# Thread: Simple eigenvalue problem (x'=[2x2]x)

1. ## Simple eigenvalue problem (x'=[2x2]x)

Hi, I've got a homework problem that should be very simple, the previous problem was identical (different numbers) and was a breeze, but this one is giving me problems. Here's what we're asked for:
Apply the eigenvalue method to find the particular solution to the system of differential equations
$\displaystyle x'=\begin{bmatrix} 4 & 4\\ 3 & 5 \end{bmatrix}x$
which satifies the initial conditions
$\displaystyle x(0)=\begin{bmatrix} -4\\ 2 \end{bmatrix}$

So I begin by finding the eigenvalues from the characteristic equation:
$\displaystyle (4-\lambda)(5-\lambda)-12=(\lambda^2-9\lambda+8)=(\lambda-1)(\lambda-8)$
$\displaystyle \lambda=1,8$
To get the matrices:
$\displaystyle \begin{bmatrix} 3 & 4\\ 3 & 4 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=0$
and
$\displaystyle \begin{bmatrix} -4 & 4\\ 3 & -3 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=0$
To get
$\displaystyle \begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} -4\\ 3 \end{bmatrix}$ and $\displaystyle \begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 1\\ 1 \end{bmatrix}$

Which gives the particular solution $\displaystyle x_p=a\begin{bmatrix} -4\\ 3 \end{bmatrix}e^t+b\begin{bmatrix} 1\\ 1 \end{bmatrix}e^{8t}$

But at this point, no values for a and/or b will satisfy $\displaystyle x(0)=\begin{bmatrix} -4\\ 2 \end{bmatrix}$

Can anyone help me out, please?
Thanks!

2. Don't you just need to solve the system

$\displaystyle -4a+b=-4,$

$\displaystyle 3a+b=2?$

That system has one unique solution, as far as I can tell.

3. Oh, yes, you're right. I got:
$\displaystyle (-4)6/7e^t-4/7e^{8t}$
and
$\displaystyle (3)6/7e^t-4/7e^{8t}$
Which are both correct.

Thank you!

4. You're very welcome. Have a good one!

5. there HAS to be a solution. at t = 0, the exponential functions are both 1. and since (-4,3) and (1,1) are linearly independent, they span a 2-dimensional subspace of R^2, which MUST be R^2 itself.

hint: a and b are not integers.