no of solutions of the initial value problem? $x \dfrac{dy}{dx} = y , y (0) = 0, x \geq 0  .$
My Approach :
$\dfrac{dy}{y} = \dfrac{dx}{x},$ by variable separable method, we get
$lny = ln x +c $ and then raising e to both sides will get $ y=x+c.$  Then substituting the intial condtions, we get $c=-1$
so $y = x -1 $is the required solution.
But it's given that it has uncountable number of solution. I don't understand it.

$$x\cdot\frac{\text{d}y(x)}{\text{d}x}=y(x)\Longleftrightarrow$$
$$xy'(x)=y(x)\Longleftrightarrow$$
$$y'(x)=\frac{y(x)}{x}\Longleftrightarrow$$
$$\frac{y'(x)}{y(x)}=\frac{1}{x}\Longleftrightarrow$$
$$\int\frac{y'(x)}{y(x)}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$
$$\ln|y(x)|=\ln|x|+\text{C}\Longleftrightarrow$$
$$y(x)=e^{\ln|x|+\text{C}}\Longleftrightarrow$$
$$y(x)=\text{C}e^{\ln|x|}\Longleftrightarrow$$
$$y(x)=\text{C}x$$
The given initial condition is not sufficient to solve for the
unknown constant, because:
If, $y(0)=0$:
$$0=\text{C}\cdot0\Longleftrightarrow \text{C}=\frac{0}{0}$$
And we can't divide by $0$.
