Question

The solution of the differential equation $\frac{dy}{dx}=\frac{\sin y+x}{\sin 2y-x\cos y}$ is:
(where $c$ is constant of integration)

• A. ${\sin }^{2}y=x\sin y+\frac{x^2}{2}+c$
• B. ${\sin }^{2}y=x\sin y-\frac{x^2}{2}+c$
• C. ${\sin }^{2}y=x+\sin y+\frac{x^2}{2}+c$
• D. ${\sin }^{2}y=x-\sin y+\frac{x^2}{2}+c$

Answer 1

The correct option is A ${\sin }^{2}y=x\sin y+\frac{x^2}{2}+c$

Here, $\frac{dy}{dx}=\frac{\sin y+x}{\sin 2y-x\cos y}$
$\Rightarrow \cos y\frac{dy}{dx}=\frac{\sin y+x}{2\sin y-x}$
Put $\sin y=t$
$\Rightarrow \frac{dt}{dx}=\frac{t+x}{2t-x}\cdots \left(i\right)$
Let $t=vx$
$\Rightarrow \frac{dt}{dx}=x\frac{dv}{dx}+v\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$ we have
$\frac{xdv}{dx}+v=\frac{vx+x}{2vx-x}=\frac{v+1}{2v-1}$
$\therefore x\frac{dv}{dx}=\frac{v+1}{2v-1}-v=\frac{v+1-2v^2+v}{2v-1}$
$\Rightarrow \frac{2v-1}{-2v^2+2v+1}dv=\frac{dx}{x}$
$\Rightarrow \frac{1}{2}\left(\frac{2v-1}{v^2-v-\frac{1}{2}}\right)dv=-\frac{dx}{x}$
Integrating both sides we have:
$\frac{1}{2}\ln |v^2-v-\frac{1}{2}|+\ln |x|=\ln |c_1|$
$\ln |x\cdot (v^2-v-\frac{1}{2}{)}^{0.5}|=\ln |c_1|$

On solving, squaring and substituting back the initial variables $v=\frac{t}{x}=\frac{\sin y}{x},\text{we get}$
$x^2.(\frac{{\sin }^{2}y}{x^2}-\frac{\sin y}{x}-\frac{1}{2})=c$
Here, $c_1^2=c$
${\sin }^{2}y=x\sin y+\frac{x^2}{2}+c$