# Matrix Exponentials to solve Differential Equations...

• Nov 23rd 2009, 04:11 AM
Unenlightened
Matrix Exponentials to solve Differential Equations...
Define the exponential of a matrix A to be $e^{A} = I + A + \frac{1}{2!}A^{2}+\frac{1}{3!}A^{3}+...$.

Let A be the following matrix $\left(\begin{array}{cc}3 & -2 \\ -2 & 3\end{array}\right)$. Show that the eigen values of A are $\lambda_{1} = 1$ and $\lambda_{2} = 5$ with associated eigenvectors $\left(\begin{array}{c}1\\1\end{array}\right),\left (\begin{array}{c}1\\-1\end{array}\right)$. Hence or otherwise calculate $e^{A}$.

Use previous to solve the system of differential equations (with respect to t)
$y'_{1}=3y_{1} + -2y_{2}$
$y'_{2}=-2y_{1}+3y_{2}$
with the initial conditions $y_{1}(0)=2$, and $y_{2}(0)=0$, by writing the system as
$Y'=AY, Y(0)=Y_{0}$
for the matrix A above, and $Y_{0} = \left(\begin{array}{c}2\\0\end{array}\right)$ and noting that a solution is $Y(t)=e^{tA}Y_{0}$

I can do the eigenvalues and eigenvectors part, and can calculate $e^{A}$ since $e^{A}=Xe^{D}X^{-1}$,
but I don't get the differential equations part...

Any help would be much appreciated...
• Nov 23rd 2009, 09:21 AM
Moo
Hello,

As it is said, the system can be written $Y'=AY$, where $Y=\begin{pmatrix} y_1(t) \\ y_2(t) \end{pmatrix}$
We define $Y'$ as $\begin{pmatrix}y_1'(t) \\ y_2'(t)\end{pmatrix}$

We indeed have Y'=AY.
Now, like in a normal DE, a solution to y'=ay is y=y_0 exp(at)

Here, as it's again said, a solution is $e^{tA}Y_0$

Now how to calculate $e^{tA}$ ?

$e^{tA}=\sum_{n=0}^\infty \frac{t^n A^n}{n!}=\sum_{n=0}^\infty \frac{t^nXD^nX^{-1}}{n!}=X \left(\sum_{n=0}^\infty \frac{(tD)^n}{n!} \right) X^{-1}$

In fact, what is $\sum_{n=0}^\infty \frac{(tD)^n}{n!}$ ?
$D=\begin{pmatrix}1&0\\0&5\end{pmatrix}$

So $tD=\begin{pmatrix}t&0\\0&5t\end{pmatrix}$

And then $\sum_{n=0}^\infty \frac{(tD)^n}{n!}=\begin{pmatrix}\sum_{n=0}^\infty \frac{t^n}{n!} &0 \\0& \sum_{n=0}^\infty \frac{(5t)^n}{n!} \end{pmatrix}=\begin{pmatrix} e^t & 0 \\ 0&e^{5t} \end{pmatrix}$

Does it look clearer to you ?