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

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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}}

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