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

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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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  • #2
Simon Bridge
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It's what I would have done.
[edit: I'd have probably made some algebra errors too :) ]

Though sometimes you are provided with exercises which can be done a simple way, then asked to do it the complicated way, in order to give you practice at doing it the complicated way... for when you have a harder problem.
 
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  • #3
ehild
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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]
You made a mistake. It should be

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


So the steady state solution is

[tex]y=C_1e^{-t} \cos 2t + C_2e^{-t}\sin 2t - \sin t - 2 \cos t [/tex]
That is the general solution if you correct your mistakes with the signs.
The steady-state solution (particular solution) is yp = sint+2cost. The general solution is the sum of the general solution of the homogeneous equation and the particular solution yp.

ehild
 
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@Simon Bridge -- yeah when I tried going through the whole process with the "complicated" method -- essentially following through this set of formulas --

$$\alpha_n = \frac{A_n a_n - B_nb_n}{A_n^2+B_n^2},\ A_n = k-\mu\left(\frac{n\pi}{p}\right)^2, \beta_n = \frac{A_n b_n + B_na_n}{A_n^2+B_n^2},\ A_n = k-\mu\left(\frac{n\pi}{p}\right)^2$$

and when i went through the whole routine to get the other Fourier coefficients I got a solution in terms of sin and cos but it was kind'a messy.

@echild -- thanks - I keep getting confused between which solution is which. Stupid of me.
 
  • #5
SteamKing
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Also, contrary to the title of this thread, you have an ODE, not a PDE.
 
  • #6
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well, it's in a PDE text :-) Asmar as it happens.
 

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