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**Solving a "simple" second order PDE, do I need the Fourier?**

## Homework Statement

The problem as given:

[itex]y'' + 2y' + 5y = 10\cos t[/itex]

We want to find the general solution and the steady-state solution. We're using [itex]\mu y'' + c y' + k y = F(t)[/itex] as our general form.

OK, so I first want the general solution to the 2nd order DE. That should be easy: the characteristic equation is [itex]\lambda^2 + 2\lambda + 5[/itex] and it has imaginary roots at [itex]-1 \pm \sqrt{-4}[/itex] which is [itex]-1\pm 2i[/itex].

So the general solution should be [itex]y=C_1e^{-t} \cos 2t + C_2e^{-t}\sin 2t[/itex]

The steady state solution is given by the Fourier series [itex]y_p(t) = \alpha_0 + \sum_{n=0}^\infty \left( \alpha_n\cos \frac{n\pi}{p} t + \beta_n \sin \frac {n\pi}{p}t\right)[/itex]

But we needn't be that complicated, it seems to me, since could try looking for solutions in the form of [itex]y_p = A \sin t + B \cos t[/itex].

We differentiate and we get [itex]y_p' = A \cos t - B \sin t[/itex] and [itex]y_p'' = -A \sin t - B \cos t[/itex].

Substitute in [itex]y_p[/itex] to the original equation. We get [tex]-A \sin t - B \cos t + 2(A \cos t - B \sin t)+ 5(A \sin t + B \cos t) = 10 \cos t[/tex]

And simplifying we have

[tex](4A - 2B) \sin t - (4B+ 2A) \cos t = 10 \cos t[/tex]

And that leaves us with [itex](4A - 2B) = 0[/itex] and [itex](-4B - 2A) = 10[/itex]. Solving for both we get [itex]B = -2[/itex] and [itex]A =-1[/itex].

So the steady state solution is

[tex]y=C_1e^{-t} \cos 2t + C_2e^{-t}\sin 2t - \sin t - 2 \cos t [/tex]

Now here's the part where everyone tells me I messed this up :-) I just feel I have done something way off here. But it seems ok... I just wonder if I should have been following through on the whole theorem in the text that says how to solve equations like this, which is much more complex but presumably would yield the same answer.

Anyhow, thanks for bearing with me.

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