Let $f(x)=\sin(x)-\int_{0}^{x}{(x-u)f(u)du}$ where $f(x)$ is continuous. Find $f(x)$. Let $$f(x)=\sin(x)-\int_{0}^{x}{(x-u)f(u)du}$$ where $f(x)$ is continuous. Find $f(x)$. 
Initially, I use FTC and obtain $f(x)=\sin(x)$ but in the question didn't mention $f$ is differentiable. Then I get stuck for half an hour. Can anyone guide me ?
 A: Differentiate both sides and rewrite our integral via parts:$$\begin{align*}f'(x)&=\cos x-\frac{d}{dx}\int_0^x (x-u)f(u)du\\&=\cos x-\frac{d}{dx}\left(\int_0^x\int_0^u f(v)dvdu\right)\\&=\cos x-\int_0^x f(u)du\end{align*}$$Differentiate again to yield:$$f''(x)=-\sin x-f(x)\\y''+y=-\sin x\\\text{ as we let }y=f(x)$$You can solve this one trivially. First we tackle the complementary homogeneous equation $y''+y=0$; assume a solution of the form $y=e^{\lambda x}$ so we have:$$\lambda^2 e^{\lambda x}+e^{\lambda x}=0\\e^{\lambda x}\left(\lambda^2+1\right)=0\implies\lambda=\pm i$$... which means we've found the complementary solution $y_c=c_1e^{-ix}+c_2e^{ix}$ or $y_c=c_1\cos x+c_2\sin x$ to our equation.
To finish it off, we'll have to recognize that $-\sin x$ is a solution to our complementary equation; let's attempt the method of undetermined coefficients with a solution $y=a_1x\cos x+a_2x\sin x$ (note the power of $x$ to make our ansatz linearly independent to our general solution) so $y''=-2a_1\sin x-a_1x\cos x+2a_2\cos x-a_2x\sin x$.$$-2a_1\sin x-a_1x\cos x+2a_2\cos x-a_2x\sin x+a_1x\cos x+a_2x\sin x=-\sin x\\(-a_1 x+2a_2+a_1x)\cos x+(-2a_1-a_2x+a_2x)\sin x=-\sin x\\2a_2\cos x-2a_1\sin x=-\sin x$$... equating coefficients yields:$$2a_2=0\implies a_2=0\\2a_1=1\implies a_1=\frac12$$So we've found a particular solution $y_p=\frac12x\cos x$.
Thus our general solution is $y=y_c+y_p=c_1\cos x+c_2\sin x+\frac12x\cos x$
A: Hint: Another approach in which you can find the final solution sis to use the Laplace Transformation. Take LT form the both sides of OE and note that there is a convolution form in $$\int_{0}^{x}{(x-u)f(u)du}=t*f(x)$$
A: Use the FTC two times ($f$ is differentiable as the sum of $\sin(x)$ and $-\int_{0}^{x}{(x-u)f(u)du}$) to get a differential equation of the form $y''+y=...$.
