar. ’17]

Solution:

Given diff. equation is \(\frac{d y}{d x}\) + + y tan x = cos^{3}x

It is a linear diff. equation in y of the first order.

Comparing it with \(\frac{d y}{d x}\) + Py = Q

we get P = tan x, Q = cos^{3}x

Which is the required solution.

Question 42.

Solve cos x . \(\frac{d y}{d x}\) + y sin x = sec^{2}x. [Mar. ’19 (TS); Mar. ’14]

Solution:

Given differential equation is

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

Dividing on both sides by cos x, we get

\(\frac{d y}{d x}\) + y tan x = sec^{3}x

Which is in the form of \(\frac{d y}{d x}\) + Py = Q

It is linear in ‘y’

Here, P = tan x, Q = sec^{3}x

IF = e^{∫Pdx}

= e^{∫tan x dx}

= e^{log|sec x|}

= sec x

∴ Solution is y(I. F) = ∫Q(I. F) dx + c

y sec x = ∫sec^{2}x . sec^{2}x dx + c

= ∫sec^{4}x dx + c

= ∫sec^{2}x . sec^{2}x dx + c

= ∫(1 + tan^{2}x) sec^{2}x dx + c

= ∫(1 + t^{2}) dt + c

Put tan x = t ⇒ sec^{2}x dx = dt

= ∫1 dt + ∫t^{2} dt + c

= t + \(\frac{\mathrm{t}^3}{3}\) + c

y sec x = tan x + \(\frac{1}{3}\) tan^{3}x + c

Question 43.

Solve (1 + x^{2}) \(\frac{d y}{d x}\) + y = tan^{-1}x. [(AP) May ’16]

Solution:

Question 44.

Solve (1 + x^{2}) \(\frac{d y}{d x}\) + 2xy – 4x^{2} = 0. [Mar. ’06]

Solution:

Given diff. equation is

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

It is a linear diff. equation in y.

Comparing it with \(\frac{d y}{d x}\) + Py = Q

Which is the required solution.

Question 45.

Solve \(\frac{d y}{d x}+y\left(\frac{4 x}{1+x^2}\right)=\frac{1}{\left(1+x^2\right)^2}\). [(AP) May ’17]

Solution:

Question 46.

Solve \(\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{1+x^2}{1+x^3}\)

Solution:

Question 47.

Solve \(\frac{1}{x} \frac{d y}{d x}+y e^x=e^{(1-x) e^x}\). [(AP) May ’18]

Solution:

Question 48.

Solve x(x – 1) \(\frac{d y}{d x}\) – y = x^{3}(x – 1)^{3}. [(AP) Mar. ’19]

Solution:

Model IV(b) – Problems on L.D.E in x

Question 49.

Solve (1 + y^{2}) dx = (tan^{-1}y – x) dy. [Mar. ’18 (TS); May; Mar. ’15 (AP); May ’15 (TS)]

Solution:

Question 50.

Solve (x + 2y^{3}) \(\frac{d y}{d x}\) = y. [May ’12]

Solution:

Question 51.

Solve (x + y + 1) \(\frac{d y}{d x}\) = 1. [Mar. ’17 (TS); Mar. ’13]

Solution:

Given diff. equation is \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}+\mathrm{y}+1}\)

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

It is a linear diff. equation in x.

Comparing it with \(\frac{d x}{d y}\) + Px = Q

Where P = -1, Q = y + 1

I.F. = e^{∫Pdy} = e^{∫(-1) dy} = e^{-y}

The solution of (1) is

x(I.F) = ∫Q(I.F) dy + c

Which is the required general solution.

Model V – Problems on Bernoulli’s D.E.

Question 52.

Solve \(\frac{d y}{d x}\) (x^{2}y^{3} + xy) = 1. [Mar. ’11]

Solution: