Trying to obtain the proof of this differential equation

Question

I am reading this doc.

At one point (page 5) we see these equations:

and then this at page 6

The solution of the above first order differential equation is well known and it is given by

$$f = A_1 e^{\ \alpha t}$$

I am to googling around for an hour trying to find the proof to that. No success.

Any ideas?

Answer 1

The characteristic equation of the second order ODE is given by

$$r^2+2\alpha r+\alpha^2=0$$

which yields to $r_{1,2}=-\alpha$ and so the general solution is

$$\nu_c(t)=(c_1+c_2t)e^{-\alpha t}$$

This proceeding is the normal way to solve a second order ODE with constant coefficients.

For the first order ODE you just have to rewrite it a little bit

$$\frac{df}{dt}+\alpha f=0\Leftrightarrow \frac{df}{dt}=-\alpha f\Leftrightarrow \frac{df}{f}=-\alpha dt$$

Integrating both sides yields to

$$\int\frac{df}{f}=-\int\alpha dt\Leftrightarrow \log f=-\alpha t+c\Leftrightarrow f=c~e^{-\alpha t}$$

Now by assuming that $\alpha$ is a not choosen constant you can absorp the minus sign within it and by setting $c=A_1$ we get

$$f=A_1e^{\alpha t}$$

This proceeding is known for solving first order ODE's aswell and it is called separation of variable.

Answer 2

$\frac{df}{dt} + \alpha f = 0 \rightarrow \frac{df}{dt} = -\alpha f $

So what function $f$ satisfies the condition where the derivative of the function equals the same function times a negative constant? The exponential.

$f(t) = Ae^{(-\alpha t)}$

$\frac{df}{dt} = -\alpha Ae^{(-\alpha t)} = -\alpha f$

Answer 3

To answer this question, and in the spirit of answering a question you had in a comment, I will show you how you might approach this problem as if you were the first mathematician to ever see it.

$$ f'(t) + \alpha f(t) = 0 $$

You see a derivative, so your first instinct might be to integrate:

$$ \int [f'(t) + \alpha f(t)] dt = \int f'(t)dt + \int \alpha f(t)dt = \int df + \alpha \int f(t) dx = f(t)+\alpha F(t) $$

We didn't really help ourselves much here, since we integrated an $f$ that wasn't a derivative. But, in the process, we notice that $\int f(t)dt = \int df$ made that first integral easier. After a long time of pondering, we might even think of the following fact: if all the $f$'s were in that one term, we could integrate it as-is, because it's just some kind of $\int g(f) df$.

Let's just pretend we could do that. So, we have some kind of equation that looks like

$$ g(f(t))f'(t) + h(t) = 0 $$

We can try to integrate this now:

$$ \int [g(f(t))f'(t) + h(t)] dt = \int g(f(t))f'(t)dt + \int h(t) dt = \int g(f)df+\int h(t)dt \\ = G(f) + H(t) $$

Since we integrated something that was $0$, we must have got a constant as the result, so

$$G(f(t))+H(t) = C \Rightarrow f(t) = G^{-1}(C-H(t))$$

whenever $g$ and $h$ are integrable and $G$ is invertible, so we've solved the equation! So we now know how an equation in the form $g(f(t))f'(t)+h(t) = 0$ can be solved, let's try to frame our original problem in this way.

First, note that $f(t)\equiv 0$ is an obvious solution, so let's assume that the solution we're looking for isn't that one, i.e. $f(t)\neq 0 $. Then, we can divide this whole equation by $f(t)$, and we get

$$ \frac{1}{f(t)}f'(t) + \alpha = 0 $$

If we let $g(f) = \frac{1}{f}$ and $h(t) \equiv \alpha$, then our equation has the desired form. Then, all that's left to do is integrate

$$ \int \left[ \frac{1}{f(t)}f'(t) + \alpha \right] dt = \int \frac{1}{f(t)}f'(t)dt + \alpha\int dt = \int \frac{1}{f} df + \alpha t = \log f + \alpha t $$

Again, we integrated something that was $0$, so it must be a constant, so there is some $C$ such that

$$ \log f(t) + \alpha t = C \Rightarrow f(t) = \exp(C-\alpha t) \Rightarrow f(t) = K\exp(-\alpha t) $$

where $K = \exp(C)$.

This technique got the name "separation of variables" for obvious reasons, since we had one term that had all the $f$'s and another separate term that had all the $t$'s. Unfortunately, not every first order equation can be solved like this, and higher order equations don't budge when we try this one on them. However, sometimes we can reduce higher order equations into first order equations that we can solve with this technique, so it's not too specific to not be worth knowing.