Now what? Where are you stuck?
The next step is the most difficult step, yet the most beautiful.
If the eigenvalues
λ are distinct real eigenvalues, then the solution to our matrix (or system) is:
X=cKeλt
That is:
X=c1X1+c2X2=c1K1eλ1t+c2K2eλ2t
How to find
K1 and
K2?
We know that the matrix
K1=[k1k2] belongs to the eigenvalue
λ1=−2, then we have this system to solve:
(1−λ1)k1+3k2=0
5k1+(3−λ1)k2=0
Or
3k1+3k2=0
5k1+5k2=0
If we choose
k1=1, then
k2=−1 and we have:
K1=[1−1]
We also know that the matrix
K2=[k1k2] belongs to the eigenvalue
λ2=6, then we have this system to solve:
−5k1+3k2=0
5k1−3k2=0
We see that
k1=53k2, so if we choose
k2=5, we get:
K2=[35]
Finally, we can write the general solution as:
X=c1K1eλ1t+c2K2eλ2t=c1[1−1]e−2t+c2[35]e6t (
Matrix form)
Or (
Normal form)
x(t)=c1e−2t+3c2ee6t
y(t)=−c1e−2t+5c2ee6t
We can also write the system solution in a more clean way like:
x(t)=Ae−2t+Bee6t
y(t)=Ce−2t+Dee6t
But we have to keep track of what those
4 constants are. I myself as a professional skydiver, prefer the Matrix form!

