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Hello I can't resolving this exercise:

Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ that satisfies this condition:

\begin{equation} f(t) = \int \limits_{0}^{t} f(s) ds \end{equation}

My try:

Muy first step I derivate the equation and obtain:

$f'(t) = f(t)$

Please help me with the next step.

Sorry for my grammar, actually I learn English

Also I need suggestions about how learn differential equations.

Thanks

copper.hat
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Baro
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    do you know a function equal to its derivative? – Sine of the Time Aug 02 '23 at 15:30
  • No, I am not. I think is by the fundamental theorem of calculus – Baro Aug 02 '23 at 15:33
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    You don't need differential equations to solve this. Much simpler if you do, of course. – copper.hat Aug 02 '23 at 15:38
  • Solve the differential equation and apply the boundary condition $y(0)=0$. (This boundary condition is obtained when you let $x=0$ in the original equation.) – Gonçalo Aug 02 '23 at 15:53
  • You have that $f(0)=0$. Assume $m\geq0$ is the maximum of all $x$ such that $f(x)=0$. Then there is $1>\delta>0$ such that $|f(x)|<1/2$ for all $x\in[m,m+\delta]$. For those $x$ and using triangle inequality we get $|f(x)|=|\int_{0}^{x}f(t)dt|\leq \delta/2$. This is a new bound on $f(x)$ better than delta. We can apply again and again, ..., the same idea with $\delta/2$ instead of $\delta$, to get $|f(x)|\leq \delta^2/2, \delta^3/2,...$. So $f(x)=0$ beyond $m$. Contradiction. Therefore $m$ doesn't exist and $f(x)=0$ for all $x\geq0$. For $x<0$ it is the same business. – NDB Aug 02 '23 at 16:02
  • $f(x)=0$ is its own derivative too as $e^x$ @SineoftheTime – user577215664 Aug 02 '23 at 22:29

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