no of solutions of the initial value problem?

Question

$x \dfrac{dy}{dx} = y , y (0) = 0, x \geq 0 .$

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.

Answer 1

$$x\cdot\frac{\text{d}y(x)}{\text{d}x}=y(x)\Longleftrightarrow$$ $$xy'(x)=y(x)\Longleftrightarrow$$ $$y'(x)=\frac{y(x)}{x}\Longleftrightarrow$$ $$\frac{y'(x)}{y(x)}=\frac{1}{x}\Longleftrightarrow$$ $$\int\frac{y'(x)}{y(x)}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$ $$\ln|y(x)|=\ln|x|+\text{C}\Longleftrightarrow$$ $$y(x)=e^{\ln|x|+\text{C}}\Longleftrightarrow$$ $$y(x)=\text{C}e^{\ln|x|}\Longleftrightarrow$$ $$y(x)=\text{C}x$$

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 $0$.

Answer 2

If you solve the differential equation correctly, then $$\ln |y|= \ln x + c = \ln e^c x,$$ and so $$y=\pm e^c x,$$ not $y=x+c$. Then $e^c \cdot 0 = 0$ for all constant $C$, given initial value doesn't give us a useful information.