Question

Solve the differential equation: $y\sin x\frac{dy}{dx}=\cos x(\sin x-y^2)$

• A. $y^2=\frac{2}{3}\sin x+\frac{c}{{\sin }^{2}x}$
• B. $y^2=\frac{-2}{3}\sin x+\frac{c}{{\sin }^{2}x}$
• C. $y^2=\frac{2}{3}\sin x-\frac{c}{{\sin }^{2}x}$
• D. $y^2=\frac{2}{3}\cos x+\frac{c}{{\sin }^{2}x}$

Answer 1

Answer: (A), (C)

The correct options are
A $y^2=\frac{2}{3}\sin x+\frac{c}{{\sin }^{2}x}$
C $y^2=\frac{2}{3}\sin x-\frac{c}{{\sin }^{2}x}$

Given differential eqn is
$y\frac{dy}{dx}\sin x=\cos x(\sin x-y^2)$

$y\frac{dy}{dx}+\left(\cot x\right)y^2=\cos x$
Put $y^2=z$
$\Rightarrow \frac{dz}{dx}=2y\frac{dy}{dx}$
$\Rightarrow \frac{1}{2}\frac{dz}{dx}+z\cot x=\cos x$
$\Rightarrow \frac{dz}{dx}+2z\cot x=2\cos x$
which is a linear differential eqn with z as dependent variable.
Here, $P=2\cot x,Q=2\cos x$
Integrating factor $I.F.=e^{\int Pdx}$
$=e^{\int 2\cot xdx}$
$I.F.=e^{2\log sinx}$
$\Rightarrow I.F.={\sin }^{2}x$
Solution of given differential eqn is
$z{\sin }^{2}x=\int 2\cos x{\sin }^{2}xdx$
Put $\sin x=t$
$\Rightarrow \cos xdx=dt$
$\Rightarrow z{\sin }^{2}x=2\frac{t^3}{3}+C$
$\Rightarrow y^2{\sin }^{2}x=\frac{2}{3}{\sin }^{3}x+C$
$\Rightarrow y^2=\frac{2}{3}\sin x+\frac{C}{{\sin }^{2}x}$
$C$ can be positive or negative. Hence, both options are correct