#### Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 70 textbook solution.

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

Hint               : using integration by parts

Given :    $\frac{d y}{d x}=e^{x} \sin x$

Solution : $\frac{d y}{d x}=e^{x} \sin x$

$d y=e^{x} \sin x d x$

Integrating both sides

\begin{aligned} &\int d y=\int e^{x} \sin x d x\\ &y=\int e^{x} \sin x d x \end{aligned}               ....(i)

Using integration by parts

Let          \begin{aligned} &I=\int e^{x} \sin x d x=\sin x e^{x}-\int \cos x e^{x} d x \\ \end{aligned}

\begin{aligned} &\qquad=\sin x e^{x}-\left[\cos x e^{x}-\int(-\sin x) e^{x} d x\right] \\ \end{aligned}

\begin{aligned} &I=\sin x e^{x}-\cos x e^{x}-\int \sin x e^{x} d x \\ \end{aligned}

\begin{aligned} &I=\sin x e^{x}-\cos x e^{x}-I \\ \end{aligned}

\begin{aligned} &2 I=\sin x e^{x}-\cos x e^{x} \\ \end{aligned}

\begin{aligned} &I=\frac{e^{x}(\sin x-\cos x)}{2} \end{aligned}

Put in (i), we get,

$\begin{gathered} y=\frac{e^{x}(\sin x-\cos x)}{2} \\ 2 y=e^{x}(\sin x-\cos x) \\ \qquad \begin{array}{r} y=\frac{1}{2} e^{x}(\sin x-\cos x)+c \end{array} \end{gathered}$                          ....(ii)

Given curve passes through (0,0)

Putting x = 0 , y = 0 in equation

Therefore ,          \begin{aligned} &0=\frac{1}{2} e^{0}(\sin 0-\cos 0)+c \\ \end{aligned}

\begin{aligned} &0=\frac{1}{2}(1)(0-1)+c \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[e^{0}=1\right] \\ \end{aligned}

\begin{aligned} &0=\frac{-1}{2}+c \end{aligned}

$c=\frac{1}{2}$

Putting value of C in (ii)

$\begin{gathered} y=\frac{1}{2} e^{x}(\sin x-\cos x)+\frac{1}{2} \\ y-\frac{1}{2}=\frac{1}{2} e^{x}(\sin x-\cos x) \\ \frac{2 y-1}{2}=\frac{e^{x}}{2}(\sin x-\cos x) \\ 2 y-1=e^{x}(\sin x-\cos x) \end{gathered}$