Solving separable ODEs Consider the separable IVP
$$\frac{dx}{dt}=f(x)g(t)\;\;\;\text{ with }x(0)=x_0$$
Suppose that functions $f,g,$ and $f'$ are all continuous.
We can find the particular solution using the formula
$$\int^{x(t)}_{x_0}\frac{1}{f(x)}dx=\int^t_0g(t)dt \tag{*}$$
However, this formula is not quite correct. The issue is we divide by $f(x)$, but it is possible that $f(x)=0$ for some $x$. In particular, suppose that $f(x^*)=0$ where $x^*$ is some real number, and suppose that $f(x)\neq0$ for any $x\neq x^*$.

*

*The starred formula is not valid if $x_0=x^*$. In this case, show that the constant function $x(t)=x^*$ is a solution to the IVP.

*The starred formula is also invalid if $x(t)=x^*$ for any $t$. Assume that $x_0\neq x^*$. Explain why the solution $x(t)\neq x^*$ for any time $t$. (Hint: this relies on the existence and uniqueness theorems).

*Suppose that $f$ is continuous but $f'$ is not. Show that the constant function $x(t)=x^*$ is still a solution to the IVP when $x_0=x^*$. Is the formula guaranteed to produce the correct answer when $x_0\neq x^*$?

My Attempt

*

*We will verify this by differentiating $x(t)=x^*$
$$\frac{d}{dt}(x)=\frac{d}{dt}(x^*)=0.$$
We are given that $f(x^*)=0$ so we know that the ODE is
$$\frac{dx}{dt}=f(x^*)g(t)=0\cdot g(t)=0.$$ Thus, the constant function is a a solution to the IVP.

But isn't this true for all $t$? not just for the initial condition? How do I use the initial condition in this case?


*We know that $f(x^*)=0$ just like in the first part. I am confused about why $x(t)$ is not a constant function.


*For this part, why does $f'$ not being continuous matter? We do not need to $f^1$ to evaluate the ODE.
 A: *

*The point is that if $x \equiv x^*$ then $x'=0$ from direct differentiation and also $f(x) g(t)=0$, so a constant function equal to a zero of $f$ is a solution to the ODE. For it to be a solution to the IVP then you need $x_0=x^*$.

*This needs a hypothesis that allows us to ensure that the solution is unique, so that one cannot have a constant solution and a non-constant solution to the same IVP (which would happen if any non-constant solution ever hit a zero of $f$, since then you could consider the IVP initialized when the non-constant solution hit the zero of $f$ and get two solutions). Here that hypothesis is that $f'$ is continuous. (As an aside, this particular hypothesis is not necessary for uniqueness, but merely $f$ and $g$ being continuous is not sufficient for uniqueness. So some additional hypothesis is required, this is just what they chose to use to write this problem.)

*Basically this is asking you to find a separable equation with non-unique solution to an IVP with $x(0)=x_0$ and $f(x_0)=0$. Hopefully you have already seen some examples of this otherwise it will be a bit challenging to make one up from scratch.

