Solving separable ODEs

Question

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 $x\neq x^*$.

1. 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.
2. 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).
3. 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

1. 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?

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

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

Answer 1

1. 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 $x_0=x^*$.
2. 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.)
3. 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.