Science:Math Exam Resources/Courses/MATH152/April 2022/Question A17/Solution 1

Recall from the hint that the general solution to a differential equation of this form is ${\textstyle {\vec {y}}(t)=c_{1}e^{\lambda _{1}t}{\vec {v}}_{1}+c_{2}e^{\lambda _{2}t}{\vec {v}}_{2}}$, so to to determine the general solution to the given system, we need to compute the eigenvalues and eigenvectors of ${\textstyle A}$.

Since ${\textstyle A}$ is an upper triangular matrix, we can read the eigenvalues off the diagonal, they are: ${\textstyle \lambda _{1}=-1}$ and ${\textstyle \lambda _{2}=2}$. To find the corresponding eigenvectors, we will use the fact that ${\textstyle A{\vec {v}}=\lambda {\vec {v}}}$.

• ${\textstyle (\lambda _{1}=-1)}$: Let ${\textstyle {\vec {v}}_{1}=[v_{1}\ \ v_{2}]^{T}}$, then

$$A{\vec {v}}_{1}=\lambda _{1}{\vec {v}}_{1}\Rightarrow \left[{\begin{array}{c}-v_{1}+v_{2}\\2v_{2}\end{array}}\right]=\left[{\begin{array}{c}-v_{1}\\-v_{2}\end{array}}\right].$$

We can see that me must have ${\textstyle v_{2}=0}$, and that ${\textstyle v_{1}}$ is a free variable. We choose it to be ${\textstyle v_{1}=1}$, so ${\textstyle {\vec {v}}_{1}=[1\ \ 0]^{T}}$.

• ${\textstyle (\lambda _{2}=2)}$: Let ${\textstyle {\vec {v}}_{2}=[w_{1}\ \ w_{2}]^{T}}$, then

$$A{\vec {v}}_{2}=\lambda _{2}{\vec {v}}_{2}\Rightarrow \left[{\begin{array}{c}-w_{1}+w_{2}\\2w_{2}\end{array}}\right]=\left[{\begin{array}{c}2w_{1}\\2w_{2}\end{array}}\right].$$

The first equation simplifies to ${\textstyle w_{2}=3w_{1}}$, and the second equation imposes no restrictions on what ${\textstyle w_{2}}$ has to be, so we can choose ${\textstyle w_{1}=1}$, which results in ${\textstyle w_{2}=3}$ and ${\textstyle {\vec {v}}_{2}=[1\ \ 3]^{T}}$.

Now, we can write the general solution to the system of differential equations:

$${\vec {y}}(t)=c_{1}e^{-t}\left[{\begin{array}{c}1\\0\end{array}}\right]+c_{2}e^{2t}\left[{\begin{array}{c}1\\3\end{array}}\right]$$