Find all f continuous functions that satisfy $f'(t) = f(t) + \int_{0}^{1} f(t) dt$

Question

I started this problem making its derivative and I obtained: $$ f''(t) = f'(t) $$ The thing here is that I don't know what's next or if I should try another way to solve this. Any idea/hint?

Answer 1

Answer for the question in the title: Let $c=\int_0^{1}f(t)dt$. Then the equation becomes $f'(t)=f(t)+c$. This can be written as $(e^{-t}f(t))'=e^{-t} (f'(t)-f(t))=ce^{-t}$ so we get $e^{-t}f(t)=-ce^{-t}+d$ where $d$ is a constant. Thus $f(t)=-c+de^{t}$ Now integrate this from $0$ to $1$ to see that $d=\frac {2c} {e-1}$. Conclusion: $f(t)=-c+\frac {2c} {e-1} e^{t}$ is solution for every real number $c$.

Answer 2

Like it was mentioned in the comments to the OP, by differentiation you get that $f'' - f' = 0$, a linear differential equation with general solution $$ f(t)=c_1 + c_2 e^t. $$

So, any function satisfying the proposed. relation is of this form. Conversely, if $f(t)=c_1+c_2 e^t$, then $$ f'(t)-f(t)-\int_0^1 f(t) dt =$$ $$ c_2 e^t -c_1- c_2 e^t -[c_1 t +c_2 e^t]_0^1 = c_2 e^t-c_1- c_2 e^t-c_1-c_2 e+c_2 = -2c_1 -c_2(e-1), $$

which is zero if $-2c_1-c_2(e-1)=0$, i.e. $ c_1 = \frac{1-e}{2} c_2$. finally, the solutions are of the form $$ \frac{1-e}{2} c_2 + c_2 e^t. $$