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I need to solve the following equation for $v(x)$: $$\int_0^tv(x)(x+1)dx=f(t)$$ I am given the function $f(t)$. I've done this so far:

If we derive both sides by $t$, we get $v(t)(t+1)=f'(t)$ and $\bar{v}(t)=\frac{f'(t)}{t+1}$. The problem is that I am still off by a constant, i.e., the above only guarantees that : $\int_0^t\bar{v}(x)(x+1)dx+c=f(t)$ which is not enough for me.

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Hint: consider taking the Fourier or Laplace transform from both sides, solve algebraic equation and making the inverse transform. – m0nhawk Dec 16 '12 at 19:38
@monhawk: how should that help? – Fabian Dec 16 '12 at 19:38
If your $f(0)=0$, then $c\equiv0$ for any $\nu(x)$. Isn't it? – 0x2207 Dec 16 '12 at 19:38
@0x2207: yes, the constant is in fact $f(0)$. – Fabian Dec 16 '12 at 19:39
@Fabian, from definition $\int_{0}^{0} g(x) dx \equiv 0$ – 0x2207 Dec 16 '12 at 19:40
up vote 2 down vote accepted

It is easy to see that the constant $c$ in your post is in fact $f(0)$. Furthermore, a little thought shows that your equations cannot be solved unless $f(0)=0$ (just plug in $t=0$ in your equation and you find $0=f(0)$).

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I don't see what is the problem here. Obsviously $f(0)=0$ (for the equation to have a solution) and by deriving as you said, we get $$v(t)=\frac{f^{\prime}(t)}{t+1}$$ If we substitue this back in the orginal equation we have $$f(t)=\int_{0}^{t}f^{\prime}(x)dx=f(t)-f(0)=f(t)$$ which is true

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The problem is the given $f(0)\neq 0$. – Hasanhasan Hasan Dec 16 '12 at 20:27
Then the problem has no solutions – Nameless Dec 16 '12 at 20:39

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