Science:Math Exam Resources/Courses/MATH215/December 2011/Question 06 (b)

MATH215 December 2011

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[hide] Question 06 (b)
Let $A=\left({\begin{array}{cc}3&1\\-4&-1\end{array}}\right)$ (b) Find the general solution of the system of equations $\mathbf {x} '=A\mathbf {x}$ Here $\mathbf {x} (t)={\begin{pmatrix}x_{1}(t)\\x_{2}(t)\end{pmatrix}}$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
If the system is "diagonalized", it can be solved easily with a linear change of variables.

[show] Hint 2
How does having a diagonalized matrix M such that $\displaystyle {}A=SMS^{-1}$ help us solve the problem $\displaystyle {}x'=Ax=SMS^{-1}x?$ Consider y = S-1x.

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[show] 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. To begin with we matrix M in Jordan normal form, such that A = SMS-1. To do we need to find the (generalized) eigenvector of A. Eigenvector v Knowing $\lambda =1$, we will proceed to find our eigenvectors by solving $(A-\lambda I)v=0$. $\left({\begin{array}{cc}2&1\\-4&-2\end{array}}\right)\left({\begin{array}{c}v_{1}\\v_{2}\end{array}}\right)=\left({\begin{array}{c}0\\0\end{array}}\right)$ which gives the homogeneous system $\left[{\begin{array}{cc}2&1\\-4&-2\end{array}}\right]$. Placing this in reduced-row form we get $\left[{\begin{array}{cc}2&1\\0&0\end{array}}\right]$ so that $v_{2}=-2v_{1}$. We take $\displaystyle {}v=(1,-2)^{T}$ as the eigenvector. Unfortunately, we only find one eigenvector from this, and we'll need to consider generalized eigenvectors and the Jordan canonical form in order to solve this problem. Find the generalized eigenvector w The generalized eigenvector satisfies $(A-\lambda I)w=v$ which gives us $\left({\begin{array}{cc}2&1\\-4&-2\end{array}}\right)\left({\begin{array}{c}w_{1}\\w_{2}\end{array}}\right)=\left({\begin{array}{c}1\\-2\end{array}}\right).$ Row reductions lead to $\left({\begin{array}{cc}2&1\\0&0\end{array}}\right)\left({\begin{array}{c}w_{1}\\w_{2}\end{array}}\right)=\left({\begin{array}{c}1\\0\end{array}}\right)$ which has a solution set that can be parameterized by $w_{1}$: $\left({\begin{array}{c}w_{1}\\1-2w_{1}\end{array}}\right)=w_{1}\left({\begin{array}{c}1\\-2\end{array}}\right)+\left({\begin{array}{c}0\\1\end{array}}\right).$ For simplicity, we can take $w_{1}=0$ so that our generalized eigenvector is $\displaystyle {}w=(0,1)^{T}.$ Setting up the Jordan normal form M Having found the generalized eigenvector, we get that the matrix A has Jordan Canonical form $\displaystyle {}A=SMS^{-1}$ where $S=\left({\begin{array}{cc}1&0\\-2&1\end{array}}\right)$ is the matrix with columns being the eigenvectors, and $M=\left({\begin{array}{cc}1&1\\0&1\end{array}}\right)$ is the matrix with the eigenvalues along the diagonal with 1's above them. A quick computation yields $S^{-1}=\left({\begin{array}{cc}1&0\\2&1\end{array}}\right).$ Our system states that $\displaystyle {}x'=Ax$ where $\displaystyle {}x=(x_{1},x_{2})^{T}.$ We can therefore write, ${\begin{aligned}x'&=SMS^{-1}x\\S^{-1}x'&=MS^{-1}x\\y'&=My\end{aligned}}$ where $\displaystyle {}y=S^{-1}x$. Solve the simplified system y' = My The system $y'=My=\left({\begin{array}{cc}1&1\\0&1\end{array}}\right){\begin{pmatrix}y_{1}\\y_{2}\end{pmatrix}}$ tells us ${\begin{aligned}y_{1}'&=y_{1}+y_{2}\\y_{2}'&=y_{2}\end{aligned}}$. The second equation gives us $\displaystyle {}y_{2}=ce^{t}.$ Using this in the first equation means ${\begin{aligned}y_{1}'&=y_{1}+ce^{t}\\y_{1}'-y_{1}&=ce^{t}\\e^{-t}y_{1}'-e^{-t}y_{1}&=(e^{-t}y_{1})'=c\\e^{-t}y_{1}&=ct+b\\y_{1}&=(ct+b)e^{t}\end{aligned}}$ which has been solved using integrating factors. Now we have $y=\left({\begin{array}{c}cte^{t}+be^{t}\\ce^{t}\end{array}}\right)$ Recover the solution x which immediately gives us our solution x, $x=Sy=\left({\begin{array}{cc}1&0\\-2&1\end{array}}\right)\left({\begin{array}{c}cte^{t}+be^{t}\\ce^{t}\end{array}}\right)=\left({\begin{array}{c}cte^{t}+be^{t}\\-2cte^{t}+(c-2b)e^{t}\end{array}}\right)$.