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^{-1}c where c is a constant.

Question 2.

\(\frac{dy}{dx}\) = e^{y-k}

Solution:

\(\frac{dy}{dx}=\frac{e^y}{e^x}\)

\(\frac{dy}{e^y}=\frac{dx}{e^x}\)

∫e^{-x}dx = ∫e^{-y}dy

-e^{-x} = -e^{-y} + C

e^{-y} = e^{-x} + c where c is a constant.

Question 3.

(e^{x} + 1) y dy + (y + 1) dx = 0

Solution:

(e^{x} + 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

e^{y} = c(y + 1) (e^{-x} +1)

Question 4.

\(\frac{dy}{dx}\) = e^{x-y} + x²e^{-y}

Solution:

\(\frac{dy}{dx}\) = e^{x-y} + x² . e^{-y}

= \(\frac{e^x}{e^y}=\frac{x^2}{e^y}\)

∫e^{y} . dy = ∫(e^{x} + x²) dx

Solution is

e^{y} = e^{x} + \(\frac{x^3}{3}\) + 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 e^{3y}

Solution:

log(1 – e^{-3y}) = x³ + c'(c’ = 3c)

Solution is

1 – e^{-3y} = e^{x³} . k(k = e^{c’})

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