# 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
$x'=\begin{bmatrix}
4 & 4\\
3 & 5
\end{bmatrix}x$

which satifies the initial conditions
$x(0)=\begin{bmatrix}
-4\\
2
\end{bmatrix}$

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

and
$\begin{bmatrix}
-4 & 4\\
3 & -3
\end{bmatrix}\begin{bmatrix}
a\\
b
\end{bmatrix}=0$

To get
$\begin{bmatrix}
a\\
b
\end{bmatrix}=\begin{bmatrix}
-4\\
3
\end{bmatrix}$
and $\begin{bmatrix}
a\\
b
\end{bmatrix}=\begin{bmatrix}
1\\
1
\end{bmatrix}$

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

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

Can anyone help me out, please?
Thanks!

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

$-4a+b=-4,$

$3a+b=2?$

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

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