Question
$\frac{{dy}}{{dx}} = \frac{{{e^x}({{\sin }^2}x + \sin 2x)}}{{y(2\log y + 1)}}$ का हल है
==>$\int_{}^{} {(2y\log y + y)dy = \int_{}^{} {{e^x}({{\sin }^2}x + \sin 2x} )dx} $
खण्डश: समाकलन करने पर, ${y^2}(\log y) = {e^x}{\sin ^2}x + c$.
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