Write the general solution of the homogeneous problem for the system 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?
 A: 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 $x$.
A: 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?
A: In MATLAB, use the following code to find the Eigenvalues of B:
d=eig(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.
