# How can I solve this set of linear differential equations?

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

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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.
Hi Gerry! Can you elaborate please? are you saying that $y=(f,g,h,k)$ is a 1x4 matrix? If so, then $A$ can only be a 1x1 'matrix' and then I'm totally confused. – Kashif Oct 26 '12 at 6:43
$y$ is a column vector. $A$ is a $4\times4$ matrix. – Gerry Myerson Oct 26 '12 at 11:53
So it looks like my matrix A is not diagonalizable (its eigenvalues only have 2 distinct real roots). What are my next steps? $A = [0,1,0,0;-2,0,0,-9/4;-1,0,0,2;0,-2,-1,0]$ – Kashif Nov 10 '12 at 0:47
Does the matrix also have a pair of conjugate nonreal eigenvalues, $a\pm bi$? With conjugate nonreal eigenvectors, $s\pm it$? There will be solutions involving $e^{rt}\cos\omega t s$ and the like, with $r=\sqrt{a^2+b^2}$, $\omega=\arctan(b/a)$. You can find the details in any treatment of constant coefficient, homogeneous, linear differential equations --- I'm sorry, I don't have it at my finger tips at present. – Gerry Myerson Nov 10 '12 at 6:57
Hi Gerry, it does have a pair of conjugate nonreal eigenvalues: $-1.1726 + 1.0607i, -1.1726 - 1.0607i, 1.1726 + 1.0607i & 1.1726 - 1.0607i$. Do you have a link for a website that has the general solution listed out? I would greatly appreciate it. – Kashif Nov 12 '12 at 20:56