We have in the form $x'' = Ax$:

$\left[ \begin{array}{cccc} x_1''\\x_2'' \end{array} \right] = \left[ \begin{array}{cccc} -10&6\\6&-10 \end{array} \right]\left[ \begin{array}{cccc} x_1\\x_2 \end{array} \right]$

The eigenvalues of the matrix $A$ are $\lambda = -4$ and $\lambda = -16$, with corresponding eigenvectors $v_1 = \left[ \begin{array}{cccc} 1\\-1 \end{array} \right]$ and $v_2 = \left[ \begin{array}{cccc} 1\\1 \end{array} \right]$. Write the general solution to the homogeneous problem.

Why wouldn't the answer just be $x = c_1e^{-4t}\left[ \begin{array}{cccc} 1\\-1 \end{array} \right] + c_2e^{-10t}\left[ \begin{array}{cccc} 1\\1 \end{array} \right]$?

What am I not understanding about the problem that actually makes the solution $x = c_1\left[ \begin{array}{cccc} 1\\-1 \end{array} \right]\cos2t+c_2\left[ \begin{array}{cccc} 1\\-1 \end{array} \right]\sin2t+c_3\left[ \begin{array}{cccc} 1\\1 \end{array} \right]\cos4t+c_4\left[ \begin{array}{cccc} 1\\1 \end{array} \right]\sin4t$?

Anyone know how to do this?

  • $\begingroup$ What you are overlooking is that those are second derivatives. What you propose as the answer would work if they were first derivatives. $\endgroup$ – Gerry Myerson Nov 12 '12 at 5:36
  • $\begingroup$ How can I solve for the second order equations, then? I'm just confused what the values $\lambda = -4$, $\lambda = -16$ and $v_1$ and $v_2$ do for me. I know I can obtain sines and cosines if my eigenvalues are complex, but I see no way in which this would apply for this problem. $\endgroup$ – Bob John Nov 12 '12 at 5:39

Your technique of finding eigenvector/eigenvalues to compose the general solution is to the first order homogeneous problem of the form $x'=Ax$. However, here we have a second order problem, of the form $x''=Ax$.

So, to employ your technique, we will first need to rewrite the system in a first order fashion. This is a common technique of reduction of order.

Consider the variables $z_1=x_1,z_2=x_1', z_3=x_2, z_4=x_2'$, and write vector $z$ as $z=\pmatrix{z_1 \\ z_2 \\ z_3 \\ z_4}$. Then notice that its derivative is $z'=\pmatrix{z_1' \\ z_2' \\ z_3' \\ z_4'}=\pmatrix{x_1' \\ x_1'' \\ x_2' \\ x_2''}$.

We can create a matrix $B$ such that $z'=Bz$, from your original problem : Since $\pmatrix{x_1''\\x_2''}=\pmatrix{-10 & 6 \\ 6 & -10}\pmatrix{x_1 \\ x_2}$, we see that we have $\pmatrix{x_1' \\ x_1'' \\ x_2' \\ x_2''}=\pmatrix{0&1&0&0\\-10&0&6&0 \\ 0&0&0&1 \\ 6 &0&-10&0}\pmatrix{x_1 \\ x_1' \\ x_2 \\ x_2'}$. Hence $z'=Bz$ for $B=\pmatrix{0&1&0&0\\-10&0&6&0 \\ 0&0&0&1 \\ 6 &0&-10&0}$.

Now we have reduced your problem to a first order problem, $z'=Bz$, apply your technique of finding eigenvalues/eigenvectors to construct the general solution to $z$.

Note: I will leave this eigenproblem to you, but you should have found the eigenvalues of $B$ to be $+2i, -2i, +4i, -4i$.

As you write down your general solution of the form $z=c_1 v_1(t)+c_2 v_2(t)+c_3 v_3(t)+c_4 v_4(t)$, remember that we are actually not looking for $z=\pmatrix{x_1\\x_1'\\x_2\\x_2'}$ , but rather the vector $x=\pmatrix{x_1\\x_2}$, which are the first and third component of $z$. So you will just copy down the first and third row of your solution to $z$ to compose the solution for $x$.

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  • $\begingroup$ How do I find the eigenvalues for $B$? Is there any easier way than doing the det() of a 4x4 matrix? $\endgroup$ – Bob John Nov 12 '12 at 6:14
  • $\begingroup$ That is the most direct way to go, solve the roots to its characteristic polynomial $p(\lambda)=det(B-\lambda I)$. Persevere, this matrix has several zeros in the first row, so computing this determinant by cofactor expansion isn't as bad as you think. $\endgroup$ – bonsoon Nov 12 '12 at 6:19
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    $\begingroup$ To help you check your work, this characteristic polynomial turns out to be $\lambda^4+20 \lambda^2+64$, (note well that this is a quadratic in $\lambda^2$) which we factor it into $(\lambda^2+4)(\lambda^2+16)$, giving us the desired roots. $\endgroup$ – bonsoon Nov 12 '12 at 6:27
  • $\begingroup$ Why did they give eigenvalues and eigenvectors? Where did they help in your solution? $\endgroup$ – Bob John Nov 12 '12 at 7:37

Take $x = ve^{\alpha t}$.

$\alpha^2v=Av$ so that $\alpha^2 = \lambda$, the eigenvalues of the system.

Thus, $\alpha = \pm2i$ or $\alpha = \pm4i$. Separating real and imaginary parts of the solutions $x=ve^{\alpha t}$ gives four linearly independent solutions:

$x = c_1\left[ \begin{array}{cccc} 1\\-1\end{array} \right]\cos2t+c_2\left[ \begin{array}{cccc} 1\\-1\end{array} \right]\sin2t+c_3\left[ \begin{array}{cccc} 1\\1\end{array} \right]\cos4t+c_4\left[ \begin{array}{cccc} 1\\1\end{array} \right]\sin4t$

Does anyone care to elaborate on this solution? Particularly, how do we deduce $\alpha^2v = Av$ and the corresponding $\alpha$ values?

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  • $\begingroup$ The point is that if $A$ has a negative eigenvalue $\lambda$ for eigenvector $v$, then $\cos(\sqrt{-\lambda} t) v$ and $\sin(\sqrt{-\lambda} t) v$ are two linearly independent solutions of $x'' = A x$. Different $\lambda$'s will give linearly independent $v$'s, and so again linearly independent solutions. If the $n \times n$ matrix $A$ has $n$ linearly independent eigenvectors for negative eigenvalues, that gives you $2n$ linearly independent solutions, and thus a complete set of solutions. $\endgroup$ – Robert Israel Nov 12 '12 at 9:03
  • $\begingroup$ Why is the first sentence the case for second-order ODEs? $\endgroup$ – Bob John Nov 12 '12 at 9:09
  • $\begingroup$ Why is what the case? Why do you look for solutions of the form $e^{\alpha t} v$? Because that's the simplest form that has a good chance of working for a constant-coefficient linear system, and it does turn out to work here. $\endgroup$ – Robert Israel Nov 12 '12 at 9:16
  • $\begingroup$ If you want something more profound, it's because this has the right behaviour under the symmetry $t \to t + c$. $e^{\alpha (t+c)}$ is a constant times $e^{\alpha t}$, and a constant multiple of a solution is a solution for any homogeneous linear system. $\endgroup$ – Robert Israel Nov 12 '12 at 9:21
  • $\begingroup$ I'm just confused, because given the initial problem, how would I deduce that because $A$ had negative eigenvalues, the solutions to the second-order ODE had to be complex? If the eigenvalues had been positive, how would this have changed things? $\endgroup$ – Bob John Nov 12 '12 at 9:23

In MATLAB, use the following code to find the Eigenvalues of B:


This stores all the eigenvalues of B in the stack 'd' as a column vector.

or[V,D]=eig(B) will give the two matrices consisting the eigenvectors and the eigenvalues of B respectively.

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