# Explanation of proof nedeed: Why is $y'=c \cdot y$ always a exponential growth/decay function?

Good evening,

I'm struggling with understanding a proof:

I know, that a solution of $$y'=c \cdot y$$ is $$y=a \cdot e^{ct}$$ and it's clear how to calculate this.

I want to proof, that all solutions of a function describing a change of population, that is proprotional to the population, like $$y'=c \cdot y$$ is a function of exponential growth (or decay.

I found a proof which I'm not able to understand:

Let g be an other solution, whilst g is not describing an exponential growth or decay. We show $$(\frac{g}{e^{ct}})'=0$$ It's fine to me how they show it's zero. But where does $$(\frac{g}{e^{ct}})'$$ come from and why are they using it here, what does it mean?

The whole proof can be find here, but it's in German. http://www.mathe-macht-spass.de/download/Arbeitsblatt_BeweisExpDiffgl.pdf

(sorry for my english, still improving)

Thanks!

It's fine to me how they show it's zero. But where does $$(\frac{g}{e^{ct}})'$$ come from and why are they using it here, what does it mean?

I'm not sure what it $$g(t)/e^{ct}$$ means in and of itself. But it a means to an end. (excuse my play on words there).

Our goal is to show that a function satisfying a certain differential equation must take a certain form. The principal equation of this form, and its solution, is the fact that a function whose derivative is zero (on an interval) must be a constant (on that interval).

So we cook up a related function which, if it were constant, would tell us the function we are originally looking at must have the form we're claiming it to have. Then we take the derivative of the related function and show it is zero.

Because we want to show $$g(t) = ae^{ct}$$ for some $$a$$, we construct the quotient $$h(t) = g(t)/e^{ct}$$. This function is constant if and only if $$g$$ has the form we claim it does. Then we prove that $$h$$ is constant by showing $$h'(t) = 0$$.

This is not obvious to anyone seeing it for the first time, but now you know the technique: find something you want to be constant, and prove that it must be constant by showing its derivative is zero.

• Thanks! The last two passages made it very clear to me, to be honested in the two calculus lectures I've heard in university, I never had to used this type of proof. But this works out now, once I got the idea. Thanks so much! (and no worries about the word play ;D) – Adrian Schumacher Mar 2 at 19:44

Let's look for a solution in the form of $$y(t)=x(t)\exp(ct)$$ Let's assume that it satisfies the differential equation. But it means that: $$x'(t)\exp(ct)+x(t)c\exp(ct)=cx(t)\exp(ct)$$ $$x'(t)\exp(ct)=0$$ But the $$\exp$$ function is always positive, so we must have that $$x'(t)=0$$ So $$x$$ is a constant function (because of the MVT).

The solution in the pdf is quite similar, but I will present it in a little but different way:
Let's assume that $$g(t)$$ is another solution to the differential equation and it's not exponential. Now let's define a new function $$f$$: $$f(t):=\frac{g(t)}{a \exp(ct)}=\frac{1}{a} g(t) \exp(-ct)$$ Now let's differentiate $$f$$: $$f'(t)=\frac{1}{a}\left(g'(t)\exp(-ct)-cg(t)\exp(-ct)\right)$$ We know that $$g$$ satisfies the differential equation: it means that $$g'(t)=c g(t)$$. Let's use it: $$f'(t)=\frac{1}{a}\left(c g(t)\exp(-ct)-c g(t)\exp(-ct)\right)=0$$ This means that $$f$$ is constant, so $$g$$ is exponential. But this is a contradiction.

• That x must be constant is absolutely clear to me. What I just don't get is, why they're starting with $\frac{g}{\exp{(ct)}}$, when I'm assuming that g is a solution of the differential equation and g is not describing exponential growth/decay. Why $x(t)$ must be constant in your proof is obvious to me. But just don't why it must be $x(t)\exp(ct)$. Because I want to proof it will always be a fucntion including $\exp(ct)$ Maybe I just don't get the point. – Adrian Schumacher Mar 2 at 19:36
• @AdrianSchumacher I extended my solution. I used the product rule instead of the quotient rule, because it's simpler. – Botond Mar 2 at 19:42
• @AdrianSchumacher It don't need to be $x(t)\exp(ct)$. You can try it with, for example $x(t)\sin(t^2)\log(t^4+2)\exp(ct)$ as well. But in that case you will get that $x(t)\sin(t^2)\log(t^4+2)$ must be a constant. – Botond Mar 2 at 19:44
• thanks @Botond Now I got it after the edit you made! – Adrian Schumacher Mar 2 at 19:47

The motivation behind doing $$\frac{g(t)}{e^x}$$ is because we think that for any $$g(t)$$ that satisfies $$g(t)=g'(t)$$, $$g(t)$$ must be of the form $$ce^t$$, but we're not sure. So if we are correctly thinking that $$g(t)=ce^t$$, then $$\frac{g(t)}{e^x}$$ must be constant. If it's not, then $$\frac{g(t)}{e^x}$$ will not be constant. This way, we have a foolproof test for determining whether or not $$g(t)=ce^t$$ is true:

$$g(t)=ce^t$$ if and only if $$\frac{g(t)}{e^x}$$ is constant

And we prove that $$\frac{g(t)}{e^x}$$ is constant by finding the derivative $$=0$$.

This method of proving that some function must be some other function because they both satisfy some differential equation (in our case $$y=y'$$) is actually quite common, and can be used to prove that some Taylor series equals some function.

As an endnote: we can be sure that there is only one function that satisfies the differential equation $$y=y'$$ because of the Existence Uniqueness Theorem: https://faculty.math.illinois.edu/~tyson/existence.pdf