Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c)

I.

Question 1.
Express x dy – ydx = \(\sqrt{x^2+y^2}\) dx in the form F (\(\frac{y}{x}\)) = (\(\frac{dy}{dx}\)).
Solution:
x. dy – y dx = \(\sqrt{x^2+y^2}\) dx
\(\frac{dy}{dx}\) – y = \(\sqrt{x^2+y^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 1

Question 2.
Express (x – y Tan-1 \(\frac{y}{x}\))dx + x Tan-1\(\frac{y}{x}\) dy = 0 in the form F(\(\frac{y}{x}\)) = \(\frac{dy}{dx}\).
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 2

Question 3.
Express x\(\frac{dy}{dx}\) = y(log y – log x + 1) in the from F (\(\frac{y}{x}\)) = \(\frac{dy}{dx}\)
Solution:
x . \(\frac{dy}{dx}\) = y(log y – log x + 1)
\(\frac{dy}{dx}\) = \(\frac{y}{x}\)(log\(\frac{y}{x}\) + 1)

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{x-y}{x+y}\)
Solution:
\(\frac{dy}{dx}=\frac{x-y}{x+y}\)
Put y = vx
\(\frac{dy}{dx}\) = v + x\(\frac{dv}{dx}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 3

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 2.
(x² + y²) dy = 2xy dx
Solution:
(x² + y²) dy = 2xy dx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 4
1 + v² = A(1 – v²) + Bv(1 – v) + Cv(1 + v)
v = 0 ⇒ = A
v = 1 ⇒ 1 + 1 = C(2) ⇒ c = 1
v = -1 ⇒ 1 + 1 = B(-1) (2) ⇒ 2 = -2B
B = -1
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 5

Question 3.
\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)
Solution:
\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 6
Multiplying with (v + 1)³
3 + v² = A(v + 1)² + B(v + 1) + C
v = – 1 ⇒ 3 + 1 = C ⇒ C = 4
Equating the co-efficients of v²
A = 1
Equating the co-efficients of v
0 = 2A + B
B = -2A = -2
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 7

Question 4.
y²dx + (x² – xy)dy = 0
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 8
v – log v = log x + log k
v = log v + log x + log k
= log k (vx)
\(\frac{y}{x}\) = log ky
Solution is ky = ey/x

Question 5.
\(\frac{dy}{dx}=\frac{(x+y)^2}{2x^2}\)
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 9

Question 6.
(x² – y²)dx – xy dy = 0
Solution:
(x² – y²)dx – xy dy = 0
(x² – y²)dx = xy . dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 10
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 11

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 7.
(x²y – 2xy²)dx = (x³ – 3x²y)dy
Solution:
(x²y – 2xy²)dx = (x³ – 3x²y)dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 12
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 13

Question 8.
y²dx + (x² – xy + y²) dy = 0
Solution:
y²dx = – (x² – xy + y²)dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 14
1 – v + v² = A(1 + v²) + (Bv + C)v
v  =  0 ⇒ 1 = A
Equating the co-efficients of v²
1 = A + .B ⇒ B = 0
Equating the co-efficients of v
-1 = C
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 15

Question 9.
(y² – 2xy)dx + (2xy – x²)dy = 0
Solution:
(y² – 2xy)dx + (2xy – x²)dy = 0
(2xy – x²)dy = – (y² – 2xy)dx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 16
2v – 1 = A(1 – v) + Bv
v = 0 ⇒ -1 = A ⇒ A = -1
v = 1 ⇒ 1 = B ⇒ B = 1
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 17

Question 10.
\(\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:
\(\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 18
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 19

Question 11.
x dy – y dx = \(\sqrt{x^2+y^2}\)
Solution:
x dy – y dx = \(\sqrt{x^2+y^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 20

Question 12.
(2x – y)dy = (2y – x)dx
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 21
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 22
⇒ (y – x)² = c²(y + x)²(y² – x²)
⇒ y – x = c²(y + x)³
⇒ (x + y)³ = c(x – y) where c = \(\frac{-1}{c^2}\)
∴ (x + y)³ = c(x – y)

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 13.
(x² – y²)\(\frac{dy}{dx}\) = xy
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 23
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 24

Question 14.
2\(\frac{dy}{dx}=\frac{y}{x}+\frac{y^2}{x^2}\)
Solution:
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 25
(y – x)² = y²c²x
Solution is y²x = c(x – y)²

III.

Question 1.
Solve : (1 + ex/y)dx + ex/y(1 – \(\frac{x}{y}\))dy = 0.
Solution:
Put x = vy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 26

Question 2.
Solve : x sin \(\frac{y}{x}.\frac{dy}{dx}\) = y sin \(\frac{y}{x}\) – x
Solution:
Dividing with x, we have
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 27

Question 3.
Solve : x dy = (y + x cos² \(\frac{y}{x}\))dx
Solution:
x.\(\frac{dy}{dx}\) = y + xcos² \(\frac{y}{x}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 28

Question 4.
Solve : (x – y log y + y log x)dx + x(log y – log x)dy = 0
Solution:
Dividing with x. dx we get
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 29
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 30

Question 5.
Solve : (y dx + x dy) x cos \(\frac{y}{x}\) = (x dy – y dx) y sin \(\frac{y}{x}\).
Solution:
The given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 31
∴ This is a homogeneous equation
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 32
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 33

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 6.
Find the equation of a curve whose gradient is \(\frac{dy}{dx}=\frac{y}{x}-\cos^2\frac{y}{x}\), where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)).
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 34
tan v = – log |x| + c
This curve passes through (1, \(\frac{\pi}{4}\))
tan \(\frac{\pi}{4}\) = c – log 1
c = 1
∴ Equation of the curve is
tan v = 1 – log |x|
tan \(\frac{y}{x}\) = 1 – log |x|