# Thread: Using eigenvectors and eigenvalues to solve DE in matrix form

1. ## Using eigenvectors and eigenvalues to solve DE in matrix form

Hi, my friend asked for help with a question and I'm wondering if there's a better method than mine by the question wording.

We have the matrix
$A =\left( \begin{array}{cc}
2 & 1 \\
1 & 2
\end{array}\right)$

and have to find the eigen values and vectors which are
3 with $\left( \begin{array}{c}
1 \\
1
\end{array} \right)$

and
1 with $\left( \begin{array}{c}
-1 \\
1
\end{array}\right)$

The next part is to solve the set of equations

$u_1'' = 2 u_1' + u_2'$
$u_2'' = u_1' + 2 u_2'$
$u_1(0) = 0$
$u_1'(0) = 1$
$u_2(0) = 1$
$u_2'(0) = 0$
using the eigenvalues and eigenvectors.

As this is of the matrix form I solve this by writing characteristic equations with the eigenvalues as the exponentials
$u_1 = A e^{t} + B e^{3t} + C$
$u_2 = D e^{t} + E e^{3t} + F$

and substituting into the above equations to get six equations in A,B,C,D,E and F thus leading to
$u_1 = \frac{1}{6}\left( 3e^{t} + e^{3t} -4 \right)$
$
u_2 = \frac{1}{6}\left( -3e^{t} + e^{3t} +8 \right)
$

which is right!

However, I haven't used the eigenvectors only the eigenvalues, is there another (better ? ) method that I should be using?

Thanks!

2. Hi

You can write the set of equations as
$U'' = AU'$ where $U = \left( \begin{array}{c}
u_1 \\
u_2
\end{array} \right)$

A is diagonalisable into the diagonal matrix $D =\left( \begin{array}{cc}
3 & 0 \\
0 & 1
\end{array}\right)$
through a transition matrix P such that $P^{-1}AP = D$

Let $V = \left( \begin{array}{c}
v_1 \\
v_2
\end{array} \right) = P^{-1}U \Rightarrow U = PV$

$U'' = AU'$ is transposed into $PV'' = APV' \Rightarrow V'' = P^{-1}APV' = DV'$

Since D is diagonal $V'' = DV'$ is easily solved.

Then $U = PV$ is used to find $u_1$ and $u_2$