Question
Find all the solutions for the equation $$y'(x)+y(x)=\int_0^x y(t) dt$$ defined on $[0,1]$.
Attempt
I have calculated the complete solution as $$y(x)=Ae^{\frac{-1+\sqrt5}{2}x}+Be^{\frac{-1-\sqrt5}{2}x}$$ However, I cannot figure out what to do with the provided interval since the equation make sense for all values of $x$ in $(-\infty,\infty)$.
Can somebody explain what the question intends me to do with the interval?
Edit
As seen by answers below I have realised the interval acts provides the initial. Proceeding under the provided information, I got the answer as follows
$$y'(0)=-y(0)$$ $$y(0)= Ae^{0}+Be^{0}=A+B$$
$$y'(0)= \frac{-1+\sqrt5}{2}Ae^{0}+\frac{-1-\sqrt5}{2}Be^{0}= \frac{-1+\sqrt5}{2}A+\frac{-1-\sqrt5}{2}B$$
$$-\left( \frac{-1+\sqrt5}{2}A+\frac{-1-\sqrt5}{2}B \right)=A+B$$
$$\frac{1-\sqrt5}{2}A+\frac{1+\sqrt5}{2}B =A+B$$
$$\frac{-1-\sqrt5}{2}A=\frac{1-\sqrt5}{2}B $$ $$\frac{1+\sqrt5}{2}A=\frac{-1+\sqrt5}{2}B $$ $$A=\frac{-1+\sqrt5}{1+\sqrt5}B $$
Inputting this in the final equation
$$y(x)=\frac{-1+\sqrt5}{1+\sqrt5}Be^{\frac{-1+\sqrt5}{2}x}+Be^{\frac{-1-\sqrt5}{2}x} = B\left(\frac{-1+\sqrt5}{1+\sqrt5}e^{\frac{-1+\sqrt5}{2}x}+e^{\frac{-1-\sqrt5}{2}x} \right)$$
Is this correct?
Hint. Let $Y(x)=\int_0^x y(t) dt$, then $Y(0)=0$ (see Ian's comment!) and $$Y''(x)+Y'(x)=y'(x)+y(x)=\int_0^x y(t) dt=Y(x)$$ which is a linear differential equation whose general solution over $\mathbb{R}$ (not just on the interval $[0,1]$) is the one you have already found : $$Y(x)=ae^{\frac{-1+\sqrt5}{2}x}+be^{\frac{-1-\sqrt5}{2}x}$$ In order to have $Y(0)=0$ we need that $a=-b$ and $$y(x)=Y'(x)=a\frac{-1+\sqrt5}{2}e^{\frac{-1+\sqrt5}{2}x}+a\frac{1+\sqrt5}{2}e^{\frac{-1-\sqrt5}{2}x}$$ or $$y(x)=Ae^{\frac{-1+\sqrt5}{2}x}+Be^{\frac{-1-\sqrt5}{2}x}\quad \text{with $A=\frac{3-\sqrt{5}}{2}B$}.$$