Second order inhomogeneous equation: $y''-2xy'-11y=e^{-ax}$ My question relates to an a second order inhomogeneous equation:
$$y''-2xy'-11y=e^{-ax}$$
First I need to investigate the homogeneous equation:
$$y''-2xy'-11y=0$$
$$y''-2xy'=11y$$
Forms Hermite's Equation where $\lambda = 11$
So I need a general solution of the homogeneous equation
$Ly(x) = 0$.
To do this I need to linearly independent solutions, $y_1(x)$ and $y_2(x)$ say, and
then the general solution of $Ly(x) = 0$.
becomes:
$$Ay_1(x) + By_2(x)$$
where $A$ and $B$ are arbitrary constants.
I am struggling to find two independent solutions of the hermite equation above:
My attempt
If I take a solution to be if the form $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
Then putting this into the ODE I get the following:
$$\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=1}^{\infty}na_{n}x^{n}-11\sum_{n=0}^{\infty}a_{n}x^{n}=0$$
Which I can reduce to:
$$\sum_{n=0}^{\infty}[(n+2)(n+1)a_{n+2}-2(n+\frac{11}{2})a_{n}]x^{n}=0$$
And I have found that rearranging gives:
$$a_{n+2} = \frac{-2(n+\frac{11}{2})a_{n}}{(n+2)(n+1)}$$
This can be used to develop a recurrence relation...
How can I use this to find my two independent solutions $y_1(x)$ and $y_2(x)$ which I need in order to calculate the inhomogeneous solution?
Maybe there are more efficient ways of calculating this ODE that I am unaware of.
Edit
I displayed the equation incorrectly and have since edited it.
Should my solution be in the form of an infinite sum rather than a finite polynomial because I believe that the solution could only be constructed as a series solution which terminates if and only if $\lambda = −2n$ where $n \in \mathbb N$.
In my case $\lambda$ isn't of the above form hence an infinte series is required.
 A: You have correctly found the recurrent relation which gives $a_{n+2}$ as function of $a_{n}$
Hence, if you start with $a_0$ you get all the even coefficients and if you start with $a_1$ you get all the odd coefficients. Hence we get
$$y_1(x)=\sum_{k=0}^\infty a_{2k}x^{2k}\quad \text{and}\quad y_2(x)=\sum_{k=0}^\infty a_{2k+1}x^{2k+1}.$$
This solutions are defined up to a constant ($a_0$ and $a_1$ respectevely) that you can arbitrarly choose (but not null, otherwise we get the trivial solution).
Using this ansatz we are implicitly supposing that the solution is analytic in some interval. Hence you can check that $y_1-c y_2 \equiv 0$ means that each coefficient of the series is $0$ and hence both $y_1$ and $y_2$ should be zero. Hence if we choose $a_0,a_1 \neq 0$ we get independent solutions.

We could use the 2 homogeneus solutions with the method of variation of parameters to get the inhomogeneous solution.
Otherwise we could expand $e^{-a x}=\sum_{k=0}^\infty \frac{(-a)^k x^k}{k!}$ and we find the coefficients $a_i$ terms by terms.
A: Approach $1$: intrgral solution
First consider $y_c''-2xy_c'-11y_c=0$ :
Similar to Help on solving an apparently simple differential equation,
Let $y_c=\int_Ce^{xs}K(s)~ds$ ,
Then $(\int_Ce^{xs}K(s)~ds)''-2x(\int_Ce^{xs}K(s)~ds)'-11\int_Ce^{xs}K(s)~ds=0$
$\int_Cs^2e^{xs}K(s)~ds-2x\int_Cse^{xs}K(s)~ds-11\int_Ce^{xs}K(s)~ds=0$
$\int_C(s^2-11)e^{xs}K(s)~ds-\int_C2se^{xs}K(s)~d(xs)=0$
$\int_C(s^2-11)e^{xs}K(s)~ds-\int_C2sK(s)~d(e^{xs})=0$
$\int_C(s^2-11)e^{xs}K(s)~ds-[2se^{xs}K(s)]_C+\int_Ce^{xs}~d(2sK(s))=0$
$\int_C(s^2-11)e^{xs}K(s)~ds-[2se^{xs}K(s)]_C+\int_Ce^{xs}(2sK'(s)+2K(s))~ds=0$
$-~[2se^{xs}K(s)]_C+\int_C(2sK'(s)+(s^2-9)K(s))e^{xs}~ds=0$
$\therefore2sK'(s)+(s^2-9)K(s)=0$
$2sK'(s)=(9-s^2)K(s)$
$\dfrac{K'(s)}{K(s)}=\dfrac{9}{2s}-\dfrac{s}{2}$
$\int\dfrac{K'(s)}{K(s)}ds=\int\left(\dfrac{9}{2s}-\dfrac{s}{2}\right)ds$
$\ln K(s)=\dfrac{9\ln s}{2}-\dfrac{s^2}{4}+c_1$
$K(s)=cs^\frac{9}{2}e^{-\frac{s^2}{4}}$
$\therefore y_c=\int_Ccs^\frac{9}{2}e^{-\frac{s^2}{4}+xs}~ds$
But since the above procedure in fact suitable for any complex number $s$ ,
$\therefore y_{c,n}=\int_{a_n}^{b_n}c_n(m_nt)^\frac{9}{2}e^{-\frac{(m_nt)^2}{4}+xm_nt}~d(m_nt)={m_n}^\frac{9}{2}c_n\int_{a_n}^{b_n}t^\frac{9}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}~dt$
For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:
$\lim\limits_{t\to a_n}t^\frac{11}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}=\lim\limits_{t\to b_n}t^\frac{11}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}$
$\int_{a_n}^{b_n}t^\frac{9}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}~dt$ converges
For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm1$
$\therefore y_c=C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\cosh xt~dt$ or $C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\sinh xt~dt$
Hence $y_c=C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\sinh xt~dt+C_2\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\cosh xt~dt$
