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

$\displaystyle

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

$

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- Jul 26th 2009, 01:13 PMdiddledabbleGeneral Solution of x'(t)=Ax(t)
General Solution of x'(t)=Ax(t)

$\displaystyle

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

$ - Jul 26th 2009, 02:13 PMpickslides
You need to solve these DEs

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

$\displaystyle x_2'(t) = 12x_1(t)+x_2(t)$ - Jul 26th 2009, 11:52 PMMoo
Hello,

Treat A as a constant (which it is).

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

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

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, 01:13 AMCaptainBlack
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, 02:53 PMdiddledabbleStill don't get it
I still am not sure what to do with the matrix.

- Jul 31st 2009, 04:56 PMHavenEigenValue Method
Okay here's an easy way to solve it:

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

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

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

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

$\displaystyle 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 $\displaystyle c_1 $

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

Now we plug our values of r back in

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

Similarly for $\displaystyle r =7 $

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

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

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

Where $\displaystyle c_1, c_2 $ are arbitrary constants