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Explain Solution R.D. Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 32 Maths Textbook Solution.

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Answer:y^{2}log\: y=x\: \sin +C

Hint: you must know the rules of solving differential equation and integrations.

Given: \frac{dy}{dx}=\frac{\sin x+x\cos x}{y\left ( 2logy+1 \right )}

Solution:\frac{dy}{dx}=\frac{\sin x+x\cos x}{y\left ( 2logy+1 \right )}

y\left ( 2log\: y+1 \right )dy=\left ( \sin x+x\cos x \right )dx

integrating both sides,

\int 2\: y\: log \: y dy+1\int y \: dy=\int \sin x\: dx+\int x\cos x\: dx

Integrating by parts,

\begin{aligned} &2 \log y \int y d y-2 \int\left(\frac{d}{d y}(\log y) x \int y d y\right) d y+\int y d y=-\cos x+x \int \cos x d x-\int \frac{d}{d x}(x) \cdot \int \cos x d x \\ &\quad \Rightarrow 2 \log y\left(\frac{y^{2}}{2}\right)-\int y d y+\int y d y=-\cos x+x \sin x+\cos x+c \\ &\quad \Rightarrow y^{2} \log y=x \sin x+c \end{aligned}

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