Solution of the Integral equation $y(x) = 1 - 2x -4x^2 +\int_{0}^x \left[ 3 + 6(x-t)-4(x-t)^2 \right]y(t)dt$

Question

I want to find the solution for the integral equation: $$y(x) = 1 - 2x -4x^2 +\int_{0}^x \left[ 3 + 6(x-t)-4(x-t)^2 \right]y(t)dt $$ I tried finding the resolvent kernel of the equation and it is getting more complicated at each new iteration. Is there any other "clean" method to find the solution? Any help is highly appreciated.

Answer 1

$$y(x) = 1 - 2x -4x^2 +\int_{0}^x \left[ 3 + 6(x-t)-4(x-t)^2 \right]y(t)dt \tag 1$$ In the integral the function is multiplied by a polynomial of second degree of $x$. So three successive differentiations are necessary in order to get rid of the integral. $$y'(x)=2-8x+3y(x) +\int_{0}^x \left[ 6 - 8(x-t) \right]y(t)dt \tag 2$$ $$y''(x)=-8+3y'(x)+6y(x) -8\int_{0}^x (x-t)y(t)dt \tag 3$$ $$y'''(x)=3y''(x)+6y'(x)-8y(x)\tag 4$$ The integral equation is transformed into a third order ODE : $$y'''-3y''-6y'+8y=0\tag 5$$ which general solution is : $$y(x)=c_1e^{-2x}+c_2e^x+c_3e^{4x} \tag 6$$ One have to determine the coefficients in putting (6) into (1) : $$c_1e^{-2x}+c_2e^x+c_3e^{4x} = 1 - 2x -4x^2 +\int_0^x \left[ 3 + 6(x-t)-4(x-t)^2 \right] (c_1e^{-2x}+c_2e^x+c_3e^{4x}) dt $$ After simplification $$(-2c_1+4c_2+c_3-4)x^2+(5c_1+2c_2-c_3-2)x+(-c_1-c_2-c3+1)=0$$ $$\begin{cases} -2c_1+4c_2+c_3-4=0\\ 5c_1+2c_2-c_3-2=0\\ -c_1-c_2-c3+1=0 \end{cases} \quad\implies\quad \begin{cases} c_1=0 \\ c_2=1 \\ c_3=0 \end{cases}$$ The solution is : $$y(x)=e^x$$