#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (iii) maths textbook solution.

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

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 Q I f 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] \end{aligned}

$=y e^{2 x}=-\cos x+C \ldots(i)$

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

\begin{aligned} &=0 e^{2(0)}=-\cos 0+C \\ &=0=-1+C \\ &=C=1 \end{aligned}

By (i)

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