Question

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

$\frac{dy}{dx}-\frac{y}{x}+cosec\left(\frac{y}{x}\right)=0,y=0whenx=1.$

Answer 1

Given, $\frac{dy}{dx}=\frac{y}{x}-cosec\left(\frac{y}{x}\right)$ ...(i)
Thus, the given differential equation is homogeneous.
So, put $\frac{y}{x}=vi.e.,y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Then, Eq.(i) becomes, $v+x\frac{dv}{dx}=v-cosecv\Rightarrow x\frac{dv}{dx}=-cosecv\Rightarrow sinvdv=-\frac{1}{x}dx$
On integarting both sides, we get
$\int sinvdv=-\int \frac{dx}{x}\Rightarrow -cosv=-log|x|+C\Rightarrow log|x|-cos\left(\frac{y}{x}\right)=C$ ...(iii)
when x=1, then y=0, therefore $log1-cos0=C\Rightarrow C=0-1=-1$
On putting the value of C in Eq. (ii), we get $log|x|-cos\left(\frac{y}{x}\right)=-1$
$\Rightarrow log|x|+1=cos\frac{y}{x}\Rightarrow log|x|+loge=cos\left(\frac{y}{x}\right)$ $[\because loge=1]$
$log|ex|=cos\left(\frac{y}{x}\right)$
This is the required solution of the given diffrential equation.