#### Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 19

Answer: $y^{2} \log y=e^{x} \sin ^{2} x+c$

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

Given: $\frac{d y}{d x}=\frac{e^{x}\left(\sin ^{2} x+\sin 2 x\right)}{y(2 \log y+1)}$

Solution: $\frac{d y}{d x}=\frac{e^{x}\left(\sin ^{2} x+\sin 2 x\right)}{y(2 \log y+1)}$

\begin{aligned} &\Rightarrow y(2 \log y+1) d y=e^{x}\left(\sin ^{2} x+\sin 2 x\right) d x \\\\ &\Rightarrow(2 y \log y+y) d y=\left(e^{x} \sin ^{2} x+e^{x} \sin 2 x\right) d x \\\\ &\Rightarrow 2 y \log y d y+y d y=e^{x} \sin ^{2} x d x+e^{x} \sin 2 x d x \end{aligned}

Integrating both sides and using integrating by parts

$\Rightarrow 2\left[\log y \int y d y-\int\left\{\frac{d}{d y}(\log y) \int y d y\right\}\right] d y+\int y d y$ $=\sin ^{2} x \int e^{x} d x-\int\left[\frac{d}{d x} \sin ^{2} x \int\left(e^{x} d x\right)\right] d x+\int e^{x} \sin 2 x d x$

$\Rightarrow 2\left[\log y\left(\frac{y^{2}}{2}\right)-\int\left(\frac{1}{y}\right)\left(\frac{y^{2}}{2}\right) d y\right]+y d y$ $=\sin ^{2} x e^{x}-\int\left[2 \sin x \cos x e^{x}\right] d x+\int e^{x} \sin 2 x+c$

\begin{aligned} &\Rightarrow y^{2} \log y-\int y d y+\int y d y=-\int e^{x} \sin 2 x d x+\int e^{x} \sin 2 x d x+e^{x} \sin ^{2} x \\\\ &y^{2} \log y=e^{x} \sin ^{2} x+c \end{aligned}