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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!

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1 Answer 1

up vote 1 down vote accepted

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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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

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