Mathematics - Differential Equation Question with Solution | TestHub

The given equation can be written as $2y\sin x\frac{dy}{dx}+y^2\cos x=\sin 2x$.

This implies:

$$\frac{d}{dx}(y^2\sin x)=\sin 2x$$

Integrating both sides, we get:

$$y^2\sin x=-\frac{1}{2}\cos 2x+C$$

To find the constant $C$, we use the initial condition $y(\pi /2)$. Substituting $x=\pi /2$:

$$(y(\pi /2){)}^2\sin (\pi /2)=-\frac{1}{2}\cos (2\cdot \pi /2)+C$$

$$(y(\pi /2){)}^2\cdot 1=-\frac{1}{2}\cos (\pi )+C$$

$$(y(\pi /2){)}^2=-\frac{1}{2}(-1)+C$$

$$(y(\pi /2){)}^2=\frac{1}{2}+C$$

If we assume $y(\pi /2)=1$, then $1=\frac{1}{2}+C$, which gives $C=\frac{1}{2}$.

Therefore, the equation becomes:

$$y^2\sin x=\frac{1}{2}(1-\cos 2x)$$

Using the trigonometric identity $1-\cos 2x=2{\sin }^{2}x$:

$$y^2\sin x=\frac{1}{2}(2{\sin }^{2}x)$$

$$y^2\sin x={\sin }^{2}x$$

Dividing by $\sin x$ (assuming $\sin x\ne 0$):

$$y^2=\sin x$$