#### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 52 maths

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

Hint: Separate the terms of x and y and then integrate them.

Given: Curve passing through (0,0) with differential equation

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

Solution:

\begin{aligned} &\frac{d y}{d x}=e^{x} \sin x \\\\ &\Rightarrow d y=e^{x} \sin x d x \end{aligned}

Integrating both sides

$\int d y=\int e^{x} \sin x d x \Rightarrow y=\int e^{x} \sin x d x$            .................(*)

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

The curve passes through (0,0)           [∴ of given]

\begin{aligned} &0=e^{0}(\sin 0-\cos 0)+c\\\\ &\Rightarrow 0=1(0-1)+c \Rightarrow c=1\\\\ &\operatorname{Put\; in}(1)\\\\ &2 y=e^{x}(\sin x-\cos x)+1 \end{aligned}