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

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

I.

Question 1.
Find the general solution of $$\sqrt{1-x^2}$$ dy + $$\sqrt{1-y^2}$$ dx = 0.
Solution:
Given differential equation is

sin-1 y = – sin-1 x + c
Solution is sin-1 x + sin-1 y = c, where c is a constant.

Question 2.
Find the general solution of $$\frac{dy}{dx}=\frac{2y}{x}$$.
Solution:
$$\frac{dy}{dx}=\frac{2y}{x}$$
∫$$\frac{dy}{dx}$$ = 2∫$$\frac{2y}{x}$$
log c + log y = 2 log x
log cy = log x²
Solution is cy = x², where c. is a constant.

II. Solve the following differential equations.

Question 1.
$$\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$$
Solution:
$$\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$$
∫$$\frac{dy}{1+y^2}$$ = ∫$$\frac{dx}{1+x^2}$$
tan-1 y = tan-1 x + tan-1c where c is a constant.

Question 2.
$$\frac{dy}{dx}$$ = ey-k
Solution:
$$\frac{dy}{dx}=\frac{e^y}{e^x}$$
$$\frac{dy}{e^y}=\frac{dx}{e^x}$$
∫e-xdx = ∫e-ydy
-e-x = -e-y + C
e-y = e-x + c where c is a constant.

Question 3.
(ex + 1) y dy + (y + 1) dx = 0
Solution:
(ex + 1 )y. dy = – (y + 1) dx

y – log (y + 1) = log (e-x + 1) + log c
⇒ y – log (y + 1) = log c (e-x + 1)
⇒ y = log (y + 1) + log c (e-x + 1)
y = log c (y + 1) (e-x +1)
Solution is
ey = c(y + 1) (e-x +1)

Question 4.
$$\frac{dy}{dx}$$ = ex-y + x²e-y
Solution:
$$\frac{dy}{dx}$$ = ex-y + x² . e-y
= $$\frac{e^x}{e^y}=\frac{x^2}{e^y}$$
∫ey . dy = ∫(ex + x²) dx
Solution is
ey = ex + $$\frac{x^3}{3}$$ + c

Question 5.
tan y dx + tan x dy = 0
Solution:
tan y dx = – tan x dy

log sin x = – log sin y + log c
log sin x + log sin y = log c
log (sin x . sin y) = log c
⇒ sin x . sin y = c is the solution

Question 6.
$$\sqrt{1+x^2}$$dx + $$\sqrt{1+y^2}$$dy = 0
Solution:
$$\sqrt{1+x^2}$$dx = –$$\sqrt{1+y^2}$$dy
Integrating both sides we get
∫$$\sqrt{1+x^2}$$dx = -∫$$\sqrt{1+y^2}$$dy
Integrating both sides we get

Question 7.
y – x$$\frac{dy}{dx}$$ = 5(y² + $$\frac{dy}{dx}$$)
Solution:
y – 5y² = (x + 5)$$\frac{dy}{dx}$$
$$\frac{dx}{x+5}=\frac{dy}{y(1-5y}$$
Integrating both sides

Question 8.
$$\frac{dy}{dx}=\frac{xy+y}{xy+x}$$
Solution:

III. Solve the following differential equations.

Question 1.
$$\frac{dy}{dx}=\frac{1+y^2}{(1+x^2)xy}$$
Solution:

log (1 + y²) = log x² – log (1 + x²) + log c
log (1 + x²) + log (1 + y²) = log x² + log c
Solution is (1 + x²) (1 + y²) = cx²

Question 2.
$$\frac{dy}{dx}$$ + x² = x² e3y
Solution:

log(1 – e-3y) = x³ + c'(c’ = 3c)
Solution is
1 – e-3y = e . k(k = ec’)

Question 3.
(xy² + x)dx+(yx²+y)dy = 0.
Solution:
(xy² + x) dx + (yx² + y) dy = 0
x(y² + 1) dx + y (x² + 1) dy = 0
Dividing with (1 + x²) (1 + y²)
$$\frac{x dx}{1+x^2}+\frac{y dy}{1+y^2}$$ = 0
Integrating
∫$$\frac{x dx}{1+x^2}$$ + ∫$$\frac{y dy}{1+y^2}$$ = 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 4.
$$\frac{dy}{dx}$$ = 2y tanh x
Solution:
$$\frac{dy}{dx}$$ = 2y tanh x
$$\frac{dy}{y}$$ = 2 tanh x dx
Integrating both sides we get
∫$$\frac{dy}{y}$$ = 2 ∫ tanh x dx
log y = 2 log |cosh x| + log c
lny = 2ln cosh x + In c
y = c cos²h x

Question 5.
sin-1 $$\frac{dy}{dx}$$ = x + y
Solution:
$$\frac{dy}{dx}$$ = sin(x + y)
x + y = t
1 + $$\frac{dy}{dx}=\frac{dt}{dx}$$
$$\frac{dt}{dx}$$ – 1 = sin t
$$\frac{dt}{dx}$$ = 1 + sin t
Integrating both sides we get
∫$$\frac{dt}{1+\sin t}$$ = ∫dx
∫$$\frac{1-\sin t}{\cos^2 t}$$ dt = x + c
∫sec² t dt – ∫tan t . sec t dt = x + c
tan t – sec t = x + c
⇒ tan (x + y) – sec (x + y) = x + c

Question 6.
$$\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}$$ = 0
Solution:
$$\frac{-dy}{y^2+y+1}=\frac{dx}{x^2+x+1}$$
Integrating both sides dy

Question 7.
$$\frac{dy}{dx}$$ = tan² (x + y)
Solution:
$$\frac{dy}{dx}$$ = tan² (x + y)
put v = x + y

2v + sin 2v = 4x + c’
2(x + y) + sin 2(x + y) = 4x + c’
x – y – $$\frac{1}{2}$$sin [2(x + y)] = c