2nd order linear differential equation with non-constant coefficients Considering the equation $2y''+(x+1)y'+3y=0$ where $X_0=2$. Find the general term in each solution. That is, the general term for Y1,Y2 where $y=A_0(Y_1)+A_1(Y_2)$
I've solved this as
$$y=A_0[1-3/4(x-2)+3/8(x-2)^3+1/64(x-2)^4+...] +A_1[(x-2)-3/4(x-2)^2+1/24(x-2)^3+9/64(x-2)^4+...]$$
I did this using the recurrence relationship:
$$A_{n+2}= -3A_{n+1}/2(n+2) - (n+3)A_n/2(n+1)(n+2)$$  where $n=1,2,3,...$
Furthermore, I confirmed Y1,Y2 as a fundamental set of solutions by verifying that $W(Y_1,Y_2)(X_0)$ does not equal zero.
I'm not sure how to find the general term of Y1 and Y2. I've read my book again and didn't find any information to clarify. I appreciate any help!
 A: Similar to Help on solving an apparently simple differential equation,
Let $y=\int_Ce^{xs}K(s)~ds$ ,
Then $2(\int_Ce^{xs}K(s)~ds)''+(x+1)(\int_Ce^{xs}K(s)~ds)'+3\int_Ce^{xs}K(s)~ds=0$
$2\int_Cs^2e^{xs}K(s)~ds+x\int_Cse^{xs}K(s)~ds+\int_Cse^{xs}K(s)~ds+\int_Ce^{xs}K(s)~ds=0$
$\int_C(2s^2+s+3)e^{xs}K(s)~ds+\int_C2se^{xs}K(s)~d(xs)=0$
$\int_C(2s^2+s+3)e^{xs}K(s)~ds+\int_C2sK(s)~d(e^{xs})=0$
$\int_C(2s^2+s+3)e^{xs}K(s)~ds+[se^{xs}K(s)]_C-\int_Ce^{xs}~d(sK(s))=0$
$\int_C(2s^2+s+3)e^{xs}K(s)~ds+[se^{xs}K(s)]_C-\int_Ce^{xs}(sK'(s)+K(s))~ds=0$
$[se^{xs}K(s)]_C-\int_Ce^{xs}(sK'(s)-(2s^2+s+2)K(s))~ds=0$
$\therefore sK'(s)-(2s^2+s+2)K(s)=0$
$sK'(s)=(2s^2+s+2)K(s)$
$\dfrac{K'(s)}{K(s)}=2s+1+\dfrac{2}{s}$
$\int\dfrac{K'(s)}{K(s)}ds=\int\left(2s+1+\dfrac{2}{s}\right)~ds$
$\ln K(s)=s^2+s+2\ln s+c_1$
$K(s)=cs^2e^{s^2+s}$
$\therefore y_c=\int_Ccs^2e^{s^2+(x+1)s}~ds$
But since the above procedure in fact suitable for any complex number $s$ ,
$\therefore y_n=\int_{a_n}^{b_n}c_n(m_nt)^2e^{(m_nt)^2+(x+1)m_nt}~d(m_nt)=m_n^3c_n\int_{a_n}^{b_n}t^2e^{m_n^2t^2+m_n(x+1)t}~dt$
For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:
$\lim\limits_{t\to a_n}t^3e^{m_n^2t^2+m_n(x+1)t}=\lim\limits_{t\to b_n}t^3e^{m_n^2t^2+m_n(x+1)t}$
$\int_{a_n}^{b_n}t^2e^{m_n^2t^2+m_n(x+1)t}~dt$ converges
For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm i$
$\therefore y=C_1\int_0^\infty t^2e^{-t^2}\cos((x+1)t)~dt$ or $C_1\int_0^\infty t^2e^{-t^2}\sin((x+1)t)~dt$
Hence $y=C_1\int_0^\infty t^2e^{-t^2}\sin((x+1)t)~dt+C_2\int_0^\infty t^2e^{-t^2}\cos((x+1)t)~dt$
