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The general solution of $\left(x \frac{d y}{d x}-y\right) \sin \frac{y}{x}=x^3 e^x$ is
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Verified Answer
The correct answer is:
$\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+\cos \frac{\mathrm{y}}{\mathrm{x}}+\mathrm{c}=0$
$$
\begin{aligned}
& \left(x \frac{d y}{d x}-y\right) \sin \frac{y}{x}=x^3 e^x \\
& \text { Put } \frac{y}{x}=t \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=\frac{d t}{d x}
\end{aligned}
$$
Put $\frac{y}{x}=t \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=\frac{d t}{d x}$
Hence eq. (i) becomes
$$
\begin{aligned}
& \mathrm{x}^2\left(\frac{d t}{d x}\right) \sin t=x^3 e^x \Rightarrow\left(\frac{d t}{d x}\right) \sin t=x e^x \\
& \therefore \int \sin t d t=\int x^x d x \\
& \therefore-\cos t=x e^x-\int e^x d x=x e^4-e^x+c \\
& \therefore-\cos \left(\frac{y}{x}\right)=e^x(x-1)+c
\end{aligned}
$$
\begin{aligned}
& \left(x \frac{d y}{d x}-y\right) \sin \frac{y}{x}=x^3 e^x \\
& \text { Put } \frac{y}{x}=t \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=\frac{d t}{d x}
\end{aligned}
$$
Put $\frac{y}{x}=t \Rightarrow \frac{x \frac{d y}{d x}-y}{x^2}=\frac{d t}{d x}$
Hence eq. (i) becomes
$$
\begin{aligned}
& \mathrm{x}^2\left(\frac{d t}{d x}\right) \sin t=x^3 e^x \Rightarrow\left(\frac{d t}{d x}\right) \sin t=x e^x \\
& \therefore \int \sin t d t=\int x^x d x \\
& \therefore-\cos t=x e^x-\int e^x d x=x e^4-e^x+c \\
& \therefore-\cos \left(\frac{y}{x}\right)=e^x(x-1)+c
\end{aligned}
$$
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