Recall from the hint that the general solution to a differential equation of this form is
, so to to determine the general solution to the given system, we need to compute the eigenvalues and eigenvectors of
.
Since
is an upper triangular matrix, we can read the eigenvalues off the diagonal, they are:
and
. To find the corresponding eigenvectors, we will use the fact that
.
: Let
, then
![{\displaystyle A{\vec {v}}_{1}=\lambda _{1}{\vec {v}}_{1}\Rightarrow \left[{\begin{array}{c}-v_{1}+v_{2}\\2v_{2}\end{array}}\right]=\left[{\begin{array}{c}-v_{1}\\-v_{2}\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e24be1c873b2099978d870a3419ced424d76f1ea)
We can see that me must have
, and that
is a free variable. We choose it to be
, so
.
: Let
, then
![{\displaystyle A{\vec {v}}_{2}=\lambda _{2}{\vec {v}}_{2}\Rightarrow \left[{\begin{array}{c}-w_{1}+w_{2}\\2w_{2}\end{array}}\right]=\left[{\begin{array}{c}2w_{1}\\2w_{2}\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/88fd6675ecf003f33d8c866b9cf04c1fa1b2efdc)
The first equation simplifies to
, and the second equation imposes no restrictions on what
has to be, so we can choose
, which results in
and
.
Now, we can write the general solution to the system of differential equations:
![{\displaystyle {\vec {y}}(t)=c_{1}e^{-t}\left[{\begin{array}{c}1\\0\end{array}}\right]+c_{2}e^{2t}\left[{\begin{array}{c}1\\3\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/876226ad31b8200716b69690a1bdcd423f040138)