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Question:

$\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 ; y=0$ when $x=1$

Solution:

$\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)$        ...(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:

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.