Slope field for differential equation, how many different solutions

Question

I am tasked to draw the slope field of the differential equation: $$x^5y'=y^2$$ and decide how many solutions it with initial condition $(x_0,y_0)=(a,b)$

Ive plotted the slope field of $y'= \frac{y^2}{x^5} $ using geogebra: https://www.geogebra.org/m/z9c3ndaf which gives me this plot:

My thinking is that $y(a)=b$ has no solutions if $a=0$ (as $x=0$ has no solution curve), one unique solution if $b=0$ and $a≠0$ (as there is only one solution curve which passes through (a,0) and has infinite solutions if $a≠0$ and $b≠0$ (though this i am not very certain about). Is my thinking correct?

Edit: I think im supposed to use the uniqueness and existence theorem, so i check when the DE and its partial derivate in respect to y is continuous. I then find that $y'= \frac{y^2}{x^5} $ and its partial derivative $\frac{2y}{x^5} $ are both continous as long as x≠0. According to the theorem, i think this means that the DE has one unique solution if the initial condition is $y(a)=0$ or $y(a)=b$ but has no solutions for $$y(0)=b$$ (where $a≠0, b≠0$)

Is my thinking correct?

Answer 1

You're on the right track.

Hint By analogy, consider the simpler equation $$x y'(x) = y(x)$$ that exhibits the same phenomena. Evaluating both sides at $x = 0$ gives that for any solution $y(x)$, $y(0) = (0) y'(0) = 0$; in particular there are no solutions satisfying an initial condition $y(0) = b$ for $b \neq 0$. On the other hand, separating variables gives $$\frac{dy}{y} = \frac{dx}{x},$$ and integrating gives that $$y(x) = C x,$$ is a solution for all $C \neq 0$, as is the limiting function $y(x) = \lim_{C \to 0} C x = 0$, and all of these solutions satisfy $y(0) = 0$.

Remark Notice that the higher-order initial conditions $y(0) = 0$, $y'(0) = C$ do determine distinct solutions for distinct $C$.

Similarly, evaluating $x^5 y'(x) = y(x)$ at $x = 0$ gives that $y(0) = 0$, so there no solutions satisfying an initial condition $y(0) = b$ for $b \neq 0$. On the other hand, separation of variables leads to the solutions $$y(x) = \frac{4 x^4}{C x^4 + 1}, \qquad C \in \Bbb R,$$ of the o.d.e, and computing gives that for all these solutions satisfy $y(0) = 0$, as does as the limiting solution $y(x) = \lim_{C \to \infty} \frac{4 x^4}{C x^4 + 1} = 0$.