Solution to $y'=y^2, y(1)=0$

Question

Solution to $y'=y^2$ is $y = \frac{1}{C-x}$.

If we have extra condition that $y(1)=0$ than we get $0 = \frac{1}{C-1}$. C then has no solutions. My text book says that the solution to this is $y=0$, but why?

score 3 · Answer 1 · answered May 05 '23 at 17:20

Well, just plug $y=0$ into the equation. Does it solve it? Yes. Does it satisfy the initial condition? Also yes. That's it, that's a solution, and that's the only solution because you have a general uniqueness theorem that guarantees that.

Forget about your general formula. These formulas very often have exceptions. In this case, for example, you could explain the solution $y=0$ as a limiting case of your general formula, corresponding to letting $C\to \infty$.

score 3 · Accepted Answer · answered May 05 '23 at 18:01

Whenever you do something that's not always legal (like dividing by an expression that may be zero), you should make sure to note those exceptions and handle them separately.

In this case, you divide by $y$, which is only legal when $y \neq 0$. You move forward with your general solution method, but eventually hit a wall, and you don't get any solution out of it. At best, you've shown that there doesn't exist any nonzero solution, but since you excluded $y=0$ from the very beginning, none of the proceeding work has anything to say about that case.

So, you check the $y=0$ case separately, and you see that it does solve the equation and satisfy the initial value, so it a solution.

What you've done is the equivalent of solving $x^3 = -x$ over the reals by dividing by $x$ and then arriving at the (incorrect) conclusion that there are no solutions.

Exactly: dividing by zero. This is an example of a mistake that occurs when you divide by zero while separating variables in a ODE. It is subtle, I found it on a good textbook as a student. (This is why in general I recommend to not trust general formulas too much, when studying nonlinear differential equations). — Giuseppe Negro, May 06 '23 at 17:13
@GiuseppeNegro That's a great example. Many elementary texts fail to consider these cases when making examples. One text I know that covers these kinds of examples (where Lipschitz fails so uniqueness fails) is Grimshaw; in the very first chapter, the example $x' = |x|^{2/3}$ is given. — Alex Jones, May 06 '23 at 20:32

score -3 · Answer 3 · answered May 05 '23 at 17:27

$y^{'} = y^{2} \\ \frac{dy}{dx} = y^{2} \\ dy\cdot y^{-2} = dx \\ \text{integrating both sides} \\ \frac{y^{-1}}{-1} = x + C \\ \frac{-1}{y} = x + C \\ \text{the solution for the above differential equation satisfies (1,0)} \\ \frac{-0}{1} = 1 + C \\ C = -1 \\ \frac{-1}{y} = x - 1 \\ y = \frac{1}{1-x} \\ \text{this is the "General Solution" we took out} \\ \text{but there is also another solution y = 0, meaning} \frac{dy}{dx} = 0 \text{ and it even gets satisfied in the differential equation given, these types of solution are known as "Particular Solution" and can only be known by observation/experience in this chapter.} $

Solution to $y'=y^2, y(1)=0$

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