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I am interested in solving the general solution for the following set of equations:

$f'(t)=g(t)$ $g'(t)=-2f(t)-\frac{9}{4}k(t)$ $h'(t)=-f(t)+2k(t)$ $k'(t)=-2g(t)-h(t)$

How can I get the general solution here?

So far I get to $f(t)=\int \! g'(t)dt=\int-2f(t)-\frac{9}{4}k(t)dt$ And then I'm lost

Thanks for all your help!

1 Answers 1

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Let $y=(f,g,h,k)$. Then you can write your system as $y'=Ay$ for a certain matrix $A$. If $A$ is diagonalizable, say it has eigenvectors $u,v,w,x$ with eigenvalues $a,b,c,d$, respectively, then your system has general solution $y=r_1e^{at}u+r_2e^{bt}v+r_3e^{ct}w+r_4e^{dt}x$ where $r_1,\dots,r_4$ are arbitrary constants.

If $A$ is not diagonalizable, things get messier, but it's still doable. The "complex eigenvalue" case is discussed in some detail at this link, also at this link.

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    Ah, I did not pay attention to that. Thank you!2012-11-13