Find the number of unique solutions of the following nonlinear ODE $$y' = \frac{10}{3} x y^\frac25, \qquad y(0)=-1$$
My attempt
Via the method of separation of variables, I got
$$y = \left( x^2 - 1 \right)^\frac53$$
Via uniqueness theorem, the above solution is unique in $(-1,1)$. By branching it off at the end points of this interval with $y=0$, I could construct four distinct solutions. Can we somehow construct more solutions, specifically infinitely many solutions to this ODE?
You cannot use separation of variables to solve a second-order differential equation. As Moo noted in the comments to your post, you can differentiate your function twice, and verify that it does not satisfy the equation. Also, there is no uniqueness theorem that can apply to your equation, because you only provided one boundary condition, not two.
Let $z=y^{\frac15},$ hence $$y''(x)=(z^5)''(x)=\frac{10}3xz(x)^2=(5z^4z')'(x)=20z^3(x)z'(x)^2+5z(x)^4z''(x),$$ which is equivalent to $$5z(x)^2\left[z(x)^2z''(x)+4z(x)^3z'(x)^2-\frac23x\right]=0,$$ implying $z(x)=0$ or $$z(x)^2z''(x)+4z(x)^3z'(x)^2-\frac23x=0.$$ I do not believe this latter equation can be solved analytically.
Given the way you phrased the problem, and given how you went about mistakenly solving it, I assume you actually meant to write the differential equation as $$y'(x)=\frac{10}3xy(x)^{\frac25},$$ instead. Then having only one boundary condition is justified in this situation. This can be solved with separation of variables. Hence $y(x)=0$ or $$y(x)^{-\frac25}y'(x)=\frac{10}3x\implies\left[\frac53y^{\frac35}\right]'(x)=\left(\frac53x^2\right)'\implies\left[y^{\frac35}\right]'(x)=\left(x^2\right)'.$$ By Riemann integrating, you obtain $$y(t)^{\frac35}-y(0)^{\frac35}=t^2.$$ So in fact, you do obtain $y(t)=(t^2-1)^{\frac53},$ and this is the only solution everywhere in $\mathbb{R}.$