Question

Differential equation help! trying to follow how the textbook solves this problem

http://tinypic.com/r/2vrs96x/8

I don't see how substituting equation 5.85 and 5.86 into 5.84 will help solve the differential equation and lead to 5.87

Answer 1

It looks to me like an attempt at using a variation of undetermined coefficients, but they leave out all the details.

The nice part about this method is that you can use the superposition principle to your advantage. As an example, consider \(y''+y=x+x^2\). When solving for the nonhomogeneous solution, you can find the coefficients of your guess solutions by considering them separately; in this case, \(y_1=A_1x+A_2\) and \(y_2=B_1x^2+B_2x+B_3\), or whatever you would prefer to use.

That's what (5.86) tells me, at any rate. The RHS of the initial equation contains a term of \(\cos2\phi\), so there must be a term of \(\cos2\phi\) in \(x_1\).

In other words, they consider
\[\frac{d^2x_1}{d\phi^2}+x_1=\color{red}{\frac{1}{2}e^2}\cos2\phi\]
and they suppose \(\hat{x}=\cos2\phi\) is a solution. Substituting \(\hat{x}\) for \(x_1\) yields (5.86):
\[\frac{d^2}{d\phi^2}(\cos2\phi)+\cos2\phi=\color{red}{-3}\cos2\phi\]
For the RHS's to match up, we have to adjust \(\hat{x}\):
\[\frac{d^2}{d\phi^2}\left(-\frac{1}{3}\cos2\phi\right)-\frac{1}{3}\cos2\phi=\cos2\phi\\
\frac{d^2}{d\phi^2}\left(-\frac{1}{6}e^2\cos2\phi\right)-\frac{1}{6}e^2\cos2\phi=\frac{1}{2}e^2\cos2\phi\]
Now the new \(\hat{x}\) is exactly what you see in (5.87), and the RHS matches up with the RHS \(\cos2\phi\) term in (5.84).

I think (5.85) could be explained similarly. If you don't trust or like this method, try solving for the nonhomogeneous part as you normally would with UC.

Answer 2

Oh, I deliberately left out \(\dfrac{3G^2M^2}{L^2}\) because it's a constant; just remember to factor it into the final solution.

Answer 3

Thanks it seems straightforward now that I am looking at the problem correctly