27.1 Initial value problem

Nov 2009
717
133
Wahiawa, Hawaii
27.1b.png
did a c/p just avoid typos

ok the example did this
$$A=\begin{pmatrix}1&2\\3&2 \end{pmatrix}$$
and

$$y'=Ay$$

with initial value $$y(0)=\begin{pmatrix}y_1(0)\\y_2(0) \end{pmatrix}$$


ok I'm ????
 
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romsek

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Nov 2013
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$y = c_1 e^{\lambda_1 t}v_1 + c_2 e^{\lambda_2 t} v_2$

where $\lambda_k, v_k$ are the $kth$ eigenvalues/vectors of $A$

Then use the initial conditions to solve for $c_1, c_2$
 
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romsek

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$A=\begin{pmatrix}1 &2\\3 &2\end{pmatrix}$

$y(0)=\begin{pmatrix}1\\3\end{pmatrix}$

Find the eigensystem.

$\left|\begin{pmatrix}1-\lambda &2\\3 &2-\lambda\end{pmatrix}\right| = \lambda^2-3\lambda -4$

$\lambda^2-3\lambda-4 = 0\\
(\lambda-4)(\lambda+1) = 0\\
\lambda = 4,~-1
$

$\lambda_1=4 \Rightarrow v_1 = \begin{pmatrix}2\\3\end{pmatrix}$
$\lambda_2=-1 \Rightarrow v_2 = \begin{pmatrix}1\\-1\end{pmatrix}$

$y(t) = c_1 e^{4t}\begin{pmatrix}2\\3\end{pmatrix} + c_2 e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}$

$y(0) = c_1 \begin{pmatrix}2\\3\end{pmatrix} + c_2 \begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}1\\3\end{pmatrix}$

$(c_1,c_2) = \left(\dfrac 4 5, -\dfrac 3 5\right)$

$y(t) = \dfrac 4 5 e^{4t}\begin{pmatrix}2\\3\end{pmatrix} - \dfrac 3 5 e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}$


$y(t) = \dfrac 1 5 \begin{pmatrix}8 e^{4t}-3 e^{-t}\\12 e^{4t} +3 e^{-t}\end{pmatrix}$
 
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Nov 2009
717
133
Wahiawa, Hawaii
Appreciate much
Have deal more with this tomorro
To hard with just cell phone.
 
Nov 2009
717
133
Wahiawa, Hawaii
how did you get \(\displaystyle v_1\) and \(\displaystyle v_2\)
 
Last edited:

romsek

MHF Helper
Nov 2013
6,664
3,000
California
how did you get \(\displaystyle v_1\) and \(\displaystyle v_2\)
If you've advanced to the point of solving systems of linear diff eqs via eigensystems
then you should know how to find eigenvectors given eigenvalues.

Review your notes.