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I am stumbling upon this problem: Let $$\mathcal{n} \in \mathbb{N}$$Find an $$\mathcal{u} : \mathbb{R_{>0}} \to \mathbb{R}$$ which solves the following equation

$$\int_o^1 \mathcal{u}(\mathcal{st})d\mathcal{s} = \mathcal{n} \mathcal{u}(\mathcal{t})$$

for every $\mathcal{t}$ $\in$ $\mathbb{R_{>0}}$

To this end, solve a differential equation. I know I am supposed to bring the equation into the explicit form, but I don't know how to start from here.

Any help is appreciated.

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  • $\begingroup$ Hello and welcome. I tried helping you with the typesetting, please let me know if I screwed something up. Did you mean $>0$ on the reals as lowered like I did it? $\endgroup$ – mathreadler May 1 '17 at 18:15
  • $\begingroup$ Yes of course! Thank you for your attention $\endgroup$ – fidel castro May 1 '17 at 18:17
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Let $x=st$ therefore: $$\int_{0}^{t}{u(x)\frac{ds}{dx}dx}=nu(t)$$ this results in: $$\int_{0}^{t}{u(x)dx}=nt\,u(t)\tag{1}$$ Now differentiate wrt. $t$ applying Leibniz's formula and you will get it. Can you follow from this?

In case you don't I will follow:

Differentiating under the integral sign (Leibniz's rule) gives the following expression for the derivative of the integral: $$\frac{d}{dt}\int_{0}^{t}{u(x)dx}=u(t)$$

Finally the differential equation is found equating it to the derivative wrt. $t$ of the RHS of equation $(1)$: $$u(t)=\frac{d}{dt}\left(nt\,u(t)\right)=nu(t) + nt\frac{d u}{dt}(t)$$

With a better presence: $$t\frac{du}{dt}(t) + \frac{n-1}{n}u(t)=0$$

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  • $\begingroup$ Thank for your help! Unfortunately I am new to differential equation, do you mean Leibniz Integral Rule? I am failing to see where this is leading $\endgroup$ – fidel castro May 1 '17 at 18:21
  • $\begingroup$ Yes!! I will work it out a little more for you! $\endgroup$ – HBR May 1 '17 at 18:29
  • $\begingroup$ I am impressed! Thank you for your thorough explanation $\endgroup$ – fidel castro May 1 '17 at 18:46
  • $\begingroup$ You are welcome! $\endgroup$ – HBR May 1 '17 at 18:46

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