Question

$\ast \frac{dy}{dx}-\frac{y}{x}+\cos ec\left(\frac{y}{x}\right)=0;y=0whenx=1.$

Answer 1

Given

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

Let y = vx

$\Rightarrow \frac{dy}{dx}=v+\frac{xdv}{dx}$

$\therefore$ Eq (1) changes to $v+\frac{xdv}{dx}=\frac{vx}{x}+cosec\left(\frac{vx}{x}\right)$

$v+x\frac{dv}{dx}-v-cosec\left(v\right)=0$

$\Rightarrow \frac{dv}{cosecv}=\frac{dx}{x}$

Now intergrating $\int sinvdv=\int \frac{1}{x}dx$

$\Rightarrow -cosv=logx+c$

$\Rightarrow -cos\left(\frac{y}{x}\right)=logx+c$

Given at $x=1y=0$

$\therefore -1=c$

$\therefore$ over solution is $-cos\left(\frac{y}{x}\right)=logx-1$

$1-cos\left(\frac{y}{x}\right)=logx$