Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Differential Equations Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2B Differential Equations Important Questions

Question 1.
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = ex – y + x2 e-y [Mar. 06; May 05]
Solution: Question 2.
x$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ – y = 2x2 sec22x [May 11]
Solution:  Question 3.
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y tan x = sin x. [T.S. Mar. 16]
Solution:
I.F. = $$e^{\int \tan x d x}$$ = elog sec x = sec x
y.sec x = $$\int$$ sin x . sec x dx = $$\int$$ tan x dx
= log sec x + c

Question 4.
cos x . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y sin x = sec2x [Mar. 14]
Solution:
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + tan x . y = sec3x
I.F. = e$$\int$$tan x dx = elog sec x = sec x
y . sec x = $$\int$$ sec4x dx = $$\int$$ (1 + tan2 x) sec2 x
dx = tan x + $$\frac{\tan ^{3} x}{3}$$ + c

Question 5.
(x + y + 1)$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 1.
Solution:
$$\frac{\mathrm{dx}}{\mathrm{dy}}$$ = x + y + 1
$$\frac{\mathrm{dx}}{\mathrm{dy}}$$ = x + y + 1
I.F. = e$$\int$$ -dy = e-y
x . e-y = $$\int$$ e-y (y + 1)dy
= – (y + 1) . e-y + $$\int$$ e-y . dy
= – (y + 1) e-y – e-y
= – (y + 2) e-y + c
x = – (y + 2) + c. e-y Question 6.
Find the order and degree of r
$$\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}$$ – 3 ($$\frac{\mathrm{dy}}{\mathrm{dx}}$$) – ex = 4. [Mar. 14]
Solution:
The equation is a polynomial in $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ and $$\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}$$.
The exponent of $$\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}$$ is 2.
Hence the degree is 2.
$$\frac{\mathrm{d}^{3} \mathrm{~y}}{\mathrm{dx}^{3}}$$ is the highest order derivative occuring in the equation.
Order of the equation is 3.

Question 7.
x$$\frac{1}{2}$$($$\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}$$)$$\frac{1}{3}$$ + x . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y = 0 has order 2 and degree 1. Prove. [T.S. Mar. 15]
Solution:
The given equation can be written as ∴ The order of the equation is 2 and its degree is 1.

Question 8.
Find the order and degree of $$\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}$$ = 6y [Mar. 16; May 11]
Solution:
Given equation is $$\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}$$ = 6y
i.e., $$\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}$$ + ($$\frac{\mathrm{dy}}{\mathrm{dx}}$$)3 = (6y)$$\frac{5}{6}$$
Order = 2, degree = 1 Question 9.
Solve $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{x(2 \log x+1)}{\sin y+y \cos y}$$ [Mar. 08]
Solution:
Given equation can be written as
(sin y + y cos y) dy = x(2 log x + 1) dx
$$\int$$ sin y dy + $$\int$$ y cos y dy = $$\int$$ 2x log x dx + $$\int$$ x dx
$$\int$$ sin y dy + y sin y – $$\int$$ sin y dy = x2 log x – $$\int$$ x2 . $$\frac{1}{x}$$ dx + $$\int$$ x dx + c
y sin y = x2 log x + c

Question 10.
(xy2 + x) dx + (yx2 + y) dy = 0. [A.P. Mar. 15, 07]
Solution:
(xy2 + x) dx + (yx2 + y) dy = 0
x(y2 + 1) dx + y (x2 + 1) dy = 0
Dividing with (1 + x2) (1 + y2)
$$\frac{x d x}{1+x^{2}}$$ + $$\frac{y d x}{1+y^{2}}$$ = 0
Integrating
$$\int \frac{x d x}{1+x^{2}}+\int \frac{y d y}{1+y^{2}}=0$$
$$\frac{1}{2}$$[(log (1 + x2) + log (1 + y2)] = log c
log (1 + x2) (1 + y2) = 2 log c = log c2
Solution is(1 + x2) (1 + y2) = k when k = c2.

Question 11.
sin-1 ($$\frac{\mathrm{dy}}{\mathrm{dx}}$$) = x + y [Mar. 07]
Solution:
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = sin (x + y)
x + y = t
1 + $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{\mathrm{dt}}{\mathrm{dx}}$$
$$\frac{\mathrm{dt}}{\mathrm{dx}}$$ – 1 = sin t
$$\frac{\mathrm{dt}}{\mathrm{dx}}$$ = 1 + sin t
$$\frac{d t}{1+\sin t}$$ = dx
Integrating both sides we get
$$\int \frac{d t}{1+\sin t}=\int d x$$
$$\int \frac{1-\sin t}{\cos ^{2} t} d t=x+c$$
$$\int$$ sec2 t dt = $$\int$$ tan t . sec t dt = x + c
tan t – sec t = x + c
⇒ tan (x + y) – sec (x + y) = x + c Question 12.
(x2 – y2) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = xy [May 11]
Solution: = log y + c
$$\frac{-x^{2}}{2 y^{2}}$$ = (log y + c)
-x2 = 2y2 (c + log y)
⇒ Solution is x2 + 2y2 (c + log y) = 0.

Question 13.
Solve : x dy = (y + x cos2 $$\frac{y}{x}$$) dx.
Solution: Question 14.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0 [T.S. Mar. 15]
Solution: 2v + log (v – 1) = 3x + c
2v – 3x + log (v – 1) = c
2(2x + y) – 3x + log (2x + y – 1) = c
4x + 2y – 3x + log (2x + y – 1) = c
Solution is x + 2y + log (2x + y – 1) = c Question 15.
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y tan x = cos3x [May 11]
Solution:
I.F. = e$$\int$$ tan x dx = elog sec x = sec x
y . sec x = $$\int$$ sec x. cos3 x dx
= $$\int$$ cos2x dx
= $$\frac{1}{2}$$ $$\int$$ (1 + cos 2x) dx
= $$\frac{1}{2}$$ (x + $$\frac{sin2x}{2}$$) + c
$$\frac{2 y}{\cos x}$$ = x + sin x . cos x + c
Solution is 2y = x cos x + sin x . cos2 x + c . cos x^

Question 16.
(1 + x2) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y = etan-1 x [May 07] [A.P. Mar. 16] [T.S. Mar. 15]
Solution: Question 17.
Solve (1 + y2)dx = (tan-1y – x)dy. [A.P. Mar. 15]
Solution:
Given $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{\tan ^{-1} y-x}{1+y^{2}}$$  Question 18.
(x2 – y2)dx – xy dy = 0 [May 06]
Solution:
(x2 – y2)dx – xy dy = 0
(x2 – y2)dx = xy . dy –$$\frac{1}{4}$$ [log (x2 – 2y2) – log x2] = log x + log c
–$$\frac{1}{4}$$ log (x2 – 2y2) + $$\frac{1}{4}$$ . 2 log x = log x + log c
–$$\frac{1}{4}$$ log (x2 – 2y2) = $$\frac{1}{2}$$ log x + log c
– log (x2 – 2y2) = – 2 log x – 4 log c
log (x2 – 2y2) = – 2 log x + log k where
k = $$\frac{1}{c^{4}}$$ = log $$\frac{\mathrm{k}}{\mathrm{x}^{2}}$$
x2 – 2y2 = $$\frac{\mathrm{k}}{\mathrm{x}^{2}}$$
Solution is x2 (x2 – 2y2) = k Question 19.
$$\frac{d y}{d x}=\frac{3 y-7 x+7}{3 x-7 y-3}$$ [T.S. Mar. 16]
Solution:  = 3ln (v – 1) – 3ln (v + 1) – 7ln (v + 1) – 7ln (v – 1)
14ln x – ln c = – 10 ln (v + 1) – 4 ln (v – 1)
ln (v + 1)5 + ln (v – 1)2 + ln x7 = ln c
(v + 1)5 . (v – 1)2 . x7 = c
($$\frac{y}{x}$$ + 1)5 ($$\frac{y}{x}$$ – 1)2 . x7 = c
(y – x)2 (y + x)5 = c
[y – (x – 1)]2 (y + x – 1)5 = c
Solution is [y – x + 1]2 (y + x – 1)5 = c. Question 20.
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ (x2y3 + xy) = 1 [Mar. 11]
Solution:
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = xy + x2y3
This is Bernoulli’s equation
x-2 . $$\frac{\mathrm{dx}}{\mathrm{dy}}$$ – $$\frac{1}{x}$$ . y = y3  Question 21.
Form the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [AP Mar. 15]
Solution:
We have y = A cos 3x + B sin 3x
Differentiating w.r.to x
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = -3A sin 3x + 3B cos 3x
Differentiating again w.r.to. x
$$\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}$$ = -9A cos 3x – 9B sin 3x
= – 9(A cos 3x + B sin 3x)
= -9y .
is $$\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}$$ + 9y = 0.
Alternate method:
Eliminating A, B from the equation
y = A cos 3x + B sin 3x
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = – 3A sin 3x + 3B sin cos 3x This is the required differential equation.

Question 22.
Solve (x2 + y2) dx = 2xy dy [A.P. Mar. 16]
Solution:
Given equation can be written as log cx(1 – v2 = log 1
cx (1 – v2) = 1
cx (1 – $$\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}$$) = 1
c(x2 – y2) = x is the required solution. Question 23.
Give the solution of x sin2 $$\frac{y}{x}$$ dx = y dx – x dy which passes through the point (1, $$\frac{\pi}{4}$$). [Mar. 14]
Solution:
The given equation can be written as The given curve passes through (1, $$\frac{\pi}{4}$$)
cot $$\frac{\pi}{4}$$ = log 1 + c
1 = 0 + c ⇒ c = 1
Solution is cot $$\frac{y}{x}$$ = log x + 1

Question 24.
Find the order and degree of the differential equation $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ = – p2y.
Solution:
The given equation is a polynomial equation in $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$
Hence the degree is 1
$$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ is the highest order derivative occuring in the equation.
Its order is 2.

Question 25.
Find the order and degree of
($$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}$$)2 – 3 ($$\frac{\mathrm{dy}}{\mathrm{dx}}$$)2 – ex = 4 [Mar. 14]
Solution:
The equation is a polynomial in $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ and $$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}$$.
The exponent of $$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}$$ is 2.
Hence the degree is 2.
$$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}$$ is the highest õrder derivative occuring in the equation.
Order of the equation is 3. Question 26.
x$$\frac{1}{2}$$($$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$)$$\frac{1}{3}$$ + x . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y = 0 has order 2 and degree 1. Prove. [T.S. Mar. 15]
Solution:
The given equation can be written as
x$$\frac{1}{2}$$($$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$)$$\frac{1}{3}$$ = -[x . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y]
Cubing both sides
x$$\frac{3}{2}$$($$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$) = -[x . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y]3
∴ The order of the equation is 2 and its degree is 1.

Question 27.
Find the order and degree of $$\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}$$ = 6y [A.P. Mar. 16; May 11]
Solution:
Given equation is $$\left(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}\right)^{\frac{6}{5}}$$ = 6y
i.e., $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ + ($$\frac{\mathrm{dy}}{\mathrm{dx}}$$)3 = (6y)$$\frac{5}{6}$$
Order = 2; degree = 1

Question 28.
Find the order of the differential equation corresponding to y = c(x – c)2, where c is an arbitrary constant.
Solution:
The given differential equation is
y = c(x – c)2
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 2c(x – c)
∴ Order of the differential equation is 1. Question 29.
Find the order of the differential equation corresponding to y = Aex + Be3x + Ce5x; (A, B, C being parameters) is a solution.
Solution:
Required differential equation is obtained by eliminating A, B, C from y,
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$, $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$, $$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}$$
Highest order deviation = $$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}$$
Order of the differential equation = 3.

Question 30.
Form the differential equation corresponding to y = cx – 2c2, where c is a parameter.
Solution:
Given y = cx – 2c2 ………………. (1)
Differentiating (1) w.r.to
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = c
Substituting in (1), required differential equation is
y = x . ($$\frac{\mathrm{dy}}{\mathrm{dx}}$$) – 2($$\frac{\mathrm{dy}}{\mathrm{dx}}$$)2

Question 31.
Form the differential equation corresponding to y = A cos 3x + B sin 3x, where A and B are parameters. [A.P. Mar. 15]
Solution:
We have y = A cos 3x + B sin 3x
Differentiating w.r.to x
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = -3A sin 3x + 3B cos 3x
Differentiating again w.r.to. x
$$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ = -9A cos 3x – 9B sin 3x
= -9(A cos 3x + B sin 3x)
= -9y
is $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ + 9y = 0 Alternate Method:
Eliminating A, B from the equation
y = A cos 3x + B sin 3x
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = -3A sin 3x = 3B sin cos 3x
$$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ = -9A cos 3x – 9B sin 3x
We get $$\left|\begin{array}{ccc} y & -\cos 3 x & -\sin 3 x \\ \left(\frac{d y}{d x}\right) & 3 \sin 3 x & -3 \cos 3 x \\ \left(\frac{d^{2} y}{d x^{2}}\right) & 9 \cos 3 x & 9 \sin 3 x \end{array}\right|$$ = 0
y(27 sin2 3x + 27 cos2 3x) – (-9 sin 3x. cos 3x + 9 cos 3x. sin 3x) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + (3 cos2 3x + 3 sin2 3x) $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ = 0
= 27y + 3 .$$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ = 0 or $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ + 9y = 0
This is the required differential equation.

Question 32.
Form the differential equation corresponding to the family of circles of radius r given by (x – a)2 + (y – b)2 = r2, where a and b are parameters.
Solution:
We have (x – a)2 + (y – b)2 = r2 ………………… (1)
Differentiating (1) w.r.to x
2(x – a) + 2(y – b) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 0 ……………….. (2)
Differentiating (2) w.r.to. x
1 + (y – b) $$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ + ($$\frac{\mathrm{dy}}{\mathrm{dx}}$$)2 = 0 ……………… (3)
From (2) (x – a) = -(y – b) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$
Substituting in (1), we get i.e., $$r^{2}\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{3}$$
Which is the required differential equation.

Question 33.
Form the differential equation corresponding to the family of circles passing through the origin and having centres on Y-axis.
Solution:
The equation of the family of all circles passing through the origin and having centres on Y—axis is
x2 + y2 + 2hy = 0 …………………. (1)
Where h is a parameter
Differentiating (1) w.r. to x
2x + 2y . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + 2h . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 0
or x + y . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + h . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 0
-(x + y . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$) = h. $$\frac{\mathrm{dy}}{\mathrm{dx}}$$
h = $$\frac{-\left(x+y \cdot \frac{d y}{d x}\right)}{\frac{d y}{d x}}$$
Substituting in (1)
We get x2 + y2 – 2y $$\frac{\left(x+y \cdot \frac{d y}{d x}\right)}{\frac{d y}{d x}}$$ = 0
x2 . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y2 . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ – 2xy – 2y2 . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 0
or (x2 – y2) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ – 2xy = 0
This is the required differential equation. Question 34.
Express the following differential equations in the form f(x) dx + g(y) dy = 0.
i) $$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+\mathrm{y}^{2}}{1+\mathrm{x}^{2}}$$
Solution:
⇒ $$\frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}}$$
$$\frac{d x}{1+x^{2}}-\frac{d y}{1+y^{2}}$$ = 0

ii) y – x $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = a (y2 + $$\frac{\mathrm{dy}}{\mathrm{dx}}$$)
Solution:
y – x . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = ay2 + a. $$\frac{\mathrm{dy}}{\mathrm{dx}}$$
y – ay2 = (x + a) . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$
$$\frac{d x}{x+a}=\frac{d y}{y-a y^{2}}$$

iii) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = ex-y + x2 e-y
Solution:
Multiplying in the ey
ey . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = ex + x2
ey . dy = (ex + x2) dx
(ex + x2) dx – ey . dy = 0

iv) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + x2 = x2 e3y
Solution:
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = x2 . e3y – x2 = x2(e3y – 1)
$$\frac{d y}{e^{3 y}-1}$$ = x2 dx ⇒ x2 dx – $$\frac{d y}{e^{3 y}-1}$$ = 0
or x2dx + $$\frac{1}{\left(1-e^{3 y}\right)}$$ . dy = 0

Question 35.
Find the general solution of
x + y$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 0.
Solution:
Given equation is x + y . $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 0
x . dx + y . dy = 0
Integrating $$\frac{x^{2}}{2}$$ + $$\frac{y^{2}}{2}$$ = c
or x2 + y2 = 2 c = c’

Question 36.
Find the general solution of $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = ex+y.
Solution:
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = ex+y = ex . ey
$$\frac{d y}{e^{y}}$$ = ex dx
$$\int$$ e-y dy = $$\int$$ ex dx
e – e-y = ex
or ex + e-y = c is the required solution. Question 37.
Solve y2 – x $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = a(y + $$\frac{\mathrm{dy}}{\mathrm{dx}}$$)
Solution:  Question 38.
Solve $$\frac{d y}{d x}=\frac{y^{2}+2 y}{x-1}$$
Solution: Question 39.
Solve $$\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}$$ [Mar. 08]
Solution:
Given equation can be written as
(sin y + y cos y) dy = x(2 log x + 1) dx
$$\int$$ sin y dy + $$\int$$ y cos y dy = $$\int$$ 2x log x dx + $$\int$$ x dx
$$\int$$ sin y dy + y sin y – $$\int$$ sin y dy = x2 log x – $$\int$$ x2 . $$\frac{1}{x}$$ dx + $$\int$$ x dx + c
y sin y = x2 log x + c Question 40.
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: Question 41.
Solve y(1 + x) dx + x(1 + y) dy = 0
Solution:
The given equation can be written as
$$\frac{(1+x)}{x}$$ dx + $$\frac{(1+y)}{y}$$ . dy = 0
$$\int$$ (1 + $$\frac{1}{x}$$)dx + $$\int$$ (1 + $$\frac{1}{y}$$) dy = 0
x + log x + y + log y = c
x + y + log (xy) = c is the required solution.

Question 42.
Solve $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = sin (x + y) + cos (x + y)
Solution:
Put x + y = t x = log (1 + tan $$\frac{t}{2}$$) + c
But t = x + y
Solution is x = log (1 + tan $$\frac{x+y}{2}$$) + c

Question 43.
Solve (x – y)2 $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = a2
Solution:
Put x – y = t   Question 44.
Solve $$\sqrt{1+x^{2}} \sqrt{1+y^{2}}$$ dx + xy dy = 0
Solution:
Given equation can be written as   Question 45.
Solve $$\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}$$
Solution:
Put x – 2y = t Question 46.
Solve $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\sqrt{y-x}$$
Solution:
Put y – x = t2 Question 47.
Solve $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + 1 = ex+y
Solution:
Put t = x + y
$$\frac{\mathrm{dt}}{\mathrm{dx}}$$ = 1 + $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = et
$$\int \frac{d t}{e^{t}}=\int d x$$
$$\int$$ e-t dt = $$\int$$ dx
-e-t = x + c
x + e-t + c = 0
Solution is x + e-(x+y) + c = 0 Question 48.
Solve $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = (3x + y + 4)2
Solution:
Put t = 3x + y + 4
$$\frac{\mathrm{dt}}{\mathrm{dx}}$$ = 3 + $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = 3 + t2
$$\frac{\mathrm{dt}}{\mathrm{t}^{3}+3}$$ = dx
$$\int \frac{d t}{t^{2}+3}=\int d x$$
$$\frac{1}{\sqrt{3}}$$ tan-1 ($$\frac{t}{\sqrt{3}}$$) = x + c
Solution is $$\frac{1}{\sqrt{3}}$$ tan-1 ($$\frac{3 x+y+4}{\sqrt{3}}$$) = x + c

Question 49.
Solve $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ – x tan (y – x) = 1
Solution:
Put y – x = t
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ – 1 = $$\frac{\mathrm{dt}}{\mathrm{dx}}$$
$$\frac{\mathrm{dt}}{\mathrm{dx}}$$ = x tan t + 1 – 1 = x tan t
$$\frac{\mathrm{dt}}{\tan \mathrm{t}}$$ = x dx
$$\int$$ cot dt = $$\int$$ x dx
log |sin | = $$\frac{x^{2}}{2}$$ + c
Solution is log |sin (y – x)| = $$\frac{x^{2}}{2}$$ + c

Question 50.
Show that f(x, y) = 1 + ex/y is a homogeneous function of x and y.
Solution:
f(kx, xy) = 1 + ekx/ky = 1 + ex/y = f(x, y)
f(x, y) is a homogeneous function degree 0.

Question 51.
Show that f(x, y) = x$$\sqrt{x^{2}+y^{2}}$$ – y2 is a homogeneous function of x and y.
Solution:
f(kx, ky) = kx$$\sqrt{k^{2} x^{2}+k^{2} y^{2}}$$ – k2y2.
= k2 (x$$\sqrt{x^{2}+y^{2}}$$ – y2) = k2 f(x, y)
f(x, y) is a homogeneous function of degree 2. Question 52.
Show that f(x, y) = x – y log y + y log x is a homogeneous function of x and y.
Solution:
f(kx, ky) = kx – ky. log ky + ky log (kx)
= k(x – y log (ky) + y log kx)
= k(x – y log k – y log y + y log k + y log x)
= k(x – y log y + y log x)
= k. f(x, y)
f(x, y) is a homogeneous function of degree 1.

Question 53.
Express (1 + ex/y)dx + ex/y (1 – $$\frac{x}{y}$$) dy = 0 in the form $$\frac{\mathrm{dx}}{\mathrm{dy}}$$ = F ($$\frac{x}{y}$$)
Solution: Question 54.
Express (x$$\sqrt{x^{2}+y^{2}}$$ – y2) dx + xy dy = 0 in the form $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = F ($$\frac{x}{y}$$)
Solution:
Given equation is Question 55.
Express $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{y}{x+y e^{\frac{-2 x}{y}}}$$ in the form $$\frac{\mathrm{dx}}{\mathrm{dy}}$$ = F ($$\frac{x}{y}$$)
Solution:  Question 56.
Solve $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{y^{2}-2 x y}{x^{2}-x y}$$
Solution:
The given equation is a homogeneous equation.
Put y = vx $$\log v \sqrt{2 v-3}=-3 \log \frac{x}{c}=\log \frac{c^{3}}{x^{3}}$$ Question 57.
Solve (x2 + y2) dx = 2xy dy [A.P. Mar. 16]
Solution:
Given equation can be written as
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{x^{2}+y^{2}}{2 x y}$$
This is a homogeneous function
Put y = vx -log (1 – v2) = log x + log c
= log cx
log cx + log (1 – v2) = 0
log cx(1 – v2) = log 1
cx (1 – v2) = 1
cx (1 – $$\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}$$) = 1
c(x2 – y2) = x is the required solution. Question 58.
Solve xy2 dy – (x3 + y3) dx = 0.
Solution:
Given equation is xy2 dy = (x3 + y2) dx
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{x^{3}+y^{3}}{x y^{2}}$$
This is a homogeneous equation.
Put y = vx Question 59.
Solve $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{x^{2}+y^{2}}{2 x^{2}}$$
Solution:
This is a homogeneous equation. Question 60.
Solve x sec ($$\frac{y}{x}$$) . (y dx + x dy) = y cosec ($$\frac{y}{x}$$) . (x dy – y dx)
Solution:
This given equation can be written as  log($$\frac{\sin v}{v}$$) = log cx2
$$\frac{\sin v}{v}$$ = cx2
$$\frac{x}{y}$$ sin ($$\frac{y}{x}$$) = cx2
Solution is sin($$\frac{y}{x}$$) = cxy. Question 61.
Give the solution of x sin2 $$\frac{y}{x}$$ dx = y dx – x dy which passes through the point (1, $$\frac{\pi}{4}$$). [Mar. 14]
Solution:
The given equation can be written as The given curve passes through (1, $$\frac{\pi}{4}$$)
cot $$\frac{\pi}{4}$$ = log 1 + c
1 = 0 + c ⇒ c = 1
Solution is cot $$\frac{y}{x}$$ = log x + 1

Question 62.
Solve (x3 – 3xy2) dx + (3x2y – y3) dy = 0
Solution:
(x3 – 3xy2) dx = -(3x2y – y3) dy  Question 63.
Transform the following two differential equations Into linear form.
x log x $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y = 2 log x
Solution:
Given equation can be written as
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + $$\frac{1}{x \log x}$$ . y = $$\frac{2}{x}$$
This is of the form $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + Py = Q Question 64.
(x + 2y3) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = y
Solution:
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ = $$\frac{x+2 y^{3}}{y}$$ = $$\frac{x}{y}$$ + 2y2
$$\frac{\mathrm{dx}}{\mathrm{dy}}$$ – $$\frac{1}{y}$$ . x = 2y2
This is of the form $$\frac{\mathrm{dx}}{\mathrm{dy}}$$ + Px = Q.

Question 65.
(cos x) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y sin x = tan x
Solution:
Given equation can be written as
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y(tan x) = (tan x) (sec x)
P = tan x ⇒ $$\int$$ P dx = $$\int$$ tan x dx = log sec x
I.F. = e$$\int$$ log sec x = sec x

Question 66.
Solve (2x – 10y3) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y = o
Solution: Question 67.
Solve (1 + x2) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + 2xy – 4x2 = 0
Solution:
Given equation can be written as  Question 68.
Solve $$\frac{1}{x}$$ $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y . ex = e(1 – x)ex
Solution:
Given equation can be written as
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + (x.ex) . y = x . e(1 – x)ex
I.F. = e$$\int$$.ex dx = e(x – 1) ex
y . e(x – 1) ex = $$\int$$ x dx
= $$\frac{\mathrm{x}^{2}}{2}$$ + c
2y. e(x – 1) ex = x2 + 2c is the required solution.

Question 69.
Solve sin2 x. $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y = cot x
Solution:
Given equation can be written as
$$\frac{\mathrm{dy}}{\mathrm{dx}}$$ + y cosec2 x = cot x . cosec2 x
I.F. = e$$\int$$ cosec2 x dx = e-cot x
y . e-cot x = $$\int$$ e-cot x . cosec2 x. cot x dx ……………… (1)
Consider $$\int$$ e-cot x . cosec2 x . cot x dx
Put -cot x = t ⇒ cosec2 x dx = dt
(1) becomes y . et = $$\int$$ -t. et dt
= -(t – 1) et + c
y . e-cot x = -(-cot x – 1) e-cot x + 1
= (cot x + 1) e-cot x + c is the required solution. Question 70.
Find the solution of the equation
x(x – 2) $$\frac{\mathrm{dy}}{\mathrm{dx}}$$ – 2(x – 1)y = x3(x – 2)
which satisfies the condition that y = 9 when x = 3.
Solution:
The given equation can be written as  Question 71.
Solve (1 + y2)dx = (tan-1y – x)dy. [A.P. Mar. 15]
Solution: is the solution.