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The general solution of the differential equation $(y \sin x+y) \frac{d y}{d x}-\cos ^2 x=0$
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Verified Answer
The correct answer is:
$y^2=2 x+2 \cos x+c$
We have, $(y \sin x+y) \frac{d y}{d x}-\cos ^2 x=0$
$$
\begin{aligned}
& \Rightarrow \quad \int y d y=\int \frac{\cos ^2 x}{1+\sin x} d x \\
& \Rightarrow \quad \int y d y=\int(1-\sin x) d x \\
& \Rightarrow \quad \frac{y^2}{2}=x+\cos x+c \\
& \Rightarrow \quad y^2=2 x+2 \cos x+c
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad \int y d y=\int \frac{\cos ^2 x}{1+\sin x} d x \\
& \Rightarrow \quad \int y d y=\int(1-\sin x) d x \\
& \Rightarrow \quad \frac{y^2}{2}=x+\cos x+c \\
& \Rightarrow \quad y^2=2 x+2 \cos x+c
\end{aligned}
$$
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