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

MATH152 April 2022

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Question A17
Let $$A=\left[{\begin{array}{cc}-1&1\\0&2\end{array}}\right].$$ Find the general solution to the system of differential equations ${\textstyle {\frac {d}{dt}}{\vec {y}}(t)=A{\vec {y}}(t)}$. (Note: this problem requires that you first find the eigenvalues and eigenvectors of the matrix.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that the form for the general solution to a ${\textstyle 2\times 2}$ differential equation system is ${\textstyle {\vec {y}}(t)=c_{1}e^{\lambda _{1}t}{\vec {v}}_{1}+c_{2}e^{\lambda _{2}t}{\vec {v}}_{2}}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $${\vec {y}}(t)=c_{1}e^{-t}\left[{\begin{array}{c}1\\0\end{array}}\right]+c_{2}e^{2t}\left[{\begin{array}{c}1\\3\end{array}}\right]$$