Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(b) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(b)

Solve the following differential equations.

Question 1.

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

Solution:

Given equation is \(\frac{d y}{d x}\) + y = e^{-x} … (1)

This is a linear differential equation.

Here P = 1, Q = e^{-x}

So the integrating factor

I.F. = e^{∫P dx} = e^{∫dx} = e^{x}

The solution of (1) is given by

ye^{x} = ∫e^{-x} . e^{x} dx = ∫dx = x + C

⇒ y – xe^{-x} + Ce^{-x}

Question 2.

(x^{2} – 1)\(\frac{d y}{d x}\) + 2xy = 1

Solution:

Given equation is (x^{2} – 1)\(\frac{d y}{d x}\) + 2xy = 1

Question 3.

(1 – x^{2})\(\frac{d y}{d x}\) + 2xy = x \(\sqrt{1-x^2}\)

Solution:

Given equation is

Question 4.

x log x \(\frac{d y}{d x}\) + y = 2 log x

Solution:

Given equation is

Question 5.

(1 + x^{2})\(\frac{d y}{d x}\) + 2xy = cos x

Solution:

Question 6.

\(\frac{d y}{d x}\) + y sec x = tan x

Solution:

Given equation is

\(\frac{d y}{d x}\) + y sec x = tan x

This is a linear equation where

P = sec x, Q = tan x

I.F. = e^{∫sec dx}

= e^{(sec x + tan x)} = sec x + tan x

The solution is y . (sec x + tan x)

= ∫(sec x + tan x) tan x dx

= ∫(sec x tan x + tan^{2} x) dx

= ∫(sec x . tan x + sec^{2} x – 1) dx

= ∫(sec x + tan x) – x + C

⇒ (y – 1) (sec x + tan x) + x = C

Question 7.

(x + tan y) dy = sin 2y dx

Given equation can be written as

Solution:

Question 8.

(x + 2y^{3})\(\frac{d y}{d x}\) = y

Solution:

Given equation can be written as

Question 9.

sin x\(\frac{d y}{d x}\)+ 3y = cos x

Solution:

Question 10.

(x + y + 1)\(\frac{d y}{d x}\) = 1

Solution:

Question 11.

(1 + y^{2}) dx + (x – \(e^{-\tan ^{-1} y}\)) dy = 0

Solution:

Given equation can be written as

Question 12.

x\(\frac{d y}{d x}\) + y = xy^{2}

Solution:

Given equation can be written as

⇒ z = -x ln x + Cx

⇒ \(\frac{1}{y}\) = -x ln x + Cx

⇒ 1 = -xy ln x + Cxy

∴ The solution is (C – ln x) xy = 1

Question 13.

\(\frac{d y}{d x}\) + y = y^{2} log x

Solution:

Question 14.

(1 + x^{2})\(\frac{d y}{d x}\) = xy – y^{2}

Solution:

The given equation can be written as

Question 15.

\(\frac{d y}{d x}\) + \(\frac{y}{x-1}\) = \(x y^{\frac{1}{2}}\)

Solution:

The given equation can be written as

Question 16.

\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x^{2}, y(1) = 1

Solution:

The given equation can be written as

\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x^{2}, y(1) = 1 … (1)

This is a linear equation.

Question 17.

\(\frac{d y}{d x}\) + 2y tan x = sin x, y\(\left(\frac{\pi}{3}\right)\) = 0.

Solution: