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

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 \(\frac{\mathbf{d y}}{\mathbf{d x}}\) = ex-y + x2 (e-y). [(AP) May ’18, (TS) ’17]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q1

Question 2.
Solve (ex + 1) y dy + (y + 1) dx = 0. [May ’10, mar. ’04]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q2

Question 3.
Solve tan y sec2x dx + tan x sec2y dy = 0.
Solution:
Given D.E is tan y sec2x dx + tan x sec2y 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

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

Question 4.
Solve (xy2 + x) dx + (yx2 + y) dy = 0. [(TS) Mar. ’20, (AP) ’15; May ’13]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q4

Question 5.
Solve \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\). [Mar. ’05]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q5

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:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q6
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q6.1
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:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q7
⇒ ∫sec2t 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 \(\frac{d y}{d x}\) = tan2(x + y).
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q8
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q8.1

Question 9.
Solve \(\frac{d y}{d x}\) – x tan(y – x) = 1. [(TS) May ’15]
Solution:
Put y – x = z, then
\(\frac{d y}{d x}-1=\frac{d z}{d x}\)
⇒ \(\frac{d y}{d x}=1+\frac{d z}{d x}\)
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q9
∴ 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:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q10
⇒ 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.

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

Question 11.
Solve \(\sqrt{1+x^2} d x+\sqrt{1+y^2} d y=0\). [Mar. ’09]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q1

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:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q3

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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q4
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q4.1

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’
\(3+\frac{d y}{d x}=\frac{d z}{d x}\)
⇒ \(\frac{d y}{d x}=\frac{d z}{d x}-3\)
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q5

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

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 = \(\frac{y}{x^2}\)
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q6

Question 17.
Solve (y2 – 2xy) dx + (2xy – x2) dy = 0. [May ’01]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q7
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q7.1

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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q8
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q8.1
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q8.2
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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q9
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q9.1

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

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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q10
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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q10.1
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q10.2
which is the required particular solution.

Model II – Problems on Homogeneous D.E.

Question 21.
Solve (x2 – y2) dx – xy dy = 0. [(AP) May ’17; Mar. ’09]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q11
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q11.1
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q11.2

Question 22.
Solve \(\frac{d y}{d x}=\frac{(x+y)^2}{2 x^2}\). [Mar. ’05]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q12
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q12.1

Question 23.
Solve (x2 – y2) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q13
⇒ -x2 = 2y2 log(cy)
⇒ x2 + 2y2 log(cy) = 0
Which is the required general solution.

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

Question 24.
Solve (x2 + y2) 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.
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q14
which is the required general solution.

Question 25.
Solve y2 dx + (x2 – xy + y2) dy = 0.
Solution:
Given D.E is y2 dx + (x2 – xy + y2) dy = 0
(x2 – xy + y2) dy = -y2 dx
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q15
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q15.1
Which is the required general solution.

Question 26.
Solve \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q16
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q16.1

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 + cos2v – v = cos2v
sec2v dv = \(\frac{\mathrm{dx}}{\mathrm{x}}\)
Integrating on both sides
∫sec2v dv = ∫\(\frac{\mathrm{dx}}{\mathrm{x}}\)
tan v = log x + c
⇒ tan(\(\frac{y}{x}\)) = log x + c
Which is the required general solution.

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

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.
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q18
Which is the required general solution.

Question 29.
Solve xy2 dy – (x3 + y3) dx = 0. [(TS) May ’18]
Solution:
Given D.E. is xy2 dy – (x3 + y3) dx = 0
xy2 dy = (x3 + y3) 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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q19
Which is the general solution of the given equations.

Question 30.
Solve (x3 – 3xy2) dx + (3x2y – y3) dy = 0. [(AP) May ’18; May ’14]
Solution:
Given D.E is (x3 – 3xy2) dx + (3x2y – y3) dy = 0
(3x2y – y3) dy = -(x3 – 3xy2) 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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q20
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q20.1
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q20.2
Which is the required solution of the given equation.

Question 31.
Solve (x2y – 2xy2) dx = (x3 – 3x2y) dy. [Mar. ’18 (AP)]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q21
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q21.1

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 + y2 – 3y + x2 – x = c
Which is the required solution.

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

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 + 3y2 + 10y + 3x2 – 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 – 3y2 – 2x2 + 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.
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q25

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:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q26

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

Question 37.
Solve (1 + x2) \(\frac{d y}{d x}\) + y = \(e^{\tan ^{-1} x}\). [(AP) Mar. ’18, ’16; May ’16, (TS) Mar. ’15 (TS); May ’13]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q27

Question 38.
Solve \(\frac{d y}{d x}\) = y tan x + ex sec x. [(TS) May ’19; Mar. ’09]
Solution:
Given diff. equation is
\(\frac{d y}{d x}\) – y tan x = ex 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 = ex sec x
I.F. = e2∫Pdx
= e-∫tan x dx
= elog|cos x|
= cos x
The solution of (1) is,
y (I.F) = ∫Q(I.F) dx + c
y (cos x) = ∫ex sec x (cos x) dx + c = ∫ex + c
y (cos x) = ex + 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
= elog|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 + ∫tan2x dx + c
= ∫sec x tan x dx + ∫(sec2x – 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.

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

Question 41.
Solve \(\frac{d y}{d x}\) + y tan x = cos3x. [(AP) May ’19, ’18; Mar. ’17]
Solution:
Given diff. equation is \(\frac{d y}{d x}\) + + y tan x = cos3x
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 = cos3x
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q31
Which is the required solution.

Question 42.
Solve cos x . \(\frac{d y}{d x}\) + y sin x = sec2x. [Mar. ’19 (TS); Mar. ’14]
Solution:
Given differential equation is
cos x . \(\frac{d y}{d x}\) + y sin x = sec2x
Dividing on both sides by cos x, we get
\(\frac{d y}{d x}\) + y tan x = sec3x
Which is in the form of \(\frac{d y}{d x}\) + Py = Q
It is linear in ‘y’
Here, P = tan x, Q = sec3x
IF = e∫Pdx
= e∫tan x dx
= elog|sec x|
= sec x
∴ Solution is y(I. F) = ∫Q(I. F) dx + c
y sec x = ∫sec2x . sec2x dx + c
= ∫sec4x dx + c
= ∫sec2x . sec2x dx + c
= ∫(1 + tan2x) sec2x dx + c
= ∫(1 + t2) dt + c
Put tan x = t ⇒ sec2x dx = dt
= ∫1 dt + ∫t2 dt + c
= t + \(\frac{\mathrm{t}^3}{3}\) + c
y sec x = tan x + \(\frac{1}{3}\) tan3x + c

Question 43.
Solve (1 + x2) \(\frac{d y}{d x}\) + y = tan-1x. [(AP) May ’16]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q33
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q33.1

Question 44.
Solve (1 + x2) \(\frac{d y}{d x}\) + 2xy – 4x2 = 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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q34
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:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q11

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

Question 46.
Solve \(\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{1+x^2}{1+x^3}\)
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q12

Question 47.
Solve \(\frac{1}{x} \frac{d y}{d x}+y e^x=e^{(1-x) e^x}\). [(AP) May ’18]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q13
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q13.1

Question 48.
Solve x(x – 1) \(\frac{d y}{d x}\) – y = x3(x – 1)3. [(AP) Mar. ’19]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q14

Model IV(b) – Problems on L.D.E in x

Question 49.
Solve (1 + y2) dx = (tan-1y – x) dy. [Mar. ’18 (TS); May; Mar. ’15 (AP); May ’15 (TS)]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q35
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q35.1

Question 50.
Solve (x + 2y3) \(\frac{d y}{d x}\) = y. [May ’12]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q36

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

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
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L2 Q15
Which is the required general solution.

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

Question 52.
Solve \(\frac{d y}{d x}\) (x2y3 + xy) = 1. [Mar. ’11]
Solution:
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q37
TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type L1 Q37.1

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