solution of ordinary differential equations How to solve the following ordinary differential equation
\begin{equation}
f_{xx}^{^{\prime \prime }}+2axf_{x}^{^{\prime }}-2sf=0
\end{equation}
with $a$ and $s$ are arbitrary real numbers and the boundary conditions
\begin{equation}
f\left( \delta \right) =f\left( -\delta \right) =1
\end{equation}
After checking some books, I got the general solution
\begin{equation}
f\left( x\right) =C_{1}\Phi \left( -\frac{s}{2a},\frac{1}{2},-ax^{2}\right)
+C_{2}\Psi \left( -\frac{s}{2a},\frac{1}{2},-ax^{2}\right)
\end{equation}
where $\Phi$ and $\Psi$ are degenerate hypergeometric functions. Are there any other forms of the solution, such as double integral or serious?
 A: To maintain the meaning of this question, $a\neq0$ and $s\neq0$ should be restricted 
Let $f(x)=\int_Ce^{xu}K(u)~du$ ,
Then $(\int_Ce^{xu}K(u)~du)''+2ax(\int_Ce^{xu}K(u)~du)'-2s\int_Ce^{xu}K(u)~du=0$
$\int_Cu^2e^{xu}K(u)~du+2ax\int_Cue^{xu}K(u)~du-2s\int_Ce^{xu}K(u)~du=0$
$\int_C(u^2-2s)e^{xu}K(u)~du+\int_C2aue^{xu}K(u)~d(xu)=0$
$\int_C(u^2-2s)e^{xu}K(u)~du+\int_C2auK(u)~d(e^{xu})=0$
$\int_C(u^2-2s)e^{xu}K(u)~du+[2aue^{xu}K(u)]_C-\int_Ce^{xu}~d(2auK(u))=0$
$\int_C(u^2-2s)e^{xu}K(u)~du+[2aue^{xu}K(u)]_C-\int_Ce^{xu}(2auK'(u)+2aK(u))~du=0$
$[2aue^{xu}K(u)]_C-\int_Ce^{xu}(2auK'(u)-(u^2-2a-2s)K(u))~du=0$
$\therefore 2auK'(u)-(u^2-2a-2s)K(u)=0$
$2auK'(u)=(u^2-2a-2s)K(u)$
$\dfrac{K'(u)}{K(u)}=\dfrac{u}{2a}+\left(-\dfrac{s}{a}-1\right)\dfrac{1}{u}$
$\int\dfrac{K'(u)}{K(u)}du=\int\left(\dfrac{u}{2a}+\left(-\dfrac{s}{a}-1\right)\dfrac{1}{u}\right)du$
$\ln K(u)=\dfrac{u^2}{4a}+\left(-\dfrac{s}{a}-1\right)\ln u+c_1$
$K(u)=cu^{-\frac{s}{a}-1}e^{\frac{u^2}{4a}}$
$\therefore f(x)=\int_Ccu^{-\frac{s}{a}-1}e^{\frac{u^2}{4a}+xu}~du$
But since the above procedure in fact suitable for any complex number $u$ ,
$\therefore f_n(x)=\int_{a_n}^{b_n}c_n(m_nt)^{-\frac{s}{a}-1}e^{\frac{(m_nt)^2}{4a}+xm_nt}~d(m_nt)={m_n}^{-\frac{s}{a}}c_n\int_{a_n}^{b_n}t^{-\frac{s}{a}-1}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}~dt$
For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:
$\lim\limits_{t\to a_n}t^{-\frac{s}{a}}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}=\lim\limits_{t\to b_n}t^{-\frac{s}{a}}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}$
$\int_{a_n}^{b_n}t^{-\frac{s}{a}-1}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}~dt$ converges
Case $1$ : $a>0$ and $s<0$
For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm i$
$\therefore f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\cos xt~dt$ or $C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\sin xt~dt$
Hence $f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\sin xt~dt+C_2\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\cos xt~dt$
Case $2$ : $a<0$ and $s>0$
For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm1$
$\therefore f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\cosh xt~dt$ or $C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\sinh xt~dt$
Hence $f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\sinh xt~dt+C_2\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\cosh xt~dt$
Compare with $\Phi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)$ and $\Psi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)$ , according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_functions#Integral_representations,
$\Phi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)\propto\int_0^1t^{-\frac{s}{2a}-1}(1-t)^{\frac{s}{2a}-\frac{1}{2}}e^{-ax^2t}~dt$ when $\dfrac{1}{2}>-\dfrac{s}{2a}>0$
$\Psi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)\propto\int_0^\infty t^{-\frac{s}{2a}-1}(1+t)^{\frac{s}{2a}-\frac{1}{2}}e^{ax^2t}~dt$ when $-\dfrac{s}{2a}>0$
A: Since this is a linear second order ODE of the type:
\begin{equation}
y^{''} + A_1(x) y^{'} +A_0(x) y = 0
\end{equation}
it is known that it is reduced to a simpler form with the middle term missing by a substitution $y = \exp(-1/2 \int a_1(x) dx ) u(x)$ with the new function satisfying the following equation:
\begin{equation}
u^{''} - \underbrace{(\frac{A_1(x)^2}{4} + \frac{A_1^{'}(x)}{2} - A_0(x))}_{v(x)} \cdot u=0
\end{equation}
The equation above is called the ${\bf input}$ equation.
Now the idea is to seek solutions of the above equation in terms of known specials functions whose arguments have been changed appropriately. In other words we make a following ansatz $u(x) = m(x) F_{1,2} (\xi(x))$ where 
\begin{equation}
\left(\frac{d^2}{d x^2} + a_1(x) \frac{d}{d x} +a_0(x)\right) F_{1,2}(x)=0
\end{equation}
is the ${\bf target}$ equation whose solutions are assumed to be known. Now by inserting the ansatz into the input equation and setting coefficients at $F$ and at $F^{'}$ to zero we get the follwoing equations for the changed variables:
\begin{eqnarray}
 \frac{m^{'}}{m} &=& \frac{1}{2} \left( a_1 \cdot \xi^{'} - \frac{\xi^{''}} {\xi}\right) \quad (i)\\
v&=& \left(\frac{a_1^2}{4} + \frac{a_1^{'}}{2}-a_0\right)\cdot (\xi^{'})^2 + \left( \frac{3(\xi^{''})^2}{4(\xi^{'})^2} - \frac{\xi^{'''}}{2\xi^{'}}\right)\quad (ii) 
\end{eqnarray}
Now back to our example. We have $v(x) = a+2 s+a^2 x^2$. I suggest to try to map our equation to the confluent hypergeometric equation or to the Whittaker equation which is its symmetric form. Here we have 
\begin{eqnarray}
a_1(x)&=&0 \\
a_0(x)&=&-1/4+\mu/x+(1/4-\nu^2)/x^2
\end{eqnarray}
Now it turns out that if the function $v(x)$ is a rational function equation (ii) has solutions in form of rational functions as well. Theorem 1 on page 3 in [1] gives a recipe on constructing the such solutions. Since our function $v(x)$ is a polynomial of order two it turns out that the solution has to be a polynomial of order two, i.e. $\xi(x) = c_0 + c_1 x + c_2 x^2$. Inserting this into (ii) we get a system of non-linear equations for the parameters $c_0$,$c_1$,$c_2$ which gives a following solution $c_0=0$,$c_1=0$,$c_2=a$,$\mu=-(a+2 s)/(4 a)$,$\nu=1/4$.
Therefore the solutions of the original equation read:
\begin{eqnarray}
y(x) &=& C_1 \cdot \exp(-\frac{a}{2} x^2) \frac{1}{\sqrt{x}} W_{-\frac{a+2 s}{4 a},\frac{1}{4}}(a x^2) + C_1 \cdot \exp(-\frac{a}{2} x^2) \frac{1}{\sqrt{x}} M_{-\frac{a+2 s}{4 a},\frac{1}{4}}(a x^2)\\
&=& C_1 \exp(-a x^2) x U(\frac{3}{4}+\frac{a+2 s}{4 a},\frac{3}{2},a x^2) + 
C_2 \exp(-a x^2) x F_{1,1}(\frac{3}{4}+\frac{a+2 s}{4 a},\frac{3}{2},a x^2)
\end{eqnarray}
Here $W$ and $M$ are the Whittaker functions and $U$ is the confluent hypergeometric function.
[1] M. Bronstein & S. Lafaille, ``Solutions of linear ordinary differential equations in terms of special functions'', Proceedings of ISSAC'2002, Lille, ACM Press, 23-28. https://www-sop.inria.fr/cafe/Manuel.Bronstein/
