Let f(x)=_0^x e^tf(t),dt+e^x be differentiable for all x.

Question

Let $f(x)=\int_0^x e^tf(t)\,dt+e^x$ be differentiable for all $x$. Then $f(x)$ equals:

Accepted Answer

$f'(x)=e^xf(x)+e^x \implies \dfrac{f'(x)}{f(x)+1}=e^x \implies \ln(f(x)+1)=e^x+C$.

At $x=0$: $f(0)=1$ (from the equation), so $\ln2=1+C \Rightarrow C=\ln2-1$.

$$f(x)+1 = 2e^{e^x-1} \implies f(x) = 2e^{e^x-1}-1.$$

Solutions · 1