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

Answer : $y=\frac{1}{5}(2 \sin 2 x-\cos 2 x)+C e^{x}$

Give : $\frac{d y}{d x}-y=\cos 2 x$

Hint : Using integration by parts

Explanation :  $\frac{d y}{d x}-y=\cos 2 x$

$\frac{d y}{d x}+(-1)y=\cos 2 x$

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=-1 \text { and } Q=\cos 2 x \end{aligned}

The integrating factor $If$ of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-1 d x} \\ &=e^{-\int 1 d x} \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int d c=x+C\right] \\ &=e^{-x} \end{aligned}

Hence, the solution of differential equation is

\begin{aligned} &y(I f)=\int Q I f d x+C \\ &\Rightarrow y\left(e^{-x}\right)=\int(\cos 2 x) e^{-x} d x+C \\ &\Rightarrow y e^{-x}=\int e^{-x} \cos 2 x d x+C \end{aligned}

$\Rightarrow y e^{-x}=\int\left(e^{-x}\right)(\cos 2 x) d x+C$

$\text { Let } I=\int\left(e^{-x}\right)(\cos 2 x) d x+C$

\begin{aligned} &\Rightarrow I=e^{-x} \int \cos 2 x d x-\int \frac{d}{d x}\left(e^{-x}\right)\left(\int \cos 2 x d x\right) d x+C \\ &\Rightarrow I=e^{-x}\left(\frac{\sin 2 x}{2}\right)-\int-e^{-x}\left(\frac{\sin 2 x}{2}\right) d x+C \\ &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2} \int e^{-x} \sin 2 x d x+C \end{aligned}

\begin{aligned} &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[e^{-x}\left(\int \sin 2 x d x\right)\right]-\int \frac{d}{d x} e^{-x}\left(\int \sin 2 x d x\right) d x+C\right\} \\ &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[e^{-x}\left(\frac{-\cos 2 x}{2}\right)\right]-\int-e^{-x}\left(-\frac{\cos 2 x}{2}\right) d x+C\right\} \\ &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} \int-e^{-x}(-\cos 2 x) d x+C\right\} \\ &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} \int e^{-x} \cos 2 x d x+C\right\} \end{aligned}

\begin{aligned} &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left\{\left[-\frac{1}{2} e^{-x} \cos 2 x\right]-\frac{1}{2} I\right\} \\ &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x+\frac{1}{2}\left(-\frac{1}{2} e^{-x} \cos 2 x\right)+\frac{1}{2}\left(-\frac{1}{2} I\right) \\ &\Rightarrow I=\frac{1}{2} e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x-\frac{1}{4} I \\ &\Rightarrow I+\frac{1}{4} I=\frac{1}{2} e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x \end{aligned}

\begin{aligned} &\Rightarrow \frac{5}{4} I=\frac{1}{4} 2 e^{-x} \sin 2 x-\frac{1}{4} e^{-x} \cos 2 x \\ &\Rightarrow \frac{5}{4} I=\frac{1}{4} e^{-x}(2 \sin 2 x-\cos 2 x) \\ &\Rightarrow 5 I=e^{-x}(2 \sin 2 x-\cos 2 x) \\ &\therefore I=\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x) \end{aligned}

By substituting the value of $I$ in the original integral we get

\begin{aligned} &\Rightarrow y e^{-x}=\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x)+C \\ &\Rightarrow y e^{-x} e^{x}=e^{x}\left[\frac{e^{-x}}{5}(2 \sin 2 x-\cos 2 x)+C\right] \end{aligned}

\begin{aligned} &\Rightarrow y e^{-x} e^{x}=\frac{e^{x-x}}{5}(2 \sin 2 x-\cos 2 x)+C e^{x} \\ &\therefore y=\frac{1}{5}(2 \sin 2 x-\cos 2 x)+C e^{x} \end{aligned}