How do you solve $y''+\frac{x}{2}y'+y=0$ a 2nd order homogenous equation? How do you solve $y''+\frac{x}{2}y'+y=0$ where $y'=\frac{\partial y}{\partial x}$
What is confusing me is that once I say $g=y'$ I'm left with :
$g'+\frac{x}{2}g+y=0$
I can't use seperation by parts here. How would one proceed here?
Thank you.
Best
 A: In fact this question is similar to solution of ordinary differential equations:
Let $y=\int_Ce^{xu}K(u)~du$ ,
Then $(\int_Ce^{xu}K(u)~du)''+\dfrac{x}{2}(\int_Ce^{xu}K(u)~du)'+\int_Ce^{xu}K(u)~du=0$
$\int_Cu^2e^{xu}K(u)~du+\dfrac{x}{2}\int_Cue^{xu}K(u)~du+\int_Ce^{xu}K(u)~du=0$
$\int_C(u^2+1)e^{xu}K(u)~du+\int_C\dfrac{ue^{xu}K(u)}{2}d(xu)=0$
$\int_C(u^2+1)e^{xu}K(u)~du+\int_C\dfrac{uK(u)}{2}d(e^{xu})=0$
$\int_C(u^2+1)e^{xu}K(u)~du+\left[\dfrac{ue^{xu}K(u)}{2}\right]_C-\int_Ce^{xu}~d\left(\dfrac{uK(u)}{2}\right)=0$
$\int_C(u^2+1)e^{xu}K(u)~du+\left[\dfrac{ue^{xu}K(u)}{2}\right]_C-\int_Ce^{xu}\dfrac{uK'(u)+K(u)}{2}du=0$
$\int_C(u^2+1)e^{xu}K(u)~du+\left[\dfrac{ue^{xu}K(u)}{2}\right]_C-\int_Ce^{xu}\dfrac{uK'(u)-(2u^2+1)K(u)}{2}du=0$
$\therefore uK'(u)-(2u^2+1)K(u)=0$
$uK'(u)=(2u^2+1)K(u)$
$\dfrac{K'(u)}{K(u)}=2u+\dfrac{1}{u}$
$\int\dfrac{K'(u)}{K(u)}du=\int\left(2u+\dfrac{1}{u}\right)du$
$\ln K(u)=u^2+\ln u+c_1$
$K(u)=cue^{u^2}$
$\therefore y=\int_Ccue^{u^2+xu}~du$
But since the above procedure in fact suitable for any complex number $u$ ,
$\therefore y_n=\int_{a_n}^{b_n}c_n(k_nt)e^{(k_nt)^2+xk_nt}~d(k_nt)={k_n}^2c_n\int_{a_n}^{b_n}te^{{k_n}^2t^2+k_nxt}~dt$
For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $k_n$ such that:
$\lim\limits_{t\to a_n}t^2e^{{k_n}^2t^2+k_nxt}=\lim\limits_{t\to b_n}t^2e^{{k_n}^2t^2+k_nxt}$
$\int_{a_n}^{b_n}te^{{k_n}^2t^2+k_nxt}~dt$ converges
For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $k_1=\pm i$
$\therefore y=C_1\int_0^\infty te^{-t^2}\cos xt~dt$ or $C_1\int_0^\infty te^{-t^2}\sin xt~dt$
Hence $y=C_1\int_0^\infty te^{-t^2}\sin xt~dt+C_2\int_0^\infty te^{-t^2}\cos xt~dt$
