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.
Best Answer
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 $y = 0$.