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I have an equation like this:

$$te^{t} = \int\nolimits_0^t e^\tau u(\tau)d\tau$$

I don't really know how to solve it.. Would it be possible to differentiate both sides of the equation? If so, how can I do it - is it like differentiate one side and then the other or is it more complex? And if it is possible what conditions need to be satisfied in order to do it?

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  • $\begingroup$ Thank you for editing, I didn't really know why this was not rendering properly. Help for MathJax is very poor on this site. $\endgroup$
    – kubal5003
    Feb 17 '11 at 11:21
  • $\begingroup$ You were very close to having the right syntax. All I did was change your ampersands (&) to double dollar signs ($$). $\endgroup$ Feb 17 '11 at 14:17
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If we differentiate the left hand side we get $e^t+t e^t$, if we differentiate the right hand side we get $e^t u(t)$ by the fundamental theorem of calculus. Then we see

$$u(t)=t+1$$

An we are done. I think we need to demand that $u(t)$ is continuous to get uniqueness. In general $u(t)=t+1$ almost everywhere.

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    $\begingroup$ Thank you. So does this mean I don't have to care about the integral boundaries (I mean 0 to t, I'm not sure if it's called boundaries in English) ? What about Newton-Leibniz $$\int_a^bf(x)dx= F(b) - F(a)$$ ? $\endgroup$
    – kubal5003
    Feb 17 '11 at 11:44
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    $\begingroup$ No, you don't need to worry about the limits of integration since you're differentiating. As @Thomas mentioned, we use the Fundamental Theorem of Calculus (second part, in the Wikipedia article) to differentiate. Then the $F(a)$ is inconsequential; though perhaps troubling at first, it makes sense since $F(a)$ is constant and the derivative of a constant is zero. $\endgroup$
    – user4689
    Feb 17 '11 at 12:45
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    $\begingroup$ You only have proven that if there is a solution then it has to be $u(t):=t+1$. You now have to check that this function actually solves the original equation. If the left side of this equation would have been $1+t e^t$ instead of $t e^t$ then your computation would still have produced $u(t):=t+1$, but in fact in this case there would be no solution at all. $\endgroup$ Feb 17 '11 at 12:54
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    $\begingroup$ Before reading this comment I tried differentiating using both F(b) and F(a). It did produce a solution: $$1 + t + \frac{u(0)}{e^t}$$ so I checked if it was ok and found out that u(0) needs to be equal to 0 in order to satisfy the initial equation. Thank you all for help. $\endgroup$
    – kubal5003
    Feb 17 '11 at 16:21

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