#### Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (vii) maths textbook solution.

Answer : $y=\frac{1}{2} e^{\sin x}+\frac{C}{e^{\sin x}}$

Give : $\frac{d y}{d x}+y \cos x=e^{\sin x} \cos x$

Hint : Using $\int \cos\; x\; dx$

Explanation : $\frac{d y}{d x}+\cos x y=e^{\sin x} \cos x$

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\cos x \text { and } Q=e^{\sin x} \cos x \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &I f=e^{\int \cos x d x} \\ &=e^{\sin x} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \sin x d x=\cos x+C\right] \end{aligned}

Hence, the solution of different equation is

\begin{aligned} &y I f=\int \text { QIfd } d x+C \\ &=y e^{\sin x}=\int e^{\sin x} \cos x e^{\sin x} d x+C \\ &=y e^{\sin x}=\int e^{2 \sin x} \cos x d x+C \end{aligned}

\begin{aligned} &=y e^{\sin x}=\int e^{2 t} d t+C \quad[\sin x=t, \cos x d x=d t] \\ &=y e^{\sin x}=\left[\frac{e^{2 t}}{2}\right]+C \\ &=y e^{\sin x}=\frac{e^{2 \sin x}}{2}+C \end{aligned}

\begin{aligned} &=y=\frac{e^{2 \sin x}}{2 e^{\sin x}}+\frac{C}{e^{\sin x}} \\ &=y=\frac{e^{\sin x}}{2}+\frac{C}{e^{\sin x}} \end{aligned}