Answer
\[y=\frac{\sqrt{1+\sin x}}{x}\]
Work Step by Step
$x^2y^2-\sin x=c$ ______(1)
Differentiate (1) with respect to $x$ treating $y$ as function of $x$
\[2xy^2+2x^2y\frac{dy}{dx}-\cos x=0\]
\[\frac{dy}{dx}=\frac{\cos x-2xy^2}{2x^2y}\]
Hence (1) ia implict solution of the differential equation $y'=\frac{\cos x-2xy^2}{2x^2y}$
$\circ y(π)=\frac{1}{π}$ _____(2)
Using (2) in (1)
$\frac{π^2}{π^2}-\sin π =c$ $\Rightarrow$ $c=1$
From (1)
\[x^2y^2-\sin x=1\]
\[y=\frac{\sqrt{1+\sin x}}{x}\]
Hence explicit solution that satisfies $y(π)=\frac{1}{π}$ is $y=\frac{\sqrt{1+\sin x}}{x}$
