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 = $$\sqrt{x^2+y^2}$$ 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: 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$$\frac{dv}{dx}$$  Question 2.
(x² + y²) dy = 2xy dx
Solution:
(x² + y²) dy = 2xy dx 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 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 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 Question 4.
y²dx + (x² – xy)dy = 0
Solution: 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: Question 6.
(x² – y²)dx – xy dy = 0
Solution:
(x² – y²)dx – xy dy = 0
(x² – y²)dx = xy . dy   Question 7.
(x²y – 2xy²)dx = (x³ – 3x²y)dy
Solution:
(x²y – 2xy²)dx = (x³ – 3x²y)dy  Question 8.
y²dx + (x² – xy + y²) dy = 0
Solution:
y²dx = – (x² – xy + y²)dy 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 Question 9.
(y² – 2xy)dx + (2xy – x²)dy = 0
Solution:
(y² – 2xy)dx + (2xy – x²)dy = 0
(2xy – x²)dy = – (y² – 2xy)dx 2v – 1 = A(1 – v) + Bv
v = 0 ⇒ -1 = A ⇒ A = -1
v = 1 ⇒ 1 = B ⇒ B = 1 Question 10.
$$\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}$$
Solution:
$$\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}$$  Question 11.
x dy – y dx = $$\sqrt{x^2+y^2}$$
Solution:
x dy – y dx = $$\sqrt{x^2+y^2}$$ Question 12.
(2x – y)dy = (2y – x)dx
Solution:  ⇒ (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) Question 13.
(x² – y²)$$\frac{dy}{dx}$$ = xy
Solution:  Question 14.
2$$\frac{dy}{dx}=\frac{y}{x}+\frac{y^2}{x^2}$$
Solution:
Put y = vx (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 Question 2.
Solve : x sin $$\frac{y}{x}.\frac{dy}{dx}$$ = y sin $$\frac{y}{x}$$ – x
Solution:
Dividing with x, we have Question 3.
Solve : x dy = (y + x cos² $$\frac{y}{x}$$)dx
Solution:
x.$$\frac{dy}{dx}$$ = y + xcos² $$\frac{y}{x}$$ 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  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 ∴ This is a homogeneous equation   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: 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|