The use of a known solution to find another

Problem:

$y'' - \frac{1}{2x}y' - \frac{1}{x^2}y = 0$

Solution:

A good guess would be $y = x^{k+2}$ with an unknown $k$, the $k + 2$ is really just to make the algebra simpler.

$\begin{eqnarray*} y'' - \frac{1}{2x}y' - \frac{1}{x^2}y &=& 0 \\ (k+2)(k+1)x^k - \frac{1}{2x}(k+2)(x^{k+1}) - \frac{1}{x^2}x^{k+2} &=& 0 \\ (k+2)(k+1) - \frac{k+2}{2} - 1 &=& 0 \\ 2(k+2)(k+1) - (k+2) - 2 &=& 0 \\ 2k^2 + 5k &=& 0 \\ k &=& 0 \text{ or } \frac{-5}{2} \\ \end{eqnarray*}$

So the answer is $y = Ax^2 + \frac{B}{\sqrt{x}}$

For the sake of exercise, let's assume we do not know the second independent answer $y = \frac{1}{\sqrt{x}}$. Can we use the answer $y = x^2$ to derive it?

Here is a technique for using an existing solution for another one. Assume an answer for $y'' + p(x)y' + q(x)y = 0$ is $y_1(x)$, we let $y_2(x) = v(x)y_1(x)$ and substitute this back to the equation, we get:

$\begin{eqnarray*} y'' - py' - qy &=& 0 \\ (v''y_1 + 2v'y_1' + vy_1'') + p(v'y_1 + vy_1') + q(vy_1) &=& 0 \\ v''y_1 + 2v'y_1' + p(v'y_1) + vy_1'' + p(vy_1') + q(vy_1) &=& 0 \\ v''y_1 + 2v'y_1' + p(v'y_1) &=& 0 \\ \frac{v''}{v'} + 2\frac{y_1'}{y_1} + p &=& 0 \\ \frac{v''}{v'} &=& - 2\frac{y_1'}{y_1} - p \\ \log v' &=& -2 \log y_1 - \int p dx \\ v' &=& e^{-2 \log y_1 - \int p dx} \\ &=& \frac{1}{y_1^2}e^{-\int p dx} \\ v &=& \int \frac{1}{y_1^2}e^{-\int p dx} dx \end{eqnarray*}$

So we will need to do is to compute the two integral for our special problem.

$\begin{eqnarray*} v &=& \int \frac{1}{y_1^2}e^{-\int p dx} dx \\ &=& \int \frac{1}{x^4}e^{-\int \frac{-1}{2x} dx} dx\\ &=& \int \frac{1}{x^4}e^{-(\frac{-1}{2}\log x)} dx \\ &=& \int \frac{1}{x^4}e^{\frac{1}{2}\log x} dx \\ &=& \int \frac{1}{x^4}e^{\log \sqrt{x}} dx \\ &=& \int \frac{1}{x^4}\sqrt{x} dx \\ &=& \int x^{-3.5} dx \\ &=& \frac{x^{-2.5}}{2.5} \\ y_2 &=& vy_1 \\ &=& \frac{x^{-2.5}}{2.5}x^2 \\ &=& \frac{x^{-0.5}}{2.5} \end{eqnarray*}$

As constant can be ignored, now we come back to full circle. In general, I guess the integrals are going to be hard to evaluate. This problem seems rather tailored for the purpose of practicing.