Problem 1 (8 points): Consider the following predator-prey system \begin{align*} \frac{dR}{dt} & = 2R - 1.2 RF \\ \frac{dF}{dt} & = -F + 0.9 RF, \end{align*} where $R = R(t)$ is the population of the prey at time $t$ and $F = F(t)$ is the population of the predator at time $t$.
a. How would you modify this system to include the effect of hunting the prey at a rate of $\alpha$ units of prey per unit time?
Solution:
$$\frac{dR}{dt} = 2R-1.2RF - \alpha$$b. How would you modify this system to include the effect of hunting of the predators at a rate proportional to the number of predators?
Solution:
$$\frac{dF}{dt} = -F+0.9RF - \alpha F$$c. Suppose the predators discover a second, unlimited source of food, but they still prefer to eat prey when they can catch them. How would you modify this system to include this assumption.
Solution:
$$\frac{dF}{dt} = \alpha F + 0.9RF$$d. Suppose the predators discover a second source of food that is limited in supply. How would you modify this system to include this assumption.
Solution:
$$\frac{dF}{dt} = \alpha F(1-\frac{F}{N})+0.9RF$$Problem 2 (15 points): Consider the equation $$ \frac{d^2 y}{dt^2} + \frac{k}{m} y = 0 $$ for the motion of a simple harmonic oscillator.
a. Consider the function $y(t) = \cos \beta t$. Under what conditions on $\beta$ is $y(t)$ a solution to the differential equation?
Solution:
In order to find where $ \beta $ is a solution to the differential equation, we must replace all the $y$ in the original differential equation with the function $cos \beta t$ from y(t).
The above action yields,$$ - \beta ^2 cos(\beta t) + \frac{k}{m}cos(\beta t) = 0$$
Now factor out the $(\frac{k}{m} - \beta^2)$ to get,
$$ (\frac{k}{m} - \beta^2)cos(\beta t) $$now solve for $\beta$,
$$ \sqrt{\frac{K}{M}} $$now find the initial conditions using what we know, $$ c_1 = 1, c_2 = 0 $$
b. Rewrite the initial value problem $y'' + y = 0$, $y(0) = 1$, and $y'(0) = 0$ as a system.
Hint: Let $v = y'$.
Solution:
The systems looks like the following.
$$ v = y' $$$$ v' + y = 0 $$and the initial conditions are $$ y(0) = 1 $$ $$ v(0) = 0 $$
c. What initial condition ($t = 0$) in the $yv$-plane corresponds to this solution?
Solution:
The inititial conditions $y(0) = 1$ and $v(0)=0$ correspond to the solution.
d. In terms of $k$ and $m$, what is the period of this solution?
Solution:
The period in terms of m and k is $$\frac{2\pi}{\sqrt{\frac{k}{m}}}$$
e. Use Sage to graph the solution curve in the $yv$-plane that corresponds to this solution when $m = 1$ and $k = 1$ (with initial conditions $y(0) = 1$ and $v(0) = 0$).
Solution:
Problem 3 (3 points): The spreading of cancer cells in the body is called {\em metastasis}. The Liotta-DeLisi model proposed in 1977 for the metastasis of malignant tumors in mice is given by the linear homogeneous system \begin{align*} x' & = - (\alpha + \beta)x \\ y' & = \beta x - \gamma y, \end{align*} where $x$ is the number of destroyed cancer cells, $y$ is the number of cells that invade the tissue, and $\alpha$, $\beta$, and $\gamma$ are positive constants depending on the type of cancer. Solve the system and give the physical interpretation for all possible choices of $\alpha$, $\beta$, and $\gamma$.
Solution: To solve the system we must first recognize that $x'$constitutes a seperated semi-autonomous differential equation. Knowing this, we can break $x'$, which is also $\frac{dx}{dt}$, apart onto both sides of the equals sign and solve.
$$ \frac{dx}{dt} = -(\alpha + \beta)x $$$$ \frac{1}{x}dx = -(\alpha + \beta)dt $$We integrate to get,
$$ ln(x) + c = -(\alpha + \beta)t $$now we apply some properties of logarithms and some fenagling of constants to get.
$$ x(t) = -c\frac{(\alpha+\beta-\gamma)}{\beta}e^{-(\alpha+\beta)t} $$Next we rewrite the second equation as $$dy + \gamma y = \beta x$$.
Once rewritten in this form, we can see that an integration factor will be neccessary. Next we find an integration factor to solve the second, partially coupled, differnetial equation. $$ \mu (x) = e^{-\gamma t} $$ Now that we have the I.F. we apply the method of the integration factor to find y(t). $$ ye^{-\gamma t} = e^{-\gamma t} \beta x - e^{-\gamma t} \gamma y $$ with some further manipulation, and after plugging x(t) in for x in the y' equation we find. $$ y(t) = ce^{-(\alpha + \beta)t} + ce^{-\gamma t} $$
Having found the solutions to the system for any values of $\alpha$, $\beta$ and $\gamma$ we can conclude that mice are dying any time those three terms are greater than 0.
Problem 4 (12 points):
Consider the system \begin{align*} x' & = y \\ y' & = x - x^3 -y \end{align*}
a. Find the equilibrium points of the system.
Solution:
To find the equilibrium points of the system, set both differential equations = 0. $$ x' = y $$ So, $$ y = 0 $$ Now we can substitute 0 for y in the second equation. $$ y' = x-x^{3}-0 $$ Rearranging and simplifying at the same time yields. $$ y' = x(1-x^{2}) $$ Thus we find the equilibrium solutions to be $y = 0$, $x = 0$, $x = 1$ yielding points (0,0) and (0,1)
b. Determine the $x$ and $y$-nullclines.
Solution:
The nullclines are x = 0, x = 1 and y = 0
c. Use Sage to plot the phase plane for the system. Be sure to draw the $x$ and $y$-nullclines, the direction field, and a sample solution.
from sage.calculus.desolvers import desolve_rk4
x, y, t = var('x y t')