$f'(x)=af(x) \Rightarrow f(x)=e^{ax} f(0)$

Question

I've been working on the following exercise: Let $f: \mathbb {R} \longrightarrow \mathbb{R}$ be differentiable. Suppose, there is an $a \in \mathbb{R}$ such that $$f'(x)=a \cdot f(x) ~~~~(*)$$ for all $x \in \mathbb{R}$.

Show that $f(x)=e^{ax} f(0)$ for all $x \in \mathbb{R}$. (HINT: Multiply $(*)$ by $e^{-ax}$ und use the product rule.)

My problems: Of course, I see that this $f$ above fulfills condition $(*)$. However, I'm not able to show that it can only be a function of this type. My major problem is that I do not understand how I am supposed to apply this given hint. Another idea was to maybe work with the defintion of the differential quotient...

Thank you for your help!

Answer 1

Following the hint you have $$ f'(x) e^{-ax} - a f(x) e^{-ax} = 0 $$ for all $x \in \Bbb R$. According to the product rule, the left-hand side is the derivative of the function $$ f(x) e^{-ax} $$ which therefore must be constant, i.e. $f(x) e^{-ax} = f(0)$.

Answer 2

$\exp(-ax)f'(x) - \exp(-ax) a f(x)=0;$

$\dfrac{d}{dx} [f(x)\exp(-ax)] =0;$

$\displaystyle \int \dfrac{d}{dx} [f(x)\exp(-ax)] dx=C;$

$f(x)= C\exp(ax);$

$f(0)=C.$

Finally: $f(x)= f(0)\exp(ax).$