How to find $f$ which satisfies $f'(x)=f(x)+\int_{0}^{2}f(t)dt$ and $f(0)=\frac{4-e^2}{3}$?

Question

$$f'(x)=f(x)+\int_{0}^{2}f(t)dt \hspace{1cm} with \hspace{1cm} f(0)=\frac{4-e^2}{3}.$$How to find $f(x)$? I tried integrating on both sides but I had no idea next.

Answer 1

I would not integrate both sides but differentiate to get a pure differential equation. (What you have is an "integro-differential equation".) Since $\int_0^2 f(t)dt$, differentiating again gives $f''(x)= f'(x)$. Now integrating both sides gives $f'(x)= f(x)+ C$ where C is the "constant of integration". We can write that as $f'(x)- f(x)= C$. That is a linear, non-homogenous, differential equation with constant coefficients. It's characteristic equation is $r- 1= 0$ which has the single solution $r= 1$. The corresponding homogeneous equation, $f'(x)- f(x)= 0$ has general solution $f(x)= Ae^x$ where A can be any constant. Since the "non-homogeneous part" is the constant C, we look for constant special solution, f(x)= K. Then f'(x)= 0 so the equation becomes $0- K= C$ or $K= -C$. The general solution to the entire equation, $f'(x)- f(x)= C$, is $f(x)= Ae^x- C$.

But $C= \int_0^2 f(t)dt= \int_0^2 Ae^t- C dt= \left[Ae^t- Ct\right]_0^2= Ae^2- 2C- Ae^0= A(e^2- 1)- 2C$. So $3C= A(e^2- 1)$ and $f(x)= Ae^x- \frac{A(e^2- 1)}{3}$.

The value of A is determined from the condition that $f(0)= A- \frac{A(e^2- 1)}{3}= \frac{4- e^2}{3}$ so that $3A- A(e^2- 1)= (4-e^2)A= 4- e^2$ so A= 1.

$f(x)= e^x- \frac{e^2- 1}{3}$.

Answer 2

The integral is some constant independent of $x$, so call it $k$ for now. Your equation is then $f'(x)=f(x)+k$, which you should be able to solve by the usual differential equation techniques. The constant of integration will multiply $e^x$. Now do the integral and you will get an equation that lets you evaluate the constant of integration

Answer 3

Differentiating reveals $f''(x)=f'(x)\implies f'(x)=Ae^x\implies f(x) =Ae^x+B$.

Next, we set $x=0$ in the integral equation and find that

$$f'(0)=A=f(0)+\int_0^2 f(t)\,dt=\frac{4-e^2}{3}+A(e^2-1)+2B \tag1$$

And we also see that

$$f(0)=A+B=\frac{4-e^2}{3}\tag 2$$

Equations $(1)$ and $(2)$ comprise a linear system of two equations with two unknowns, the solutions of which yield $A$ and $B$.

Can you finish now?