Solutions to Real Quiz 10 of Dr. Z.’s Dynamical Models in Biology class

Name: Dr. Z. Email: (print!)

1. By hand solve the system

\frac{dx}{dt} = −x − y , \frac{dy}{dt} = x + y , x(0) = 1 , y(0) = 1 .

Plot, by hand, the phase-plane diagram. Solution to 1: The system is

x′(t) = Ax(t) ,

where A is the 2 × 2 matrix

[ \begin{matrix}−1 & −1 \\ 1 & 1 \end{matrix} ] .

Let’s look for the eigenvalues

det [ \begin{matrix}−1 − λ & −1 \\ 1 & 1 − λ \end{matrix} ] = (−1 − λ)(1 + λ) + 1 = λ^{2} .

Setting λ^{2} = 0, we get a double root λ = 0. Since e^{0·t} = 1 the general solution is

x(t) = a + bt ,

for some vectors a and b . Since x(0) = [ \begin{matrix}1 \\ 1 \end{matrix} ] we have

a = [ \begin{matrix}1 \\ 1 \end{matrix} ] .

Differentiating with respect to t we get

x′(t) = b ,

Plugging-in t = 0 we get

x′(0) = b ,

But

b = x′(0) = Ax(0) = [ \begin{matrix}−1 & −1 \\ 1 & 1 \end{matrix} ] [ \begin{matrix}1 \\ 1 \end{matrix} ] = [ \begin{matrix}−2 \\ 2 \end{matrix} ] ,

so

b = [ \begin{matrix}−2 \\ 2 \end{matrix} ]

Hence the solution in vector form is

x(t) = [ \begin{matrix}1 \\ 1 \end{matrix} ] + t [ \begin{matrix}−2 \\ 2 \end{matrix} ]

Spelling it out

x(t) = 1 − 2t , y(t) = 1 + 2t .

The phase-plane diagram is simply a straight line with slope −1 crossing the y-axis at (2, 0) and the x-axis at (0, 2).

Comment: A much easier way would be to ‘cheat’ and convert it to a scalar diff. eq. as follows

Adding the two equations we have

\frac{d(x + y)}{dt} = 0 .

So x + y is a constant and using the initial conditions we get x + y = 2 so

y = 2 − x .

Plugging into the first equation we have

x′(t) = −2 ,

so x(t) = −2t+c for some constant. But x(0) = 1 so c = 1 and x(t) = −2t+1 and y(t) = 2−x(t) = 1 + 2t, much faster without linear algebra.