Question

Solution of $2ysinx\frac{dy}{dx}=2sinx\cdot cosx-y^{2}cosx$, $x=\frac{\pi }{2},y=1$ is given by

• A. $y^2=sinx$
• B. $y=si{n}^{2}x$
• C. $y^2=cosx+1$
• D. None of these

Answer 1

Answer: (A)

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, $v\cdot \sin x=\int \sin x(2\cos x)dx+C$
$\Rightarrow y^2\sin x={\sin }^{2}x+C$
When $x=\frac{\pi }{2},y=1$
then $C=0$
$\therefore y^2=\sin x$