The solution of an ODE Consider the ODE $Cg''(x)+xg'(x)+2g(x)=0$, where $g(x)$ is defined in a neighborhood of $1$ and $C$ is a nonzero constant. How to find the solution of this equation? If it is difficult to find the solution for general $C$, is there some $C$ such that we can find a solution using elementary methods?
 A: One solution is $x e^{-x^2/(2C)}$; a second fundamental solution can be obtained using reduction of order.
A: Whenever you have annoying constants in an ode, try to scale them away first.
Transform $x\mapsto Kx$ and $g\mapsto Lg$. Then the equation becomes (by dimensionality) $$CLK^{-2}g''+Lxg'+2Lg=0$$ As you can see, the $L$ does not help, but putting $K=C^{1/2}$ eliminates the $C$, $$y''+ty'+2y=0$$ where $x=C^{1/2}t$ and $y(t)=g(x)$.
You can solve this with the power series method to get $$u_1(t)=t-\frac{1}{2}t^3+\frac{1}{8}t^5-\frac{1}{48}t^7+\cdots=te^{-t^2/2}$$ $$u_2(t)=1-t^2+\frac{1}{3}t^4+\frac{1}{15}t^6+\cdots$$
A: Similar to Help on solving an apparently simple differential equation:
Let $g(x)=\int_a^be^{xs}K(s)~ds$ ,
Then $C\int_a^bs^2e^{xs}K(s)~ds+x\int_a^bse^{xs}K(s)~ds+2\int_a^be^{xs}K(s)~ds=0$
$\int_a^b(Cs^2+2)e^{xs}K(s)~ds+\int_a^bse^{xs}K(s)~d(xs)=0$
$\int_a^b(Cs^2+2)e^{xs}K(s)~ds+\int_a^bsK(s)~d(e^{xs})=0$
$\int_a^b(Cs^2+2)e^{xs}K(s)~ds+[se^{xs}K(s)]_a^b-\int_a^be^{xs}~d(sK(s))=0$
$\int_a^b(Cs^2+2)e^{xs}K(s)~ds+[se^{xs}K(s)]_a^b-\int_a^be^{xs}(sK'(s)+K(s))~ds=0$
$[se^{xs}K(s)]_a^b-\int_a^b(sK'(s)-(Cs^2+1)K(s))e^{xs}~ds=0$
$\therefore sK'(s)-(Cs^2+1)K(s)=0$
$sK'(s)=(Cs^2+1)K(s)$
$\dfrac{K'(s)}{K(s)}=Cs+\dfrac{1}{s}$
$\int\dfrac{K'(s)}{K(s)}~ds=\int\left(Cs+\dfrac{1}{s}\right)~ds$
$\ln K(s)=\dfrac{Cs^2}{2}+\ln s+k_1$
$K(s)=kse^\frac{Cs^2}{2}$
$\therefore g(x)=\int_a^bkse^{\frac{Cs^2}{2}+xs}~ds$
But since the above procedure in fact suitable for any complex number $s$ ,
$\therefore g_n(x)=\int_{a_n}^{b_n}k_nm_nte^{\frac{C(m_nt)^2}{2}+xm_nt}~d(m_nt)=m_n^2k_n\int_{a_n}^{b_n}te^{\frac{m_n^2Ct^2}{2}+m_nxt}~dt$
For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:
$\lim\limits_{t\to a_n}t^2e^{\frac{m_n^2Ct^2}{2}+m_nxt}=\lim\limits_{t\to b_n}t^2e^{\frac{m_n^2Ct^2}{2}+m_nxt}$
$\int_{a_n}^{b_n}te^{\frac{m_n^2Ct^2}{2}+m_nxt}~dt$ converges
