Show that $ f(x) = A.exp(2x) $ if $ f'(x) = 2f(x) $

Question

Show that $ f(x) = A.exp(2x) $

if

$ f'(x) = 2f(x) $ for some $ A \in \mathbb{R} $

Is it sufficient to say that if the derivative of a function contains itself, then it must be the exponential function. Is there maybe a proof to show this?

Thanks for any help.

Answer 1

If you have this differential equation:

$$f'(x)=2f(x)$$

You can separate the variables (divide both sides by $f(x)$) to get:

$$\frac{f'(x)}{f(x)}=2$$

Notice that in this case you "ignore" the solution $f(x)=0$ because your left hand side would otherwise be undefined. You need to remember that $f(x)=0$ would also be a solution to this equation. You can integrate both sides with respect to $x$:

$$\int \frac{f'(x)}{f(x)} dx=\int2dx $$

You can make a substitution: $f(x)=u$ and $f'(x)=du$

$$\int\frac{1}{u}du=2x+C\iff \ln(u)=2x+C \iff u=e^{2x+C}$$

Resubstitute $u=f(x)$

$$f(x)=e^{2x+C}$$

You can check this by plugging this result back into your initial differential equation.

$$(e^{2x+C})'=2(e^{2x+C})\iff 2e^{2x+C}=2e^{2x+C}\space \checkmark$$

Notice that: $$e^{2x+C}=e^{2x}e^C$$

You can rust "rename" your constant $e^C$ to $A\implies Ae^{2x}$

Answer 2

No, this is not enough. Obviously the object of the problem is to make you "discover" this point, probably because you have heard of logarithm before.

You have $f'(x)=2f(x)$

It means that $\forall x$ such that $f(x) \neq 0$, you have $\dfrac{f'(x)}{f(x)}=2$

But can you integrate $\dfrac{f'(x)}{f(x)}$?

Answer 3

observe that $$ \left(\frac{f(x)}{e^{2x}}\right)'=\left(\frac{f'(x)-2f(x)}{e^{2x}}\right)=0 $$ Hence the function $\frac{f(x)}{e^{2x}}$ is constant.