I want to solve the following second order ODE with non-constant coefficients: $$y^{\prime\prime} + (ax+b) y^{\prime} + (cx+d)y = 0,\tag{1}$$
where $a,b,c,d\in\mathbb{R}$. From what I understand, its solution involves confluent hypergeometric functions of the first and second kind; e.g., see this post. In particular, it appears that finding the solution of (1) involves finding the solution of the degenerate hypergeometric equation
$$xy^{\prime\prime} +\left(\frac{1}{2} -x\right)y^{\prime} -ey=0,\tag{2}$$
where $e\in\mathbb{R}$ is some appropriately determined constant (not to be confused with Euler's number). Alas, I could not quite figure out the procedure for finding the general solution from that post.
Not only would I like to understand the solution of (1), but I would like to know how to "reduce" the problem from solving (1) to solving (2). (The Wikipedia page on the confluent hypergeometric function lays out the steps for this reduction when we have a non-constant affine coefficient on the $y^{\prime\prime}$ term in (1), but not when the coefficient is a constant.) I have skimmed over the book by Abramowitz and Stegun but didn't find it particularly helpful, although I'm sure that my question is embedded in their development. Other references would be delightful.
Best Answer
The equation you posted is much more complex than the one in the linked post. $$y^{\prime\prime} + (ax+b)\,y^{\prime} + (cx+d)\,y = 0$$ The idea is first to find a transform such that the multiplying factor of $y$ does not depend on $x$.
So, let $y=z\,e^{f(x)}$. Cheating a little, I found that $$f(x)= -\frac{a }{2}x^2+ \left(\frac{c}{a}-b\right)x$$ does the job.
Replacing, we end with $$z''-(ax+\alpha)z'+\beta z=0\tag 1$$ where $$\alpha=b-\frac{2 c}{a} \qquad \text{and} \qquad \beta=\frac{-a^3+a^2 d-a b c+c^2}{a^2}$$ As you can see here changing variable, we have Hermite differential equation.
So the general solution of $(1)$ is a linear combination of a confluent hypergeometric function of the first kind and an Hermite polynomial.