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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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    Thanks for this info. The problem does look like Fredholm equation described. My only problem is that $f(0)\neq 0$, does this mean that there is no $v\in L_2[0,1]$ that satisfy the above?2012-12-16

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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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    Then the problem has no solutions2012-12-16