Question

Show that the differential equation $\left[\begin{array}{c}xsi{n}^2\left(\begin{array}{c}\frac{y}{x}\end{array}\right)-y\end{array}\right]dx+xdy=0$ is homogeneous.
Find the particular solution of this differential equation, given that $y=\frac{\pi }{4}$ when $x=1$.

Answer 1

We have $[x{\sin }^2\left(\frac{y}{x}\right)-y]dx+xdy=0$

$=\frac{dy}{dx}=-\frac{1}{x}[x{\sin }^2\left(\frac{y}{x}\right)-y]=f(x,y)$ ---- (1)

Put $x=kx,y=ky$ in equation (1),
$f(kx,ky)=(-\frac{1}{kx}[kx{\sin }^2\left(\frac{ky}{kx}\right)-ky])=k^0(-\frac{1}{x}\left[x{\sin }^2(\frac{y}{x}-y)\right])=k^{o}f(x,y)$

So it is clear that the given equation is homogenous differential equtaion of degree 0.

Now, let $y=vx$ in (1),then

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Therefore, $v+x\frac{dv}{dx}=-\frac{1}{x}\left[x{\sin }^2(\frac{vx}{x}-vx)\right]\Rightarrow v+x\frac{dv}{dx}=-{\sin }^{2}v+v$

$\Rightarrow -\int {\text{cosec}}^{2}vdv=\int \frac{dx}{x}$
$\Rightarrow \cot v=\log \left|x\right|+c$
$\Rightarrow \cot \left(\frac{y}{x}\right)=\log \left|x\right|+C$

Since given that $y=\frac{\pi }{4}$ when x = 1, so, $\cot \frac{\pi }{4}=\log \left|x\right|+C$
$\Rightarrow c=1$

The required solution is $\cot \left(\frac{y}{x}\right)=\log \left|x\right|+1$