Question

If the curve satisfying $y{y}_1\sin x=\cos x(\sin x-y^2)$ passes through $(\frac{\pi }{2},2)$, then the value of $3\left(y\right(\frac{\pi }{6}){)}^2$ is equal to

• A. $81$
• B. $61$
• C. $41$
• D. $21$

Answer 1

The correct option is C $41$

Given differential equation,
$\frac{dy}{dx}y=\cot x(\sin x-y^2)$
Take $y^2=t\Rightarrow 2ydy=dt$
$\Rightarrow \frac{dt}{2dx}=\cot x(\sin x-t)$
$\Rightarrow \frac{dt}{dx}+2t\cot x=2\cos x$
Integrating Factor,$I=e^{\int 2\cot xdx}=e^{2ln\left(\sin x\right)}$
$\Rightarrow t{e}^{2ln\left(\sin x\right)}=\int 2\cos x{e}^{2ln\left(\sin x\right)}dx$
$\Rightarrow t{e}^{2ln\left(\sin x\right)}=\frac{2{sin}^{3}x}{3}+C$
$\Rightarrow y^2{sin}^{2}x=\frac{2{sin}^{3}x}{3}+C$
Given $y\left(\pi /2\right)=2\Rightarrow C=\frac{10}{3}$
$3{\left(y\right(\pi /6\left)\right)}^2=41$