Yet another example of linear second order ODEs being reduced to hypergeometric functions.

Question

Let $A$, $B$ and $C$ be real numbers.

By generalizing the answer to Mistake in power series solution of $(x+1)y''-(2-x)y'+y=0$ . We have found a general solution to the following linear second order Ordinary Differential Equation. The equation below: \begin{equation} (x+A) y^{''}(x) + (x+B) y^{'}(x) + C y(x) = 0 \end{equation} is solved by: \begin{eqnarray} y(x) = \exp(-x/2) (A+x)^{\frac{A-B}{2}}\left(C_1 M_{\frac{(A-B+2 C)}{2},\frac{(1+A-B)}{2}}(x+A) + C_2 W_{\frac{(A-B+2 C)}{2},\frac{(1+A-B)}{2}}(x+A)\right) \end{eqnarray} where $M_{\cdot,\cdot}()$ and $W_{\cdot,\cdot}()$ are the Whittaker functions https://en.wikipedia.org/wiki/Whittaker_function .

Now, this time my question will be very humble. Is it possible to find a solution if all the three coefficients meaning at the second, first and the zeroth derivative of the function $y(x)$, if all those coefficients are arbitrary linear functions of $x$.

Answer 1

Let $A$,$A_1$,$B$,$B_1$,$C$ and $C_1$ be real numbers. We consider the following ODE: \begin{equation} \left(A+A_1 x\right) y^{''}(x) + \left( B + B_1 x\right) y^{'}(x) + \left(C+C_1 x\right) y(x)=0 \end{equation} If we follow the procedure described in my answer to Mistake in power series solution of $(x+1)y''-(2-x)y'+y=0$ we get the following solutions: \begin{eqnarray} y(x) = \exp(-\frac{B_1 x}{2 A_1})\cdot \left(A+A_1 x\right)^{\frac{-A_1 B+A B_1}{2 A_1^2}} \cdot \left( C_1 M_{\kappa,\mu}(\frac{\sqrt{\mathfrak D} (A_1 x+A)}{A_1^2}) + C_2 W_{\kappa,\mu}(\frac{\sqrt{\mathfrak D} (A_1 x+A)}{A_1^2}) \right) \end{eqnarray} Here \begin{eqnarray} {\mathfrak D} &=& B_1^2 - 4 A_1 C_1\\ \kappa &=& \frac{-A_1 (2 A C_1+B B_1)+A B_1^2+10 A_1^2}{2 A_1^2 \sqrt{{\mathfrak D}}}\\ \mu &=& -\frac{1}{2} - \frac{-A_1 B+A B_1}{2 A_1^2} \end{eqnarray} and $M_{\kappa,\mu}()$, $W_{\kappa,\mu}()$ are Whittaker functions. The following Mathematica code verifies the solution:

In[847]:= {A, A1, B, B1, CC, C1} = RandomInteger[{1, 10}, 6];
mu =.; kappa =.;
DD = B1^2 - 4 A1 C1;
{kappa, mu} = {(A B1^2 - A1 (B B1 + 2 A C1) + 2 A1^2 CC)/(
   2 A1^2 Sqrt[DD]), -1/2 - (-A1 B + A B1)/(2 A1^2)};

Clear[y]; x =.;
y[x_] = E^(-((B1 x)/(2 A1))) (A + A1 x)^((-A1 B + A B1)/(
   2 A1^2)) (C[1] WhittakerM[kappa, mu, (Sqrt[DD] (A1 x + A))/A1^2] + 
     C[2] WhittakerW[kappa, mu, (Sqrt[DD] (A1 x + A))/A1^2]);
FullSimplify[((A1 x + A) D[y[x], {x, 2}] + (B1 x + B) D[y[x], 
     x] + (C1 x + CC) y[x])]

Out[853]= 0