$$f(x) = 2a \int_{0}^{x}{f(t)\;dt} - \left(\frac{b^2}{2}\right)\int_{0}^{1}{|x-t|f(t)\;dt}$$ where $0<a<b$
My task is to solve for $f(x)$. I'm having difficulty solving this integral equation. What makes it really hard is the variable upper bound in the first integral, therefore implies that this is a Volterra integral equation. However this is a double integral equation. The second equation is a Fredholm integral equation. That's where I'm stuck. I've tried hitting it from different angles, but only to be in vain.
Split the second integral from $0$ to $x$ and from $x$ to $1$. $$f(x)=2a\int_0^xf(t)~dt-\dfrac{b^2}2\int_0^x(x-t)f(t)~dt+\dfrac{b^2}2\int_x^1(x-t)f(t)~dt$$ Now from fundamental theorem of calculus, we have $$f'(x)=2af(x)-\dfrac{b^2}2\int_0^xf(t)~dt+\dfrac{b^2}2\int_x^1f(t)~dt$$ Differentiate it again to get $$f''(x) = 2af'(x) - b^2 f(x)$$ $$f(x)=C_1e^{ax}\sin\sqrt{b^2-a^2}x+C_2e^{ax}\cos\sqrt{b^2-a^2}x$$ Now substitute the solution back to the integral equation to determine $C_1$ and $C_2$ , i.e. to determine $C_1$ and $C_2$ from the following identity $$C_1e^{ax}\sin\sqrt{b^2-a^2}x+C_2e^{ax}\cos\sqrt{b^2-a^2}x\equiv2a\int_0^x(C_1e^{at}\sin\sqrt{b^2-a^2}t+C_2e^{at}\cos\sqrt{b^2-a^2}t)~dt-\dfrac{b^2}2\int_0^1|x-t|(C_1e^{at}\sin\sqrt{b^2-a^2}t+C_2e^{at}\cos\sqrt{b^2-a^2}t)~dt$$