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I am reading a paper from George Stokes from 1851, and there is a part that I would like help on. He gives the differential equation,

[tex]

\left(\frac{d^2}{dr^2}-\frac{2}{r^2}\right)^2 f(r) = 0

[/tex]

He states that the integral of the equation is

[tex]

f(r) = Ar^{-1} + Br + Cr^2 + Dr^4

[/tex]

I am trying to figure out how he went about the integration. Do you have any suggestions? Here is my workflow so far.

[tex]

\left(\frac{d^2}{dr^2}-\frac{2}{r^2}\right)^2 f(r) = \left(\frac{d^2}{dr^2}-\frac{2}{r^2}\right)\left(\frac{d^2}{dr^2}-\frac{2}{r^2}\right)f(r) = 0

[/tex]

Multiplying the differentials,

[tex]

0 = \left(\frac{d^4}{dR^4} - \frac{d^2}{dR^2}\frac{2}{R^2} - \frac{2}{R}\frac{d^2}{dR^2} + \frac{4}{R^4}\right)f(x)

[/tex]

Simplifying the derivative,

[tex]

0 = f^{IV}(R) - \frac{4}{R^2}f^{II}(R) + \frac{8}{R^3}f^{I}(R) - \frac{8}{R^4}F(R)

[/tex]

I can plug in f(r), f'(r), f''(r), and f''''(r) into this differential equation and prove that f(r) is a solution to the differential equation. Now I tried to integrate this equation.

[tex]

\frac{8}{R^4}f(R) = f^{IV}(R) - \frac{4}{R^2}f^{II}(R) + \frac{8}{R^3}f^{I}(R)

[/tex]

[tex]

\int\frac{8}{R^4}f(R) = \int f^{IV}(R) - \int\frac{4}{R^2}f^{II}(R) + \int\frac{8}{R^3}f^{I}(R)

[/tex]

Now I am stuck with the following section of the equation

[tex]

[...] \int\frac{4}{R^2}f^{II}(R) + \int\frac{8}{R^3}f^{I}(R) [...]

[/tex]

I do not know how to integrate an unknown function f(R) that is multiplied by R. I couldn't find an inverse chain rule or inverse product rule for integrals. Is there a trick to doing this? Is there a simpler path to solve the differential equation? Thanks

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# Solving a differential equation by integration

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