$$\frac{d^2y}{dx^2}=ay\frac{dy}{dx}+by+c$$
This is autonomous ODE. One can reduce it to the first order, thanks to the usual change :
$\frac{dy}{dx}=u(y) \quad\to\quad \frac{d^2y}{dx^2}=u\frac{du}{dy}\quad\to\quad u\frac{du}{dy}=a\,y\,u(y)+b\,y+c$
This is an Abel's ODE of the second kind.
The change $\quad u(y)=\frac{1}{v(y)}\quad$ leads to an Abel's ODE of the first kind :
$$\frac{dv}{dy}=-(b\,y+c)v^3-a\,y\,v^2$$
For further progress, see :
https://www.hindawi.com/journals/ijmms/2011/387429/#sec2
By rescaling $x$ we may eliminate the constant $4$ and study instead the equation
$$
y'''(x)=xy(x)
$$
It is interesting to consider a slight generalization to
$$\tag{1}
y^{(n)}(x)=x^my(x)
$$
By applying WKB methods to (1) we find directly the large $x$ behavior of the $n$ independent solutions is
$$\tag{4}
y(x)\sim N \left(x^{m(1-n)/2n} \right)\exp\left(\omega \int\limits^x dx' {x'}^{m/n}\right)\qquad,\qquad x\to \infty
$$
Where there are $n$ possibilities for $\omega$ such that $\omega^n=1$. For $n=3$ and $m=1$ we have
$$\tag{5}
y(x)\sim N x^{-1/3}\exp\left(3\omega x^{4/3}/4 \right) \qquad,\qquad x\to\infty
$$
Where $\omega$ has values $1, \ e^{2\pi i/3}, \ e^{4\pi i/3}$. The first corresponds to an exponentially growing function of $x$, the other two are oscillating and exponentially decaying. No linear combination of solutions can go to a nonzero constant for large $x$.
Let us find the solutions to (1) in integral form, which will be possible explicitly if at least one of $m$ or $n$ are equal to $1$.
Taking the Fourier transform (henceforth FT) of (1) we have
$$\tag{6}
(ik)^n\tilde{y}(k)=2\pi (i)^m \int\frac{dk'}{2\pi}\tilde{y}(k') \delta^{(m)}(k-k')
$$
Note that $\delta^{(m)}$ is the mth distributional derivative of Dirac's delta. In obtaining (6), on the LHS we made use of the derivative of FT property. On the RHS we used the 'FT of product of functions equals convolution of their FTs' property. We now have a differential equation for the Fourier coefficients $\tilde{y}$
$$\tag{7}
\tilde{y}^{(m)}=i^{n-m}k^n \tilde{y}
$$
If (7) is easier to solve than (1), we have made progress. It is interesting that when $m=n$, solutions of (1) are equal to their Fourier transform. For $m=1$ (7) is a first order ODE
$$\tag{8}
\tilde{y}'=i^{n-1}k^n\tilde{y}
$$
The solution is
$$\tag{9}
\tilde{y}(k)=N \exp\left(i^{n-1}\frac{k^{n+1}}{n+1} \right)
$$
On taking the inverse FT we have
$$\tag{10}
y(x)=N \int\limits_{-\infty}^\infty \frac{dk}{2\pi} \exp\left(ikx+i^{n-1}\frac{k^{n+1}}{n+1} \right)
$$
Now you need to show (just by substitution) that if
$$\tag{11}
Y_\Gamma(x) =\int\limits_{\Gamma} \frac{dk}{2\pi} \exp\left(ikx+i^{n-1}\frac{k^{n+1}}{n+1} \right)
$$
Then $Y_\Gamma$ satisfies (1) (with $m=1$) for any curve $\Gamma$ in the complex $k$ plane so long as the integrand vanishes at the endpoints of $\Gamma$. Linearly independent solutions can be constructed by choosing different endpoints of $\Gamma$. I'll now set $n=3$ for clarity. Then for large $k$ the integrand behaves like $e^{-k^4/4}$. There are four wedges in the complex $k$ plane where the integrand vanishes for large $|k|$. If $k=|k|e^{i\phi}$ then the wedges are in
$$
\phi \in (-\pi/8, \ \pi/8), \quad (3\pi/8, \ 5\pi/8), \quad (7\pi/8, \ 9\pi/8),\quad (11\pi/8, \ 13\pi/8),
$$
Picking $\Gamma$ to begin in the third and end in the first wedge yields the solution
$$\tag{12}
A(x)=\int\limits_{-\infty}^\infty \frac{dk}{2\pi}e^{ikx-k^4/4}
$$
Picking $\Gamma$ to begin in the forth and end in the second wedge yields the solution
$$\tag{13}
B(x)=i\int\limits_{-\infty}^\infty \frac{dk}{2\pi}e^{-kx-k^4/4}
$$
Picking $\Gamma$ to begin in the forth and end in the first wedge yields the solution
$$\tag{14}
C(x)=\int\limits_0^\infty\frac{dk}{2\pi}(e^{ikx}+ie^{kx})e^{-k^4/4}
$$
Notes:
- The solution equation [3] in OP appears to be correct. Note that it is one particular solution for the inhomogeneous equation, and vanishes when $b=0$, so it's not really helpful for solving the homogeneous problem.
- This construction is similar to the way integral solutions of the Airy equation; $\operatorname{Ai}$ and $\operatorname{Bi}$ are found.
- Contrary to my prior assertion, the proposed solution [2] in OP behaves like
$$
y(x)\sim \sqrt{\frac{2\pi}{3}}x^{-1/3}e^{+3x^{4/3}/4}, \qquad, \qquad x\to \infty
$$
Which can be found using Laplace's method. It does not approach a constant, but grows exponentially. It is a linear combination of the $A$, $B$, and $C$ solutions.
- My FT convention is
$$
f(x)=\int\frac{dk}{2\pi}e^{ikx}\tilde{f}(k) \\
\tilde{f}(k)=\int dx \ e^{-ikx} f(x)
$$
Best Answer
$$y'' + (4x)y' + (2+4x^2)y = 0$$ Integrating factor is $\mu (x) =e^{x^2}$ $$e^{x^2}(y'' + 2xy') +e^{x^2}(2xy' +(2+4x^2)y) = 0$$ $$(e^{x^2}y')' +(2xe^{x^2}y)'=0$$ Integrate:
$$e^{x^2}y' +2xe^{x^2}y=K_1$$ Integrate again : $$(ye^{x^2})'=K \implies ye^{x^2}=K_1x+K_2$$ $$y(x)=e^{-x^2}(K_1x+K_2)$$