#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 41 Maths Textbook Solution.

Answer:$y\cos x=\frac{\cos 2x}{2}+c$

Hint: you must know the rules of solving differential equation and integrations.

Given:$\frac{dy}{dx}-y\tan x=-2\sin x$

Solution:$\frac{dy}{dx}-y\tan x=-2\sin x$

$\frac{dy}{dx}+\left ( -\tan x \right )y=-2\sin x$

The above equation look like,

$\frac{dy}{dx}+py=q$

Where p = -tan x and q = -2 sin x

Integrating factor = $e^{\int px}$

$=e^{\int -\tan x\: dx}$

$=e-^{\int \tan x\: dx}$

we have $\int \tan x\: dx=log\left ( \sec x \right )+c$

\begin{aligned} &\text { I. } F=e^{-\log (s e c x)} \\ &\begin{array}{l} \text { I.F }=e^{-\log \left(\frac{1}{\operatorname{secx}}\right)} \\ \text { I.F }=\frac{1}{\sec x} \quad\left[\because \mathrm{e}^{\log x}=\mathrm{x}\right] \\ \therefore \text { I. } F=\operatorname{Cos} \mathrm{x} \end{array} \end{aligned}

Hence, now the solution of differential equation is,

\begin{aligned} &\mathrm{y}(\mathrm{I} . \mathrm{F})=\int(\mathrm{q} \times \mathrm{I} \cdot \mathrm{F}) \mathrm{d} \mathrm{x}+\mathrm{C} \\ &\mathrm{y}(\operatorname{Cos} \mathrm{x})=\int(-2 \sin \mathrm{x} \cdot \cos \mathrm{x}) \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \operatorname{Cos} \mathrm{x}=-\int \sin 2 x \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \cos \mathrm{x}=\frac{\cos 2 x}{2}+c \end{aligned}