$\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 ; y=0$ when $x=1$
$\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0$
$\Rightarrow \frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)$ ...(1)
Let $F(x, y)=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)$.
$\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{\lambda x}-\operatorname{cosec}\left(\frac{\lambda y}{\lambda x}\right)$
$\Rightarrow F(\lambda x, \lambda y)=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)=F(x, y)=\lambda^{0} \cdot F(x, y)$
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$
$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:
$v+x \frac{d v}{d x}=v-\operatorname{cosec} v$
$\Rightarrow-\frac{d v}{\operatorname{cosec} v}=-\frac{d x}{x}$
$\Rightarrow-\sin v d v=\frac{d x}{x}$
Integrating both sides, we get:
$\cos v=\log x+\log \mathrm{C}=\log |\mathrm{C} x|$
$\Rightarrow \cos \left(\frac{y}{x}\right)=\log |\mathrm{C} x|$ ...(2)
This is the required solution of the given differential equation.
Now, y = 0 at x = 1.
$\Rightarrow \cos (0)=\log \mathrm{C}$
$\Rightarrow 1=\log \mathrm{C}$
$\Rightarrow \mathrm{C}=e^{1}=e$
Substituting C = e in equation (2), we get:
$\cos \left(\frac{y}{x}\right)=\log |(e x)|$
This is the required solution of the given differential equation.
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