Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 05 (a)

MATH152 April 2010

  •  QA 1  •  QA 2  •  QA 3  •  QA 4  •  QA 5  •  QA 6  •  QA 7  •  QA 8  •  QA 9  •  QA 10  •  QA 11  •  QA 12  •  QA 13  •  QA 14  •  QA 15  •  QA 16  •  QA 17  •  QA 18  •  QA 19  •  QA 20  •  QA 21  •  QA 22  •  QA 23  •  QA 24  •  QA 25  •  QA 26  •  QA 27  •  QA 28  •  QA 29  •  QA 30  •  QB 1(a)  •  QB 1(b)  •  QB 1(c)  •  QB 2(a)  •  QB 2(b)  •  QB 3(a)  •  QB 3(b)  •  QB 3(c)  •  QB 3(d)  •  QB 4(a)  •  QB 4(b)  •  QB 4(c)  •  QB 4(d)  •  QB 4(e)  •  QB 5(a)  •  QB 5(b)  •  QB 6(a)  •  QB 6(b)  •  QB 6(c)  •  QB 6(d)  •  QB 6(e)  •  

   Other MATH152 Exams

  •  April 2016  •  April 2015  •  April 2014  •  April 2022  •  April 2011  •  April 2010  •  April 2012  •  April 2013  •  April 2017  •  

Question B 05 (a)

Consider the differential equation system $\mathbf {y} '=A\mathbf {y} ~$ where A has an eigenvalue $\lambda _{1}=1+2i$ with corresponding eigenvector $\mathbf {k} _{1}=\left[{\begin{array}{c}2+3i\\1\end{array}}\right]$ Recall that for real matrices, eigenvalues and eigenvectors come in complex conjugate pairs. Write the general solution of the system of differential equations. This can be written in real or complex form.

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.

Hint 1
What happens if you substitute $\mathbf {v} e^{\lambda _{1}t}$ into the equation? What must $\mathbf {v}$ be?

Hint 2
One particular solution to the equation can be found using the given pair of eigenvalue and eigenvector. What is the other pair of eigenvalue and eigenvector? How can you combine these two solutions to form the general solution?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. There are 3 sections in this page but only the first section is "what is supposed to be written" in the exam. Solution Since eigenvalues and eigenvectors occur in conjugate pairs for the real matrix A, the other pair is: ${\begin{aligned}\lambda _{2}&=1-2i\\\mathbf {k} _{2}&=\left[{\begin{array}{c}2-3i\\1\end{array}}\right]\end{aligned}}$ The general solution is then: ${\begin{aligned}y&=C_{1}\mathbf {k} _{1}\exp {\lambda _{1}t}+C_{2}\mathbf {k} _{2}\exp {\lambda _{2}t}\\&=C_{1}\left[{\begin{array}{c}2+3i\\1\end{array}}\right]\exp((1+2i)t)+C_{2}\left[{\begin{array}{c}2-3i\\1\end{array}}\right]\exp((1-2i)t)\end{aligned}}$ We delay showing how to write it into real form to part (b). Motivation To see a reason why this is so. Let us recall that $e^{at}$ solves $x'=ax$ which is the single equation analogue of the given system. From this, we can make a guess that a constant vector times an exponential function might solve our system. More precisely, we substitute the guess $\mathbf {v} e^{ct}$ into the system to get (both $\mathbf {v} ,c$ are to be determined): ${\begin{aligned}(\mathbf {v} e^{ct})'&=A\mathbf {v} e^{ct}\\c\mathbf {v} e^{ct}&=A\mathbf {v} e^{ct}\\c\mathbf {v} &=A\mathbf {v} \end{aligned}}$ (The cancellation in the last step is because the exponential function is always positive). So it turns out for our guess to work, c and v as a pair must solve the eigenvalue problem $A\mathbf {v} =c\mathbf {v}$, i.e. they must be a pair of eigenvalue and eigenvector of A! Now, from the two pairs of eigenvalues and eigenvectors we know, we have the following solutions: $\mathbf {y} _{1}=\mathbf {k} _{1}e^{\lambda _{1}t}$ $\mathbf {y} _{2}=\mathbf {k} _{2}e^{\lambda _{2}t}$ To be precise, we should say they are just particular solutions of the problem. To obtain the formula of the general solution, we observe that the differential equation system is linear and so a linear combination of $\mathbf {y} _{1}$ and $\mathbf {y} _{2}$, i.e. $\mathbf {y} =C_{1}\mathbf {y} _{1}+C_{2}\mathbf {y} _{2}$ solves the system too: ${\begin{aligned}\mathbf {y} &=\left(C_{1}\mathbf {y} _{1}+C_{2}\mathbf {y} _{2}\right)'\\&=C_{1}\mathbf {y} _{1}'+C_{2}\mathbf {y} _{2}'\\&=C_{1}A\mathbf {y} _{1}+C_{2}A\mathbf {y} _{2}\\&=A\left(C_{1}\mathbf {y} _{1}+C_{2}\mathbf {y} _{2}\right)\\&=A\mathbf {y} \end{aligned}}$ It turns out this represents all possible solutions. Extended Reading The following outlines the general method to deduce the general solution systematically (in the syllabus of Math 215/255). The key is to see that the existence of the two pairs of eigenvalues and eigenvectors of A allows us to "diagonalize" the matrix A. In fact, once we have found all the eigenvalues and corresponding eigenvectors, if the number of eigenvectors equals the rank of the matrix A (in this case, it is 2), then we can rewrite A into the following form (known as the diagonalization of A): $A=VDV^{-1}~$ where $V=\left[{\begin{array}{cc}\mathbf {k} _{1}&\mathbf {k} _{2}\end{array}}\right]=\left[{\begin{array}{cc}2+3i&2-3i\\1&1\end{array}}\right]$ $D=\left[{\begin{array}{cc}\lambda _{1}&0\\0&\lambda _{2}\end{array}}\right]=\left[{\begin{array}{cc}1+2i&0\\0&1-2i\end{array}}\right].$ To solve the given linear system of differential equations, we proceed as follows exploiting the diagonalizability of A. Left multiplying $V^{-1}$ on both sides of the linear system gives: $V^{-1}\mathbf {y} '=DV^{-1}\mathbf {y}$ So if we let $\mathbf {x} =V^{-1}\mathbf {y}$ Then, we get a simple decoupled system for $\mathbf {x} =\left[{\begin{array}{c}x_{1}\\x_{2}\end{array}}\right]$ (see note): $\left\{{\begin{array}{ccc}x_{1}'&=&\lambda _{1}x_{1}\\x_{2}'&=&\lambda _{2}x_{2}\end{array}}\right.$ The general solution must then be: $\mathbf {x} =\left[{\begin{array}{c}C_{1}\exp {\lambda _{1}t}\\C_{2}\exp {\lambda _{2}t}\end{array}}\right]=\left[{\begin{array}{c}C_{1}\exp {(1+2i)t}\\C_{2}\exp {(1-2i)t)}\end{array}}\right]$: Therefore, ${\begin{aligned}\mathbf {y} &=V\mathbf {x} \\~&=\left[{\begin{array}{cc}\mathbf {k} _{1}&\mathbf {k} _{2}\end{array}}\right]\left[{\begin{array}{c}C_{1}\exp {\lambda _{1}t}\\C_{2}\exp {\lambda _{2}t}\end{array}}\right]\\~&=C_{1}\mathbf {k} _{1}\exp {\lambda _{1}t}+C_{2}\mathbf {k} _{2}\exp {\lambda _{2}t}\\~&=C_{1}\left[{\begin{array}{c}2+3i\\1\end{array}}\right]\exp {(1+2i)t}+C_{2}\left[{\begin{array}{c}2-3i\\1\end{array}}\right]\exp {(1-2i)t}\end{aligned}}$ where $C_{1},C_{2}~$ are complex constants. Note: The word decoupled means the equations can be regarded separately as single equations themselves, i.e. this system is just a collection of single equations whose solutions do not affect each other.