I am trying to solve the following ODE:
$$f''(x)-(ax+b)f(x)=0 \tag1$$
To obtain the expression for $f(x)$, I have tried to use the method described in this science article. According to that method, the general solution to equation $(1)$ is
$$f(x)= \sum _{k=0}^{\inf}g_{_k}(x) \tag2$$ where
$$g_{_0} (x)= 1$$ $$g_{_1}(x)=\int_0^x (x-t)\Big((at+b)g_{_0}\Big)dt$$ $$g_{_2}(x)=\int_0^x (x-t)\Big((at+b)g_{_1}(t)\Big)dt\tag3$$ $$...$$ $$g_{_k}(x)=\int_0^x (x-t)\Big((at+b)g_{_{k-1}}(t)\Big)dt$$ I tried to write $g_{_1}(x)$, $g_{_2}(x)$ and $g_{_3}(x)$ in a way where a pattern could occur:
$$g_{_1}(x)= {1+3\over 2}{a\over 3!}x^3 + {b\over2!}x^2$$ $$g_{_2}(x)= {1+3\over 2}{3+5\over 2}{a^2\over 6!}x^6 + \Bigg({1\over 5!} + {1\over 6!}\Bigg)abx^5+{{b^2 \over 4!} x^4} \tag4$$
$$g_{_3}(x)= {1+3\over 2}{3+5\over 2}{6+8\over 2} {a^3\over 9!}x^9 + {1+3\over 2}{3+5\over 2} {a^2 b\over 8!}x^8 + \Bigg( {{6!}\over{5!}} + {{5!}\over{5!}}\Bigg) {a^2 b\over 8!}x^8 + \Bigg( {{5!}\over{5!}} + {{5!}\over{6!}}\Bigg){a {b^2}\over 7!}x^7+ {6+4\over 2}{a {b^2}\over 7!}x^7 + {{b^3 \over 6!} x^6}$$
It can be seen in expressions $(4)$ that I failed to do that. I was not able to find the pattern.
My question is: If $f(x)$ is the general solution to equation $(1)$, what is its exact analytical expression?