Question

Show that the differential equation $x\frac{dy}{dx}\sin \frac{y}{x}+x-y\sin \frac{y}{x}=0$ is homogeneous. Find the particular solution of this differential equation given that $x=1$ when $y=\frac{\pi }{2}$.

Answer 1

$x\frac{dy}{dx}\sin \frac{y}{x}=y\sin \frac{y}{x}-x$

$\frac{dy}{dx}=\frac{y}{x}-\frac{1}{\sin \frac{y}{x}}$

Let $F(x,y)=\frac{y}{x}-\frac{1}{\sin \frac{y}{x}}$

Therefore, $F(\lambda x,\lambda y)=\frac{\lambda y}{\lambda x}-\frac{1}{\sin \frac{\lambda y}{\lambda x}}$
$={\lambda }^o(\frac{y}{x}-\frac{1}{\sin \frac{y}{x}})$
$={\lambda }^{o}F(x,y)$

Given differential equation is homogeneous
Let $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now, $x(v+x\frac{dv}{dx}\sin v+x-vx\times \sin v)=0$
$vx\sin v+x^2\sin v\frac{dv}{dx}+x-vx\sin v=0$

$\int \sin vdv=-\int \frac{dx}{x}$

$-\cos v=-\log \left|x\right|+c$
$-\cos \left(\frac{y}{x}\right)=-\log \left|x\right|+c$

When $x=0$, $y=\frac{\pi }{2}$
$-\cos \left(\frac{\frac{\pi }{2}}{0}\right)=-\log \left|0\right|+c$
$\Rightarrow c=0$

$\cos \left(\frac{y}{x}\right)=\log \left|x\right|$
$x=e^{\cos \left(\frac{y}{x}\right)}$

Therefore, $\cos \left(\frac{y}{x}\right)=\log x$