#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise  Revision Exercise Question 23 Maths Textbook Solution.

Answer:$y=log|logx|+\frac{x^{2}}{2}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+c$

Hint:  integrate both sides and then apply the formula of

Given: $\frac{dy}{dx}-x\sin ^{2}x=\frac{1}{xlogx}$

Solution: $\frac{dy}{dx}-x\sin ^{2}x=\frac{1}{xlogx}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x} \log x}+\mathrm{x} \sin ^{2} \mathrm{x}$

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{1}{\mathrm{x} \operatorname{logx}}+\frac{\mathrm{x}}{2}(1-\cos 2 \mathrm{x})\left[\because \operatorname{Cos} 2 \mathrm{x}=1-2 \sin ^{2} \mathrm{x}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x} \log x}+\frac{\mathrm{x}}{2}-\frac{x \operatorname{Cos} 2 \mathrm{x}}{2} \\$

$\Rightarrow \int \mathrm{dy}=\int\left(\frac{1}{\mathrm{xlog} \mathrm{x}}+\frac{\mathrm{x}}{2}-\frac{x \operatorname{Cos} 2 \mathrm{x}}{2}\right) \mathrm{d} \mathrm{x} \quad \text { (integrate both sides) } \\$

$\Rightarrow \int \mathrm{dy}=\int\left(\frac{1}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}+\frac{1}{2} \int \mathrm{x} \mathrm{dx}-\frac{1}{2}(\mathrm{x} \operatorname{Cos} 2 \mathrm{x}) \mathrm{d} \mathrm{x}\right. \\$

$\Rightarrow \mathrm{y}=\log |\log \mathrm{x}|+\frac{\mathrm{x}^{2}}{2}-\frac{\mathrm{x}}{2} \int \cos 2 \mathrm{x} \mathrm{dx}+\frac{1}{2} \int \frac{\sin 2 \mathrm{x}}{2} \mathrm{dx} \\$

$\Rightarrow \mathrm{y}=\log |\log \mathrm{x}|+\frac{\mathrm{x}^{2}}{2}-\frac{\mathrm{x}}{2} \cdot \frac{\sin 2 \mathrm{x}}{2}+\frac{1}{2}\left(-\frac{\cos 2 \mathrm{x}}{4}\right)+\mathrm{C} \\$

$\Rightarrow \mathrm{y}=\log |\log \mathrm{x}|+\frac{\mathrm{x}^{2}}{2}-\frac{\mathrm{x} \sin 2 \mathrm{x}}{4}-\frac{\cos 2 \mathrm{x}}{8}+\mathrm{C}$