# Thread: 27.1 Initial value problem

1. ## 27.1 Initial value problem

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 ????

2. ## Re: 27.1 Initial value problem

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

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

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

3. ## Re: 27.1 Initial value problem

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

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

Find the eigensystem.

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

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

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

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

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

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

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

$y(t) = \dfrac 1 5 \begin{pmatrix}8 e^{4t}-3 e^{-t}\\12 e^{4t} +3 e^{-t}\end{pmatrix}$

4. ## Re: 27.1 Initial value problem

Appreciate much
Have deal more with this tomorro
To hard with just cell phone.

5. ## Re: 27.1 Initial value problem

how did you get $\displaystyle v_1$ and $\displaystyle v_2$

6. ## Re: 27.1 Initial value problem

Originally Posted by bigwave
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.