Question

Solution of the differential equation $2y\sin x\frac{dy}{dx}=2\sin x\cos x-y^2\cos xsatisfying,y\left(\frac{\pi }{2}\right)=1$

is given by?

• A. $y^2=\sin x$
• B. $y={\sin }^{2}x$
• C. $y^2=\cos x+1$
• D. $y^2\sin x=4{\cos }^{2x}$

Answer 1

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

$2y\sin x\frac{dy}{dx}=2\sin x\cos x-y^2\cos x$

$\Rightarrow 2y\sin xdy=(\sin 2x-y^2\cos x)dx$

When we compare this equation with

$Mdx+Ndy=0$

then

$M=\sin 2x-y^2\cos x$

$N=-2y\sin x$

$\Rightarrow \frac{2M}{2y}=-2y\cos x$ and $\frac{2N}{2x}=-2y\cos x$

$\therefore$ It is an exact differential equation

$\Rightarrow (\sin 2x-y^2\cos x)dx-2y\sin xdy=0$

$\Rightarrow \sin 2xdx-(y^2\cos xdx+2ydy\sin x)=0$

$\Rightarrow \sin 2xdx-d\left(y^2\sin x\right)=0$

$\Rightarrow \int \sin 2xdx-y^2\sin x=c$

$\Rightarrow -\frac{\cos 2x}{2}-y^2\sin x=c$

which is satisfying $y\left(\pi /2\right)=1$

where $y^2\sin x=-\frac{\cos 2x}{2}+c$ ….$\left(1\right)$

at $y\left(\pi /2\right)=1$ we get

$(1{)}^2\times \sin (\pi /2)=-\frac{\cos \left(\pi \right)}{2}+c$

$1=\frac{1}{2}+c$

$c=\frac{1}{2}$

$\Rightarrow$ Putting value of c in equation $\left(1\right)$ we get solution

$y^2\sin x=-\frac{\cos 2x}{2}$.