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.
$ 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.
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