# Find $f(x)=?$ functional equation

I would appreciate if somebody could help me with the following problem:

Q: Find $f(x)$ ($f'(x)$: conti-function , $x \in\mathbb{R}$) $$f(x)=\sin ^2x+\int_{0}^{x}tf(t)dt$$

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Here is what google returns for the query "conti function". –  1015 Jun 1 '13 at 3:45

$f '(x)-xf(x)=2\sin(x)\cos(x)$ $\to$ $$f(x)=e^{\left(-\dfrac{x^2}{2}\right)} \left(\int_0^x e^{\left(-\dfrac{t^2}{2}\right)}\cdot2\sin(t)\cos(t)dt+c\right)$$

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Please check my edit and see if messed something up. Sorry if I did! –  Amzoti Jun 1 '13 at 3:39
That ought to be $e^{-\frac{x^2}{2}}$ in the integral, I believe. –  Zen Jun 1 '13 at 3:41
@Amzoti:thanks u so much –  Maisam Hedyelloo Jun 1 '13 at 3:52
@MaisamHedyelloo You can evaluate the value of constant $c$. Put $x =0$ in the equation to get the initial condition $f(0) = 0$. I hope this will give $c = 0$. If it is correct you may edit your answer. –  srijan Jun 1 '13 at 4:33

Differentiate both sides wrt x to get $$f'(x)=2\sin(x)\cos(x)+xf(x)$$ which is a linear ODE in f(x). Then use the integrating factor method. You'll get f(x) down to an integral in terms of x, but it's not going to be an elementary antiderivative.

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I'm very silly, I apologize for the earlier comment –  DanZimm Jun 1 '13 at 6:34

It's basically a fundamental theorem of calculus problem. I was thinking your only confusion would probably be how to get rid of $t$. Remember the fundamental theorem of calculus says if $F(x)=\int_0^t f(t)dt$, then we have $F'(x)=f(x)$. I think that's probably the only trick you need to use. The rest of this question is just standard ODE solving.