#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 40 Maths Textbook Solution.

Answer: $y\, \cos ec\: x+\cot \: x=c$

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

Given: $\frac{dy}{dx}-y\cot x=\cos ec\: x$

Solution:$\frac{dy}{dx}-y\cot x=\cos ec\: x$

The above equation is in form of

$\frac{dy}{dx}+p\: y=q$

Where p = -cot x and q = cosec x

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

Considering $\int p\: d\: x$

\begin{aligned} \Rightarrow & \int p d x=-\int \cot x d x \\ &=-\log |\sin x| &\left[\because e^{\log x}=x\right] \\ & \therefore \mathrm{e}^{\int p d x}=e^{-\log |\sin x|}=e^{\log (\sin x)^{-1}} \\ &=\sin x^{-1}=\frac{1}{\sin x}=\operatorname{cosec} x \end{aligned}

$\therefore \text { Integrating factor I.F }=\operatorname{cosec} \mathrm{x}$

Now, General solution is,

\begin{aligned} &\mathrm{y}(\mathrm{I} . \mathrm{F})=\int \mathrm{q}(\mathrm{I} . \mathrm{F}) \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=\int \operatorname{cosec} \mathrm{x} \operatorname{cosec} \mathrm{x} \mathrm{dx}+\mathrm{c} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=\int \operatorname{cosec} \mathrm{x}^{2} \mathrm{dx}+\mathrm{c} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=-\cot \mathrm{x}+\mathrm{C} \\ &\Rightarrow \mathrm{y} \operatorname{cosec} x+\cot \mathrm{x}=\mathrm{c} \end{aligned}