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Looking at Differential Equations, Dynamical Systems, and Linear Algebra, they introduce us to ODEs with this equation:

$$\frac{dx}{dt} = ax$$

This can be re-written as:

$$x'(t) = ax(t)$$

And the solution to this equation is:

$$f'(t) = aKe^{at} = af(t)$$

They claim that there are no other solutions, and they support this claim with the following equation. Let $u(t)$ be any solution and compute the derivative of $u(t) e^{-at}$:

$$\begin{align} \frac{d}{dt}(u(t) e^{-at}) &= u'(t) e^{-at} + u(t)(-ae^{-at}) \\ &= au(t)e^{-at} - au(t) e^{-at} =0\end{align}$$

I do not follow how they go from $u'(t) e^{-at}$ to $au(t)e^{-at}$

Also, why are they proving that there are no other solutions by taking the solution that we found and multiplying it by some function $u(t)$?

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The idea is that $u$ is not any kind of function, it is a solution to the differential equation, i.e. it satisfies $$u'(t)=au(t).$$ And that is why $u'(t)e^{-at} = au(t)e^{-at}$. Then they show that $$\frac{d}{dt}\left(u(t) e^{-at}\right)=0,$$ hence $u(t)e^{-at}=constant=:K$ (here $K$ just denotes this constant). Now by multiplying with $e^{at}$ you get $$u(t)=Ke^{at}$$ and since $u$ was an aribtrary solution, you obtain uniqueness.

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  • $\begingroup$ I am still perplexed. 1. Why do we differentiate a product of $u(t)$ and $e^{-at}$, i.e. how would it occur to us that to prove that $Ke^{at}$ is a unique solution, we need to differentiate those two things. 2. $u(t) e^{-at}$ it is constant because the deriv is zero, OK, but why is it equal to $K$. 3. Why do we multiply by $e^{at}$. Yeah... :) $\endgroup$ – i squared - Keep it Real Apr 11 at 21:41
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    $\begingroup$ @i squared - Keep it Real : How to get the idea: You want to show that $u(t)=constant\cdot e^{\alpha t}$, because that means that the solution is unique (up to multiplying with a constant) Then you multiply what you want to show with $e^{-\alpha t}$ and you have $u(t)e^{-\alpha t} = constant$. In other words if you can show that $u(t)e^{-\alpha t}$ is independend of $t$, i.e. a constant you are finished. Hence you derive it by $t$ to show it is zero. $\endgroup$ – humanStampedist Apr 12 at 10:08
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    $\begingroup$ @ i squared - Keep it Real : To your second point: $K$ is a generic constant. What uniqueness of the solution means in your case is uniqueness up to multiplying with a real constant. That means every $u$ that solves you differential equation is of the form $u(t)=C e^{\alpha t}$ and $C\in\mathbb{R}$ can be choosen any way you like. $\endgroup$ – humanStampedist Apr 12 at 10:11
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    $\begingroup$ Thank you. This is so neat. Although, I would ponder on this for days if I had to discover this myself. First, need to realize that something exponential will solve this. Second, need to think about showing that this is a unique solution. Third, need to come up with an idea that $u(t) e^{-\alpha t} = \text{constant}$, if it is unique. Fourth, need to think that this can be shown via a derivative using the ODE relation: $u'(t) = a u(t)$, which you actually need to realize is equivalent to $\frac{du}{dt} = au$. I love maths, but it probably doesn't love self-learners. $\endgroup$ – i squared - Keep it Real Apr 12 at 22:47

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