# General Solution of x'(t)=Ax(t)

• Jul 26th 2009, 02:13 PM
diddledabble
General Solution of x'(t)=Ax(t)
General Solution of x'(t)=Ax(t)

$
A=\begin{array}{cc}1&3\\12&1\end{array}
$
• Jul 26th 2009, 03:13 PM
pickslides
You need to solve these DEs

$x_1'(t) = x_1(t)+3x_2(t)$

$x_2'(t) = 12x_1(t)+x_2(t)$
• Jul 27th 2009, 12:52 AM
Moo
Hello,

Treat A as a constant (which it is).

We know that the general solution of this type of equation is $x(t)=e^{At}$

And when A is a matrix, $e^{At}=\sum_{n\geq 0} \frac{A^n t^n}{n!}$

See here : http://www.mathhelpforum.com/math-he...al-matrix.html for an example of computing this exponential.

If you don't understand, I can explain.
• Jul 27th 2009, 02:13 AM
CaptainBlack
Quote:

Originally Posted by diddledabble
General Solution of x'(t)=Ax(t)

$
A=\begin{array}{cc}1&3\\12&1\end{array}
$

You can also convert this into a pair of uncoupled second order homogeneous constant coefficient ODE's.

To do this write them out in full as two first order ODE's differnetiate one of them again and substitute the derivative from the other to eliminate one of the variables ...

CB
• Jul 29th 2009, 03:53 PM
diddledabble
Still don't get it
I still am not sure what to do with the matrix.
• Jul 31st 2009, 05:56 PM
Haven
EigenValue Method
Okay here's an easy way to solve it:

$x'(t) = Ax(t)$

We assume the form of the solution that $x(t) = v(t)*e^{rt}$

It turns out the r, corresponds to the eigenvalue of the matrix A and $v(t)$ are the eigenvectors

$A=\left[\begin{array}{cc}1&3\\12&1\end{array}\right]$

$det(A - rI) = \left|\begin{array}{cc}1-r&3\\12&1-r\end{array}\right| = r^2 -2r -35 = (r+5)(r-7) = 0$ for an aribitrary constant $c_1$

so the roots of this polynomial are: $r = -5, 7$

Now we plug our values of r back in
$A - (-5)I = \left[\begin{array}{cc}6&3\\12&6\end{array}\right]\\\rightarrow\\\left[\begin{array}{cc}2&1\\0&0\end{array}\right]$

So we see that $-2v_1 = v_2$, which means our eigenvector is $v_1(t) = \left[\begin{array}{cc}1\\-2\end{array}\right]$ and our first solution is $x_1(t) = \left[\begin{array}{cc}1\\-2\end{array}\right]\\e^{-5t}$

Similarly for $r =7$
we compute $A - (7)I = \left[\begin{array}{cc}-6&3\\12&-6\end{array}\right]\\\rightarrow\\\left[\begin{array}{cc}-2&1\\0&0\end{array}\right]$

eigenvector: $v_1(t) = \left[\begin{array}{cc}1\\2\end{array}\right]$

second solution $x_2(t) = \left[\begin{array}{cc}1\\2\end{array}\right]\\e^{7t}$

Our final solution is a linear combination to the two solutions $x(t) = {c_1}x_1(t) + {c_2}x_2(t) = c_1\left[\begin{array}{cc}1\\-2\end{array}\right]\\e^{-5t}\\+\\c_2\left[\begin{array}{cc}1\\2\end{array}\right]\\e^{7t}$

Where $c_1, c_2$ are arbitrary constants