Science:Math Exam Resources/Courses/MATH152/April 2022/Question B5 (c)

MATH152 April 2022

  •  QA01  •  QA02  •  QA03  •  QA04  •  QA05  •  QA06  •  QA07  •  QA08  •  QA09  •  QA10  •  QA11  •  QA12  •  QA13  •  QA14  •  QA15  •  QA16  •  QA17  •  QA18  •  QA19  •  QA20  •  QB1 (a)  •  QB1 (b)  •  QB2 (a)  •  QB2 (b)  •  QB2 (c)  •  QB3 (a)  •  QB3 (b)  •  QB4 (a)  •  QB4 (b)  •  QB4 (c)  •  QB5 (a)  •  QB5 (b)  •  QB5 (c)  •  

   Other MATH152 Exams

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

Question B5 (c)

With the variables ordered ${\textstyle I,V_{1},V_{2}}$, let the matrix of the system of differential equations be ${\textstyle A}$. (You don’t need to find ${\textstyle A}$.) Consider the following Matlab result: > [P,D] = eig(A) P = 0.5 - 0.34i 0.5 + 0.34i 0.22 + 0.00i 0.62 + 0.00i 0.62-0.00i 0.85 + 0.00i -0.43 - 0.21i -0.43 + 0.21i 0.47 + 0.00i D = Diagonal Matrix -0.15 + 0.17i 0 0 0 -0.15 - 0.17i 0 0 0 -0.77 + 0.00i There exist initial conditions that result in a non-oscillating solution (a real solution that does not involve sin and cos). Find one such solution ${\textstyle I(t)}$, ${\textstyle V_{1}(t)}$, ${\textstyle V_{2}(t)}$ that is not oscillating and not zero. Note: you don’t need the results from (a) and (b) to attempt (c).

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

Recall, again, that solutions to a differential equation of the form ${\textstyle {\frac {d}{dt}}{\vec {y}}(t)=A{\vec {y}}(t)}$ have the form ${\textstyle {\vec {y}}(t)=ce^{\lambda t}{\vec {v}}}$ (the general solution is the sum of multiple of these solutions, one for each eigenvalue). If we want a real valued solution that doesn’t involve sine and cosine, what does this mean for the eigenvalues and eigenvectors?

Hint 2
The matlab command $[P,D]=\mathrm {eig} (A)$ yields the matrix $P$, whose columns are the eigenvectors of $A$, and the diagonal matrix $D$, whose entries are the eigenvalues of $A$.

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. From the hint, we know that we may find a solution to a differential equation of the form ${\textstyle {\frac {d}{dt}}{\vec {y}}(t)=A{\vec {y}}(t)}$ as a function ${\textstyle {\vec {y}}(t)=e^{\lambda t}{\vec {v}}}$, where ${\textstyle (\lambda ,{\vec {v}})}$ is an eigenvalue-eigenvector pair of ${\textstyle A}$. A general solution to the differential equation is a linear combination of “eigensolutions” of this type. If we want a non-oscillating solution, then we will need ${\textstyle \lambda }$ to be real-valued (see below for an explanation). From the given eigendecomposition of ${\textstyle A}$ we see that the third eigenvalue is real-valued. Thus, a non-oscillating solution is $$\left[{\begin{array}{c}I(t)\\V_{1}(t)\\V_{2}(t)\end{array}}\right]=e^{-0.77t}\cdot \left[{\begin{array}{c}0.22\\0.85\\0.47\end{array}}\right].$$ To see why ${\textstyle \lambda }$ must be real-valued, consider an eigenvalue ${\textstyle \lambda =a+ib}$. Then ${\textstyle e^{\lambda t}=e^{at+ibt}=e^{at}(\cos(bt)+i\sin(bt))}$ has sine and cosine terms as long as ${\textstyle b\neq 0}$. Thus, to make sure there are no sine and cosine terms we have to take ${\textstyle b=0}$, meaning ${\textstyle \lambda }$ is real-valued.