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#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 30 Maths Textbook Solution.

$\Rightarrow$ Answer: $\frac{1}{2}\left ( log\: y \right )^{2}+\left ( 2-x^{2} \right )\cos x+2x\sin x=C$

Hint: Apply integration and then the formula of $\int u\: v\: dx$   and substitution method.

Given: $\cos cx\left ( log\: y \right )dy+x^{2}y \: d\: x=0$

Solution:$\cos cx\left ( log\: y \right )dy+x^{2}y \: d\: x=0$

\begin{aligned} &\Rightarrow \frac{\log y}{y} d y=-\frac{x^{2}}{\operatorname{cosec} x} d x \\ &\Rightarrow \int \frac{l o g y}{y} d y=-\int x^{2} \sin x d x \end{aligned}        (integrating both sides)

Let log y=t

\begin{aligned} &\Rightarrow \frac{1}{y} d y=d t \quad(\text { diff. w.r.t. } y) \\ &\Rightarrow \because \int t d t=-\int x^{2} \sin x d x \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=-x^{2}(-\cos x)+\int 2 x(-\cos x) d x \\ &\because\left[\int u v d x=u \int v d x-\int\left(\frac{d}{d x} u \int v d x\right) d x\right] \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=x^{2} \cos x-2 \int x \cos x d x \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=x^{2} \cos x-2 x \sin x+2 \int \sin x d x \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=x^{2} \cos x-2 x \sin x-2 \cos x+C \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=\left(x^{2}-2\right) \cos x-2 x \sin x+C \\ &\Rightarrow \frac{1}{2}(\log y)^{2}+\left(2-x^{2}\right) \cos x+2 x \operatorname{Sin} x=C \end{aligned}