Question
Solve the following initial value problems:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=0,\text{y}(2)=\text{x}$

Answer

$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=0,\text{y}(2)=\text{x}$
It is a homogeneous equation. put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So, $\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{\text{x}}-\sin\Big(\frac{\text{vx}}{\text{x}}\Big)$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\sin\text{v}$
$\frac{\text{dv}}{\sin\text{v}}=-\frac{\text{dx}}{\text{x}}$
$\text{cosec(v)dv}=-\frac{\text{dx}}{\text{x}}$
integrating both sides we get,
$\log(\text{cosec(v)}-\cot(\text{v}))=-\log\text{x}+\log\text{c}$
$\log\Big(\text{cosec}\Big(\frac{\text{y}}{\text{x}}\Big)-\cot\Big(\frac{\text{y}}{\text{x}}\Big)\Big)=-\log\text{x}+\log\text{c}$
Putting the values $\text{x}=2$ and $\text{y}=\pi$
$\log\Big(\text{cosec}\Big(\frac{\pi}{2}\Big)-\cot\Big(\frac{\pi}{2}\Big)\Big)=-\log2+\log\text{C}$
$\text{C}=0$
$\log\Big(\text{cosec}\Big(\frac{\text{y}}{\text{x}}\Big)-\cot\Big(\frac{\text{y}}{\text{x}}\Big)\Big)=-\log\text{x}$

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