Properties of First Order Linear ODE as it relates to the solution I am given the following image, and asked which of the following forms of ODE's could this solution match to (vertical asymptote at $t = 8$). $(1) \ y' + p(t)y=g(t)$ or $ (2) \ y' = f(y)$ or $ (3) \ y' = f(y)g(t)$.
From the graph, I can see that the slope varies with $t$ hence it can not be autonomous like in $(2)$. However, what justifications can I make to decide whether this can be a solution to $(1)$ or $(3)$?

 A: I will assume in my answer that $p$, $f$ and $g$ are continuous, otherwise the answer is trivial.
The form (1) is impossible because the solution of (1) is
$$
y(t)=e^{-{\int{p(t)\,dt}}}\left(\int g(t) e^{\int{p(t)\,dt}}\, dt + C\right).
$$
If $p(t)$ is continuous, then its antiderivative is also continuous, $e^{-{\int{p(t)\,dt}}}$ is continuous etc, thus, $y(t)$ must be continuous.
(3) is possible. It is easy to see that the function $y=\tan\frac{\pi\varphi(t)}{16}$, where $\varphi(t)$ is some differentiable function, is a solution to the equation
$$\tag{4}
y'=(1+y^2)\frac{\pi\varphi'(t)}{16}.
$$
Let us choose $\varphi(t)$ from these considerations:

*

*It oscillates;

*$\varphi(8)=8$;

*$\forall t\in[0,8)\; |\varphi(t)|<8$.

One of the suitable functions is
$$
\varphi(t)=t\cos\frac{5\pi t}4.
$$
Then the resulting equation (4) is
$$\tag{5}
y'=(1+y^2)\frac{\pi}{16}\left( \cos\frac{5\pi t}4 - \frac54t\pi\sin\frac{5\pi t}4 \right).
$$
The solution to the initial value problem for (5) with the initial condition $y(0)=0$ is $\tan\frac{\pi\varphi(t)}{16}$:

The form of the oscillations turned out to be not quite sinusoidal, as in the picture from the question, but, I think, this is not the key point.
