Differential Equation?

Question

I have a differential equation: $$\frac{dy}{dx}=\frac{y^2}{\sqrt{1-x}}\qquad y(0)=0$$ When I separate it out, I got $$-\frac1y=-2\sqrt{1-x}+C$$ My question is, since 0 cannot divide a number, is it possible for there to be no solution? Or is there a different way to solve for the equation?

Answer 1

Since the RHS of the equation is Lipschitz continuous at $y=0$ and $t=0$, the Picard-Lindelöf theorem asserts that the solution corresponding to this initial condition exists and is unique. See that the trivial solution $y(x)=0$ satisfies both the equation and the initial conditions. Therefore, that is the solution to this IVP, i.e., its only solution is the trivial solution.

If you proceed with your derivation, you will get $$ y = \frac{1}{C-2\sqrt{1-x}}, $$ applying the initial condition, $$ 0 = \frac{1}{C-2}, $$ which has no solution for $C$, reflecting the fact that the only solution is the trivial and we can't evaluate the solution that way.