Question

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

Answer: (A)

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