2nd order differential equation with missing y' I have the following 2nd order differential equation:
$$y'' + p(x) y =0, \tag{1}$$
where $p(x)$ involves only first order of $x$, for example, $p(x)=ax+b$. Any suggestion how to obtain or guess a solution for (1)? Thanks.
 A: This does not seem (for me) to be so easy to solve in a nice way. I suppose there might be a substitution, but I can't think of any. This, however, can most certainly be solved by assuming that $y(x)$ is analytic on reasonable domain, and so with that assumption we can write out $y(x)$ as a power series and $y'(x)$ as its derivative.
$$y(x)=a_0 + a_1x + a_2x^2 + ...$$
$$y''(x)=2a_2+6a_3x+12a_4x^2+...$$
$$p(x)=ax+b$$
And now we solve:
$$\left(2a_2+6a_3x+12a_4x^2+...\right)+\left(ax+b\right)\left(a_0 + a_1x + a_2x^2 + ...\right)=0$$
$$\left(2a_2+ba_0\right)+\left(6a_3+aa_0+ba_1\right)x+\left(12a_4+aa_1+ba_2\right)x^2+\left(20a_5+aa_2+ba_3\right)x^3+...=0$$
Now we have the following equations to solve:
$$\begin{align}
2a_2+ba_0&=0\\
6a_3+aa_0+ba_1&=0\\
12a_4+aa_1+ba_2&=0\\
20a_5+aa_2+ba_3&=0\\
\vdots\\
(n)(n-1)a_n+aa_{n-3}+ba_{n-2}&=0\\
\vdots
\end{align}
$$
And if you confidently plow through those equations, you will arrive at your general analytic solution for $y(x)$ in terms of $a$,$b$, and two particular $a_j$'s of your choice.
