# Get Answers to all your Questions

#### Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (ix) Maths Textbook Solution.

Answer:  \begin{aligned} &y=\tan x-1+\frac{c}{e^{\tan x}}\\ & \end{aligned}

Give:  $\frac{d y}{d x} \cos ^{2} x=\tan x-y$

Hint: Using integration by parts.

Explanation:  \begin{aligned} &\frac{d y}{d x} \cos ^{2} x=\tan x-y \\ & \end{aligned}

$=\frac{d x}{d y} \cos ^{2} x+y=\tan x$

Divide by $\cos ^{2}x$

\begin{aligned} &=\frac{d y}{d x}+\frac{y}{\cos ^{2} x}=\frac{\tan x}{\cos ^{2} x} \\ \\&=\frac{d y}{d x}+\left(\frac{1}{\cos ^{2} x}\right) y=\frac{\tan x}{\cos ^{2} x} \end{aligned}

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{1}{\cos ^{2} x} \operatorname{and} Q=\frac{\tan x}{\cos ^{2} x} \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{1}{\cos ^{2} x} d x} \\ &=e^{\int \sec ^{2} x d x} \\ &=e^{\tan x} \quad\left[\int \sec ^{2} x d x=\tan x+C\right] \end{aligned}

Hence, the solution of different equation is

\begin{aligned} &y I f=\int \text { QIfd } x+C \\ &=y e^{\tan x}=\int \frac{\tan x}{\cos ^{2} x} e^{\tan x} d x+C \ldots(i) \end{aligned}

Using integration by parts,

$=\int \tan x \sec ^{2} x e^{\tan x} d x+C$

Put  \begin{aligned} &\tan x=t, \sec ^{2} x d x=d t \\ & \end{aligned}

\begin{aligned} & \\ &\quad=\int t e^{t} d t+C \\ &\quad=t e^{t}-\int(1) e^{t} d t \\ &\quad=t e^{t}-e^{t}+C \\ &=\tan x e^{\tan x}-e^{\tan x}+C \end{aligned}

Put in  $\left ( i \right )$

$=y e^{\tan x}=\tan x e^{\tan x}-e^{\tan x}+C$

Divide by $e^{\tan x}$

$=y=\tan x-1+\frac{C}{e^{\tan x}}$

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support