Question

The solution of $2y\sin x\frac{dy}{dx}=2\sin x\cdot \cos x-y^2\cos x,x=\frac{\pi }{2},y=1$ is given by :

• A. $y^2=\sin x$
• B. $y={\sin }^{2}x$
• C. $y^2=\cos x+1$
• D. None of these

Answer 1

The correct option is A $y^2=\sin x$

On dividing by $\sin x$,
$2y\frac{dy}{dx}+y^2\cot x=2\cos x$
Put $y^2=v\Rightarrow \frac{dv}{dx}+v\cot x=2\cos x$
$IF=e^{\int \cot xdx}=e^{\log \sin x}=\sin x$
$\therefore$ Solution is,
$v\cdot \sin x=\int \sin x\left(2\cos x\right)dx+C$
$\Rightarrow y^2\sin x+{\sin }^{2}x+C$
When $x=\frac{\pi }{2},y=1$
then $C=0$
$\therefore y^2=\sin x$.