Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 05 (a)/Solution 1

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.