# Existence and Uniqueness Theorems for First-Order ODE's

There is a theorem regarding the existence and uniqueness of solutions to first-order ODE's:

Thm.: Consider the initial value problem $$y'=f(t,y)$$, $$y(t_0)=y_0$$. Suppose $$f$$ and $$\frac{∂f}{∂y}$$ are continuous on some open rectangle $$(t, y)∈(a, b)×(c, d)$$ containing the point $$(t_0, y_0)$$. Then in some interval $$t_0-h, t_0+h)⊆(a, b)$$, there exists a unique solution $$y = g(t)$$ that satisfies the initial value problem.

Now if we consider this example: $$t^2 y'+2ty-y^3=0$$, where $$t>0$$, the general solution is $$y = ±\sqrt{\frac{5t}{2+Ct^5}}$$. But $$y=0$$ is also a solution and I am wondering whether the above theorem can help me find such solutions which are not contained in the family of general solutions.

So, dividing the whole equation by $$t^2$$ gives $$y'=\frac{y^3}{t^2}-\frac{2y}{t}$$. Let $$f(t,y)=\frac{y^3}{t^2}-\frac{2y}{t}$$, then $$\frac{∂f}{∂y} = \frac{3y^2}{t^2}-\frac{2}{t}$$. As long as $$t≠0$$, then $$f$$ and $$\frac{∂f}{∂y}$$ are continuous, so I don't understand why we cannot identify anything wrong with $$y=0$$ using this theorem. To my understanding, since the solution $$y=0$$ is not contained in the general solutions, it should violate the conditions stated in the theorem, so I am quite confused. Thank you very much for answering.

• @Robert Israel Thank you for your answer. In general, is there any way we can find such particular solutions (which are not covered by the general solution) without guessing? If they exist, are they always asymptotes? Or are they always the result of some parameters going to infinity? – John Lei Sep 13 '17 at 3:14

It often happens that there are particular solutions that are not covered by the general solution, but may be a limit of the general solution as some parameter goes to $+\infty$ or $-\infty$. In this case taking $C \to \infty$ gives you $y=0$.
There is no violation of the Existence and Uniqueness Theorem here. If $t_0 > 0$ and $y_0 > 0$, there is a unique $C$ such that $y_0 = \sqrt{\frac{5 t_0}{2+C t_0^5}}$, and the unique solution with this initial condition is $y = \sqrt{\frac{5 t}{2+C t^5}}$ with that $C$. Similarly for $y_0 < 0$ with $y_0 = -\sqrt{\frac{5 t_0}{2+C t_0^5}}$. And for $y_0 = 0$, the unique solution is $y = 0$.