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If $\frac{d}{dx} e^x = e^x$ and $f:R\rightarrow R$ is a differentiable function such that $f'(x)=af(x)$ with $a \in R$. Show that there exist a real number $c$ such that $$f(x)=c*exp(ax)$$

Let $g(x)= f(x)exp(-ax)$ then $$g'(x)=f'(x)exp(-ax)-af(x)exp(-ax)=(f'(x)-af(x))*exp(-ax)=0$$ we conclude that $g(x)=c$.

I dont understant where does $g(x)= f(x)exp(-ax)$ come from. I would like to know in general how to we proceed to solve this kind of question . Thank you.

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They defined $g(x) = f(x) e^{-ax}$, because they knew in advance that the solution is $f(x) = c e^{ax}$. Therefore, $g(x) = c e^{ax}e^{-ax} = c$. That's why $g(x) = c$.

In either case... it is just yet another way of proving that $f'(x) = a f(x) \implies f(x) = c e^{ax}$.

Your problem is what is called an Ordinary Differential Equation. Or simply ODE. There is no general technique to solve them. However, there are general techniques to solve certain types of ODEs. A good read is the wikipedia page about it. It has a nice summary of analytical solutions of certain types of ODE, and the method used. You can check out the method.


And about your problem: $f'(x) = c f(x)$, there is a more elegant way that does not come up with any crazy $g(x)$ in advance. We simply divide: $$ \frac{f'(x)}{f(x)} = a $$

Now you can remmember that: $$ \frac{d}{dx}\left[\ln f(x)\right] = \frac{f'(x)}{f(x)} $$

Therefore: $$ \frac{f'(x)}{f(x)} = \frac{d}{dx}\left[\ln f(x)\right] = a = \frac{d}{dx}\left[ax + b\right] \quad\implies\quad \frac{d}{dx}\left[\ln f(x)\right] = \frac{d}{dx}\left[ax + b\right] $$

Where $b\in\mathbb{R}$ is any constant. Therefore: $$ \ln f(x) = ax + b \quad\implies\quad f(x) = e^{ax+b} = e^b e^{ax} $$

Define $c = e^b$, and then we have it: $f(x) = c e^{ax}$.

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  • $\begingroup$ That is beautiful, thank you very much! $\endgroup$ – Elina Nov 12 '16 at 21:38

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