Take the 2-minute tour ×  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.  share|improve this question 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 show 7 more comments
2 Answers  up vote 2 down vote accepted  It is easy to see that
the constant $c$ in you