So I have this question:

Solve the initial value problem:

$$\begin{align*} \frac{dx_1}{dt}&=3x_3-2x_4,\\ \frac{dx_2}{dt}&=-2x_3+3x_4,\\ \frac{dx_3}{dt}&=3x_1-2x_2,\\ \frac{dx_4}{dt}&=-2x_1+3x_2,\\ \end{align*} $$ with $$ \begin{align*} x_1(0) &= 1,\\ x_2(0) &= 2,\\ x_3(0) &= -1,\\ x_4(0) &= 3. \end{align*} $$

I know the first step is to form the coefficient matrix and then I have to find the eigenvalues....but I really just know the idea of what I'm supposed to do...I don't actually know how to start. Where do I even begin with this problem?

Edit: Okay so I know
$$A=\pmatrix{0&0&3&-2\cr0&0&-2&3\cr3&-2&0&0\cr-2&3&0&0\cr}$$ and to find the eigenvalues I find the determinant of this matrix:

which gives me the eigenvalues: $\lambda=-5,5,-1,1$

Which after pluggin in and solving gives me the eigenvectors:
$v_1=(1,-1,-1,1)\\ v_2=(-1,1,-1,1)\\ v_3=(-1,-1,1,1)\\ v_4=(1,1,1,1) $

But I just don't know WHAT to do with those vectors...Or am i on the wrong track?

Edit2: Think I figured it out, all thanks to Gerry Myerson

We have $X=\left[ \begin{array}{ccc} c_1e^{a_1t} & -c_2e^{a_2t} & -c_3e^{a_3t} & c_4e^{a_4t} \\ -c_1e^{a_1t} & c_2e^{a_2t} & -c_3e^{a_3t} & c_4e^{a_4t} \\ -c_1e^{a_1t} & -c_2e^{a_2t} & c_3e^{a_3t} & c_4e^{a_4t} \\ c_1e^{a_1t} & c_2e^{a_2t} & c_3e^{a_3t} & c_4e^{a_4t} \end{array} \right]$

Then I just use the inital values:
$X(0)=\left[ \begin{array}{ccc} c_1 & -c_2 & -c_3 & c_4 \\ -c_1 & c_2 & -c_3 & c_4\\ -c_1 & -c_2 & c_3 & c_4 \\ c_1 & c_2 & c_3 & c_4 \end{array} \right]= \left[ \begin{array}{ccc} 1 \\ 2\\ -1 \\ 3 \end{array} \right]$

Solving that gives these values to the constants: $c_1=\frac{3}{4}, c_2=\frac{5}{4}, c_3=-\frac{1}{4}, c_4=\frac{5}{4}$

And finally putting it all together, my solution is: $X=\left[ \begin{array}{ccc} \frac{3}{4}e^{a_1t} -\frac{5}{4}e^{a_2t}+ \frac{1}{4}e^{a_3t}+\frac{5}{4}e^{a_4t} \\ -\frac{3}{4}e^{a_1t}+\frac{5}{4}e^{a_2t}+\frac{1}{4}e^{a_3t}+\frac{5}{4}e^{a_4t} \\ -\frac{3}{4}e^{a_1t}-\frac{5}{4}e^{a_2t}-\frac{1}{4}e^{a_3t}+\frac{5}{4}e^{a_4t} \\ \frac{3}{4}e^{a_1t}+\frac{5}{4}e^{a_2t}-\frac{1}{4}e^{a_3t}+\frac{5}{4}e^{a_4t} \end{array} \right]$

  • $\begingroup$ You don't have any lecture notes? Textbook? $\endgroup$ Commented Nov 29, 2012 at 22:54

2 Answers 2


OK, so you've come as far as $x'=Ax$ where $$A=\pmatrix{0&0&3&-2\cr0&0&-2&3\cr3&-2&0&0\cr-2&3&0&0\cr}$$ Next step is to find the eigenvalues and eigenvectors of $A$. If there are $4$ linearly independent eigenvectors $v_1,v_2,v_3,v_4$ with corresponding eigenvalues $a_1,a_2,a_3,a_4$ then the general solution to $x'=Ax$ is $$x=c_1e^{a_1t}v_1+c_2e^{a_2t}v_2+c_3e^{a_3t}v_3+c_4e^{a_4t}v_4$$ where $c_1,\dots,c_4$ are arbitrary constants: then you can use the initial conditions you've been given to work out the values of $c_1,\dots,c_4$.

If there aren't $4$ linearly independent eigenvectors, life is more difficult. So, which is it?

  • $\begingroup$ can you please take a look at my editted post, i've shown some more work I've done. Is that right, is so where do I go from there? Thanks $\endgroup$
    – Richard
    Commented Nov 30, 2012 at 16:32
  • $\begingroup$ You have the eigenvectors and eigenvalues, and I have given you the formula for the general solution. So you write that down and let $t=0$ and use the initial conditions (which were part of the statement of the problem) to work out the values of $c_1,\dots,c_4$. $\endgroup$ Commented Nov 30, 2012 at 21:49
  • $\begingroup$ so you figure if the 4 eigenvectors are linearly independent, then use that formula $x=c_1e....$ but I don't understand how that gives me $x_1, x_2, x_3, x_4$ separately. I just not too sure on this whole process $\endgroup$
    – Richard
    Commented Dec 1, 2012 at 5:57
  • $\begingroup$ No, you don't figure if the eigenvectors are linearly independent --- eigenvectors belonging to distinct eigenvectors are guaranteed to be linearly independent. $x_1,\dots,x_4$ are the components of the vector $x$. $\endgroup$ Commented Dec 1, 2012 at 11:56
  • $\begingroup$ I think I actually got it this time. Can you verify my work in my edited post? $\endgroup$
    – Richard
    Commented Dec 1, 2012 at 16:41

This is a linear differential equation. Let

$$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}.$$

Then, you have

$$\dot{\mathbf{x}} = A \mathbf{x}$$

for some matrix $A$. Now, use the properties of matrix-vector multiplication to fill in the entries of $A$, for instance $$\dot{x}_1 = a_{11}x_1 + a_{12}x_2 + a_{13}x_3 + a_{14}x_4.$$

Can you find $A$? Then, can you find its eigenvalues? Do you know about matrix exponentials?

  • $\begingroup$ yah A is just $\left( \begin{array}{ccc} 0 & 0 & 3 & -2 \\ 0 & 0 & -2 & 3 \\ 3 & -2 & 0 & 0 \\ -2 & 3 & 0 & 0 \end{array} \right)$ but then where do I go from there? $\endgroup$
    – Richard
    Commented Nov 29, 2012 at 22:33

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