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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?

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    $\begingroup$ The interval seems irrelevant, but in differentiating you "destroyed" the information $y(0)+y'(0)=0$ which you need to recover. $\endgroup$
    – Ian
    Commented Mar 3, 2018 at 13:44
  • $\begingroup$ I am getting infinitely many possible values of A and B (1 equation 2 variables). Is that consistent with the question? $\endgroup$
    – mathnoob123
    Commented Mar 3, 2018 at 13:56
  • $\begingroup$ $$\frac{-1-\sqrt5}{2}A=\frac{-1-\sqrt5}{2}B$$ You sure about this ?$$ You sure about this ? $\endgroup$
    – user577215664
    Commented Mar 3, 2018 at 14:25
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    $\begingroup$ Is the final solution all correct now? $\endgroup$
    – mathnoob123
    Commented Mar 3, 2018 at 14:31
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    $\begingroup$ Thank you very much @Isham $\endgroup$
    – mathnoob123
    Commented Mar 3, 2018 at 14:46

1 Answer 1

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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 $$y(x)=Ae^{\frac{-1+\sqrt5}{2}x}+Be^{\frac{-1-\sqrt5}{2}x}\quad \text{with $A=\frac{3-\sqrt{5}}{2}B$}.$$

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  • $\begingroup$ So basically the interval provides me the initial condition? $\endgroup$
    – mathnoob123
    Commented Mar 3, 2018 at 13:50
  • $\begingroup$ I am getting infinitely many possible values of A and B (1 equation 2 variables). Is that consistent with the question? $\endgroup$
    – mathnoob123
    Commented Mar 3, 2018 at 13:54
  • $\begingroup$ @mathnoob123 You are right! I added a reference to his comment. $\endgroup$
    – Robert Z
    Commented Mar 3, 2018 at 13:55
  • $\begingroup$ Yes there are infinite solutions all proportional to $e^{\frac{-1+\sqrt5}{2}x}-e^{\frac{-1-\sqrt5}{2}x}$. In fact you can see directy that if $y(x)$ is a solution then also $Cy(x)$ is a solution for any $C\in\mathbb{R}$. $\endgroup$
    – Robert Z
    Commented Mar 3, 2018 at 13:56
  • $\begingroup$ I am getting $\frac{\sqrt5}{2}A =\frac{-1+\sqrt5}{2}B$ $\endgroup$
    – mathnoob123
    Commented Mar 3, 2018 at 13:59

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