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

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

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\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}

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