#### Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 subquestion (iii)

Answer:  \begin{aligned} & \cos x+y e^{2 x}=1 \\ & \end{aligned}

Give:  $\frac{d y}{d x}+2 y=e^{-2 x} \sin x, y(0)=0$

Hint: Using integrating factor

Explanation: $\frac{d y}{d x}+2 y=e^{-2 x} \sin x$

This is a linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=2 \text { and } Q=e^{-2 x} \sin x \end{aligned}

The integrating factor  $If$  of this differential equation is

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

Hence, the solution is

\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y\left(e^{2 x}\right)=\int e^{-2 x} \sin x e^{2 x} d x+C \\ &=y e^{2 x}=\int \sin x d x+C \quad\left[e^{-2 x} e^{2 x}=e^{-2 x+2 x}=e^{0}=1\right] \\ &=y e^{2 x}=-\cos x+C \ldots(i) \end{aligned}

Now  $\begin{gathered} y(0)=0, \text { When } x=0, y=0 \\ \end{gathered}$

$\quad=0 e^{2(0)}=-\cos 0+C$

\begin{aligned} &=0=-1+C \\ &=C=1 \\ & \end{aligned}

By i

$=y e^{2 x}=-\cos x+1 \\$

$=\cos x+y e^{2 x}=1$