Question

For the given differential equation find the particular solution satisfying the given conditions.

$[xsi{n}^2\left(\frac{y}{x}\right)-y]dx+xdy=0,y=\frac{\pi }{4}whenx=1.$

Answer 1

Given, $si{n}^2\left(\frac{y}{x}\right)-\frac{y}{x}+\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{y}{x}-si{n}^2\left(\frac{y}{x}\right)$ ....(i)
Thus, the given differential equation is homogeneous.
So, put
$\frac{y}{x}=v$
i.e., $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Then, Eq. (i) becomes
$v+x\frac{dv}{dx}=v-si{n}^{2}v\Rightarrow x\frac{dv}{dx}=-si{n}^{2}v\Rightarrow cose{c}^{2}vdv=-\frac{1}{x}dx$
On integrating both sides, we get
$\int cose{c}^{2}vdv=-\int \frac{dx}{x}\Rightarrow -cotv=-log|x|+C\Rightarrow log|x|-cotv=C\Rightarrow log|x|-cot\left(\frac{y}{x}\right)=C$ $(Putv=\frac{y}{x})$ ...(ii)
When x=1, then $y=\frac{\pi }{4},$ therefore,
$log|1|-cot\frac{\pi }{4}=C\Rightarrow C=0-1=-1$
On putting the value of C in Eq. (ii), we get
$log|x|-cot\left(\frac{y}{x}\right)=-1\Rightarrow log|x|-cot\left(\frac{y}{x}\right)=-loge$ $(\because 1=loge)$
$\Rightarrow cot\left(\frac{y}{x}\right)=log|ex|$ $[\because logm+logn=logmn]$
This is the required solution of the given differential equation.