Question

Solve: $\sin x\frac{dy}{dx}-y=\sin x\cdot \tan \frac{x}{2}$.

Answer 1

The given differential equations is:

$\sin x\frac{dy}{dx}-y=\sin x\cdot \tan \frac{x}{2}$ ... $\left(i\right)$

Divide the differential equation $\left(i\right)$ by $\sin x$

$\Rightarrow \frac{dy}{dx}-y\text{cosec}x=\tan \frac{x}{2}$

Now, it is a first order differential equation,

And solution is $u\left(x\right)y=\int u\left(x\right)\tan \frac{x}{2}dx+C$ ...... $\left(ii\right)$

Where $u\left(x\right)=e^{\int -\text{cosec}xdx}=e^{-(-\ln (\text{cosec}x+\cot x\left)\right)}=e^{\ln (\text{cosec}x+\cot x})=\text{cosec}x+\cot x$

But $\text{cosec}x+\cot x=\cot \frac{x}{2}$

$\therefore u\left(x\right)=\cot \frac{x}{2}$

From eq $\left(ii\right)$,

$\int u\left(x\right)\tan \frac{x}{2}dx=\int \left(\cot \frac{x}{2}\right)\tan \frac{x}{2}dx=\int 1dx=x$

Substitute these values in eq $\left(ii\right)$ we get,

$\cot \frac{x}{2}y=x+C$

Hence, $y=x\tan \frac{x}{2}+C\tan \frac{x}{2}$