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Solve for particular solution. dy/dx - y/x + cosec ( y/x) = 0; y = 0 when x = 1

Solve for particular solution.

    Q14.    \frac{dy}{dx} - \frac{y}{x} + \textup{cosec}\left (\frac{y}{x} \right ) = 0;\ y = 0 \ \textup{when}\ x = 1

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M manish

\frac{dy}{dx} = \frac{y}{x} -cosec(y/x) =F(x,y)....................................(i)

the above eq is homogeneous. So,
Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}
                                 
Substitute this value in equation (i)

\\=v+x\frac{dv}{dx}=v- cosec\ v\\ =x\frac{dv}{dx} = -cosec\ v\\ =-\frac{dv}{cosec\ v}= \frac{dx}{x}\\ =-\sin v dv = \frac{dx}{x}

on integrating both sides, we get;

\\=cos\ v = \log x +\log C =\log Cx\\ =\cos(y/x)= \log Cx.................................(ii)

now y = 0 and x =1 , we get 

C =e^{1}

put the value of C in eq 2

\cos(y/x)=\log \left | ex \right |

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