We will solve the equation
\begin{equation}
(a_1x+b_1)f''(x)+(a_2x+b_2)f'(x)+(a_3x+b_3)f(x)=g(x)\tag 1
\end{equation}
where $f$, $g\in L_2(\bf R\rm)$ and $a_1$, $a_2$, $a_3$, $b_1$, $b_2$, $b_3$ are constants in $\bf R\rm$.
Let the Fourier Transform of a function $f$ of $L_2(\bf R\rm)$ is
$$\widehat{f}(\gamma)=\int^{\infty}_{-\infty}f(t)e^{-i t\gamma}dx$$
the Inverse Fourier Transform is
$$f(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}\widehat{f}(\gamma)e^{i \gamma x}d\gamma$$
Then it is known (integration by parts)
\begin{equation}
\int^{\infty}_{-\infty}f(x)x^ne^{-ix\gamma}dx=i^n(\widehat{f})^{(n)}(\gamma) .
\end{equation}
\begin{equation}
\widehat{(f^{(n)})}(\gamma)=(i\gamma)^n\widehat{f}(\gamma) .
\end{equation}
$$
\int^{\infty}_{-\infty}f'(x)A(x)e^{-i x\gamma}dx=
$$
\begin{equation}
=\int^{\infty}_{-\infty}f(x)A'(x)e^{-ix \gamma}dx+(-i\gamma)\int^{\infty}_{-\infty}f(x)A(x)e^{-ix\gamma}dx .
\end{equation}
$$
\int^{\infty}_{-\infty}f''(x)A(x)e^{-ix\gamma}=\int^{\infty}_{-\infty}f(x)A''(x)e^{-ix\gamma}dx+
$$
\begin{equation}
+2(-i\gamma)\int^{\infty}_{-\infty}f(x)A'(x)e^{-ix\gamma}dx+(-i\gamma)^2\int^{\infty}_{-\infty}f(x)A(x)e^{-ix\gamma}dx .
\end{equation}
Theorem.
When $f$, $g\in L_2(\bf R\rm)$ and $\lim_{|x|\rightarrow \infty}|f(x)x^{2+\epsilon}|=0$, $\epsilon>0$, equation $(1)$ can reduced to
\begin{equation}
(-ia_1\gamma^2+a_2\gamma+ia_3)\frac{\widehat{f}(\gamma)}{d\gamma}+(-b_1\gamma^2-2ia_1\gamma+ib_2\gamma+a_2+b_3)\widehat{f}(\gamma)=\widehat{g}(\gamma)
\end{equation}
which is solvable.
Using the above theorem in your equation we have
$$
y''-2xy'-11y=X_{[0,\infty)}(x)e^{-a x}\tag{eq}
$$
We get
$$
-(13+s^2)Y(s)-2sY'(s)=g(s)\textrm{, }g(s)=\frac{1}{a-is}
$$
Solving this equation we get
$$
Y(s)=-e^{1/2(-s^2/2-13\log s)}\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt
$$
Hence
$$
y(x)=-\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{1/2(-s^2/2-13\log s)}\left(\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt\right)e^{isx}ds
$$
The equation
$$
y''-2xy'-11y=0
$$
Have general solution
$$
y(x)=C_1H_{-11/2}(x)+C_2\cdot {}_1F_1\left(\frac{11}{2};\frac{1}{2};x^2\right)
$$
Hence equation $(eq)$ have general solution
$$
y(x)=C_1H_{-11/2}(x)+C_2\cdot {}_1F_1\left(\frac{11}{2};\frac{1}{2};x^2\right)-
$$
$$
-\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{1/2(-s^2/2-13\log s)}\left(\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt\right)e^{isx}ds,
$$
where $H_n(x)$ is the $n-$th Hermite function and ${}_1F_1(a;b;x)$ is the ${}_1F_{1}$ hypergeometric function.
For more details in this kind of equations see here .
GENERAL NOTES
The degree of the term $x^my^{(n)}$ is $\nu=m-n$. We gather together all the terms of the DE of $\nu$ degree. In this way we can split a differential equation $A(x)y''+B(x)y'(x)+C(x)=0:(DE)$ into $N$ distinct groups of terms with $\nu_i$, $i=1,2,\ldots,n$ degree. The number $N$ is called degree of the DE.
If the degree $N$ is 2, we call the DE 2-degree. i.e
$$
(1-x^2)y''-2xy'+\lambda y=0\textrm{ (Legendre) }
$$
$$
y''-2xy'''+\lambda y=0\textrm{ (Hermite) }
$$
$$
x^2y''+xy'+(x^2-\nu^2)y=0\textrm{ (Bessel) }
$$
Every 2-degree DE have two parts: The part with the largest degree ($L_{max}$ part degree) and the part of the smaller degree ($L_{min}$ part-degree).
i) If the part of largest degree have the term $y''$ we call it DE of the first kind.
ii) If the part of largest degree have the term $y'$ we call it DE of the second kind.
iii) If the part of the largest degree have the term $y$ we call it DE of the third kind.
- Step $l$ of a 2-degree (DE) is the difference of the degree of the larger part minus the degree of the smaler part.
I) If the 2-degree (DE) is of the first kind, then its solution is
$$
y(x)=x^{\mu}\Phi(a,b;c;\lambda x^{l}),
$$
where
$$
\Phi(a,b,c;x)=c_1\cdot {}_2F_1(a,b;c;x)+c_2\cdot x^{1-c}{}_2F_1(a+1-c,b+1-c;2-c;x),
$$
where ${}_2F_1(a,b;c;,x)$ is the well known Gauss hypergeometric series.
The asymptotic behavior $y=x^s$ of (DE) around $x=0$ lead us to the starting powers $s_1,s_2$. Then $x^{s_1}=x^{\mu}(x^l)^0\Rightarrow s_1=\mu$, $x^{s_2}=x^{\mu}(x^l)^{1-c}\Rightarrow s_2=\mu+l(1-c)$.
The asymptotic behavior in infinty $y=x^{k}$, lead us to $x^{k_1}=x^{\mu}(x^l)^{-a}\Rightarrow k_1=\mu-la$, $x^{k_2}=x^{\mu}(x^l)^{-b}\Rightarrow k_2=\mu-lb$
The parameter $\lambda$ is evaluated demmanding that change of variable $t=\lambda x^l$ leaves the finite singular points of (DE) at $t=1$.
II) If the 2-degree (DE) is of the second kind, then its solution is
$$
y(x)=x^{\mu}\Phi(a;c;\lambda x^{l}),
$$
where
$$
\Phi(a;c;x)=c_1\cdot {}_1F_1(a;c;x)+c_2\cdot x^{1-c}\cdot {}_1F_1(a+1-c;2-c;x).
$$
In $x=0$, we have $x^{s_1}=x^{\mu}(x^l)^0\Rightarrow s_1=\mu$ and $x^{s_2}=x^{\mu}(x^l)^{1-c}\Rightarrow s_2=\mu+l(1-c)$ and in $x=\infty$, we have $x^{k_1}=x^{\mu}(x^l)^{-a}\Rightarrow k_1=\mu-la$. The $\lambda$ is evaluated from the asymptotic subtitution $y_{\infty}(x)\approx e^{\lambda x^l}$ in the (DE).
III) The last case is when the 2-degree (DE) is if third kind. Then
$$
y(x)=x^{\mu}Z_{k}(\lambda x^{l/2}),
$$
where
$$
Z_{k}=c_1J_{k}(x)+c_2Y_{k}(x)\textrm{, where }J_k(x)\textrm{ and }Y_k(x)\textrm{ are the Bessel... }
$$
The the asymptotic behavior in $x=0$ give us $x^{s_1}=x^{\mu}(x^{l/2})^k\Rightarrow s_1=\mu+\frac{lk}{2}$, $x^{s_2}=x^{\mu}(x^{l/2})^{-k}\Rightarrow s_2=\mu-\frac{lk}{2}$. The $\lambda$ is evaluated from the asyptotic behavior in $x=\infty$, $y_{\infty}(x)\approx \exp\left(\pm i\lambda x^{l/2}\right)$
Example 1.
$$
y''-xy=0\tag 1
$$
Obviously (1) is of 2-degree with step $l=1-(-2)=3$. We have
$$
L_{mim}=D^2\textrm{ and }L_{max}=-x.
$$
Since the largest term has no differentials this equation is a 2-degree of the third kind with step $l=3$. Hence its general solution is
$$
y(x)=x^{\mu}Z_{k}(\lambda x^{3/2}).\tag 2
$$
Setting $y=x^{s}$ in (1), we get $s(s-1)x^{s-2}=0\Leftrightarrow s_1=0$, $s_2=1$, we get from (2): $x^1=x^{\mu}(x^{3/2})^{-k}\Rightarrow 1=\mu+\frac{3}{2}k$. Also $x^{0}=x^{\mu}(x^{3/2})^{-k}\Rightarrow 0=\mu-\frac{3}{2}k$. Hence $\mu=1/2$, $k=1/3$. For the evaluation of $\lambda$, we set $y_{\infty}(x)\approx \exp(\pm i \lambda x^{3/2})$, (using the asymptotic formula: $(e^{S})''\approx (S')^2e^{S}$, where $S=\lambda x^{\rho}$, $\rho>0$, $x>>1$),
we get after inserting this into (1): $\lambda=\pm i\frac{2}{3}$. Hence the ecxact solution of (1) (Airy equation) is
$$
y(x)=x^{1/2}Z_{1/3}\left(i\frac{2}{3}x^{3/2}\right)
$$
Example 2.
$$
xy''+(2-x^2)y'-2xy=0\tag 2
$$
We rewrite (2) in the form $(xy''+2y')+(-x^2y'-2xy)=0$. Hence $L_{min}=xD^2+2D$, $L_{max}=-x^2D-2x$. Hence (2) is a two degree DE of second kind with step $l=2$. Hence the solution is of the form
$$
y(x)=x^{\mu}\Phi(a;c;\lambda x^{2})
$$
The asymptotic behavior at $x=0$ is: The starting powers are (set $y=x^s$ in $L_{min}y=0$ to get) $s_1=0$, $s_2=-1$. $x^0=x^{\mu}(x^2)^{0}\Rightarrow \mu=0$ and $x^{-1}=x^{\mu}(x^2)x^{1-c}\Rightarrow c=\frac{3}{2}$.
The asymptotic behavior at $x=\infty$ is: Solve $L_{max}x^s=0\Leftrightarrow -x^{2}sx^{s-1}-2xx^s=0\Leftrightarrow s=-2$. Hence $k_1=-2$. Hence $x^{-2}=x^{\mu}(x^2)^{-a}\Leftrightarrow a=1$. Also if we set $y_{\infty}(x)\approx\exp(\lambda x^2)$ in (2) we get $\left(e^{\lambda x^2}\right)'=2\lambda x e^{\lambda x^2}$, $\left(e^{\lambda x^2}\right)''=4\lambda^2 x^2 e^{\lambda x^2}$. Hence seting these in (2), we get $4\lambda^2x^3e^{\lambda x^2}-2\lambda x^3e^{\lambda x^2}=0\Rightarrow 4\lambda^2=2\lambda\Rightarrow \lambda=1/2$. Hence the ecxact solution of (2) is
$$
y(x)=\Phi\left(1;\frac{3}{2};\frac{x^2}{2}\right)
$$
Hence
$$
y(x)=c_1\cdot {}_1F_1\left(1;\frac{3}{2};\frac{x^2}{2}\right)+c_2 \cdot x^{-1/2}\cdot{}_1F_1\left(\frac{1}{2};\frac{1}{2};\frac{x^2}{2}\right)
$$
Example 3. Solve
$$
(1-x^4)y''+\left(n(n+1)x^2-\frac{m(m+1)}{x^2}\right)y=0
$$
Answer
$$
y(x)=x^{m+1}\Phi\left(\frac{m-n}{4},\frac{m+n+1}{4};\frac{2m+5}{4};x^4\right)
$$
Example 4. The Hermite equation is
$$
y''-2xy'+2\nu y=0
$$
This equation is 2-degree of the second kind and step $l=0-(-2)=2$...etc
$$
y(x)=\Phi\left(\frac{-\nu}{2};\frac{1}{2};x^2\right)
$$
