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I try to solve the two integro differential equations

$f(at)=\frac{df(t)}{dt}$ and $f(at)=\int_{0}^{t}f(\tau)d\tau$ $a\gt0$.

Do you have an idea to suggest to me.

Thank you very much for your kind help.

Gonçalo
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user425269
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1 Answers1

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Hint: $$\Longrightarrow \frac{df(t)}{dt} = \int_0^t f(\tau)d\tau$$ Define $F(t):=\int_0^t f(\tau)d\tau$, then $$F''(t) =F(t)\tag{1}$$ the solution of $(1)$ is $F(t) = A\cdot e^t+B\cdot e^{-t})$

Then you deduce $f(t)$

NN2
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  • @Gonçalo It's a typo. I corrected it – NN2 Apr 28 '23 at 14:03
  • @ NN2 : Sorry, but there is a parameter in the first part of the equations : f(a*t). Thus I don't think that the F(t) you propose is a solution. – user425269 Apr 28 '23 at 18:09
  • @user425269 We have $$F''(t)=:\frac{df(t)}{dt} =f(at)=\int_{0}^{t}f(\tau)d\tau :=F(t)$$ right? – NN2 Apr 28 '23 at 18:13
  • Dear NN2. No. In fact I have two different relation. The fault is mine. It would be better to write : how to solve: $f(at)=\frac{df(t)}{dt}$ and $g(at)=\int_{0}^{t}g(\tau)d\tau$ $a\gt0$. – user425269 Apr 29 '23 at 17:31
  • How to do that if now I have $f(at)=\frac{df(t)}{dt}$ and $g(at)=\int_{0}^{t}g(\tau)d\tau$ $a\gt0$. – user425269 Apr 29 '23 at 17:35
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    @user425269 In this case, I found this post than may answer your first equation. The second equation can be proved with similar method – NN2 Apr 29 '23 at 18:57
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    Yes, thank you. This is a way to solve the problem. – user425269 Apr 30 '23 at 19:43
  • @user425269 You're welcome! – NN2 Apr 30 '23 at 20:06