Solution of $\frac{d y}{d x}=\frac{x \log x^2+x}{\sin y+y \cos y}$ is

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Aicharya · Accepted Answer

We have, $\frac{d y}{d x}=\frac{x \log x^2+x}{\sin y+y \cos y}$ $\Rightarrow(\sin y+y \cos y) d y=\left(x \log x^2+xight) d x$ On integrating both sides, we get $-\cos y+y \sin y+\cos y$ $=\frac{x^2}{2} \log x^2-\frac{x^2}{2}+\frac{x^2}{2}+C$ $\Rightarrow \quad y \sin y=\frac{x^2}{2} \log x^2+C$ $	herefore \quad y \sin y=x^2 \log x+C$