Math 3C, Midterm, Fall 2000


Calculus

Remember to show all your work for partial credit! Calculators are permitted but no books, notes or homework.

1.

(20 points) Match the direction fields in Figure 1 and the ODE's listed below.

(a)

$\dot x = \sin(t)\:\:\:$ Figure 1.3.8

(b)

$\dot x = -x + 1 \:\:\:$ Figure 1.3.12

(c)

$\dot x = x + t\:\:\:$ Figure 1.3.9

(d)

$\dot x = t - x\:\:\:$ Figure 1.3.11

(e)

$\dot x = \sin(x)\:\:\:$ Figure 1.3.10

2.

(10 points) Solve the ODE

$$\dot x = \sin^2(t), \: \: x(0)=1.$$

Solution:

$$x(t) = \int (\frac{1}{2} - \frac{1}{2} \cos(2t))dt =$$

$$\frac{t}{2}-\frac{1}{4}\sin(2t) + C.$$

x(0) = C = 1.

3.

(20 points) Use four steps and Euler's Method to find the solution of the ODE

$$y\prime = x + y, \: \: y(0)=1,$$

at x = 0.4. Estimate the error in your value of y(0.4).

Solution:

$$h = {{x_4-x_0}\over n} = {{0.4 - 0}\over 4} = 0.1$$

where n is the number of steps and h is the step size.

$$y_{m+1} = y_m +f(x_m,y_m)h, \: \: y_0 =1,$$

$$\begin{eqnarray*}y_1 &=& 1 + (0+1)0.1 = 1.1\\ y_2 &=& 1.1 + (0.1+1.1)0.1 = 1.22... ...2+1.22)0.1 = 1.362\\ y_4 &=& 1.362 + (0.3+1.362)0.1 = 1.5282\\ \end{eqnarray*}$$

The error is of the order

h² = 0.01

Thus the approximation is

$$y(0.4) \approx y_4 = 1.53,$$

because the error tells us to carry only two significant digits.

4.

(10 points) Solve the ODE

$$\dot x = x^2, \: \: x(0)=1.$$

Solution:

$$\frac{dx}{x^2} = dt,$$

$$\int \frac{dx}{x^2} = \int dt,$$

$$-\frac{1}{x} = t + C,$$

$$x = -\frac{1}{t+C},$$

$$x(0) = -\frac{1}{C}, \: \: \: C = -1.$$

$$x(t) = -\frac{1}{t-1}$$

blows up to $\infty$ at t = 1.

How does the behaviour of the solution change if x(0) = -1?

If x(0) = -1, then C = 1, the solution becomes

$$x = -\frac{1}{t+1}$$

which just decays to 0 as $t \to \infty.$

5.

(40 points) Do the qualitative analysis of the population equation

$$\dot p = p(p-1)(3-p),$$

(a)

Find the stationary solutions.

p(p-1)(3-p) = 0,

p = 0, 1, 3.

(b)

Determine the stability of the stationary solutions.

f(p) = p(p-1)(3-p),

$$\frac{df}{dp} = (p-1)(3-p)+p(3-p)- p(p-1)$$

$$f\prime(0) = -3 < 0, \: f\prime(1) = 2 > 0, \: f\prime(3) = -6 < 0.$$

Thus 0 and 3 are stable stationary solutions and 1 is unstable.

(c)

Draw the extended phase portrait (in p and t space). See Figure 1.

  

Figure: The extended phase space.

(d)

Determine what happens to populations with initial values, p(0) = 1/2, p(0) = 3/2, p(0) = 5, as $t \to \infty$.

Solution:

$p(0) = \frac{1}{2}$ decays to 0 as $t \to \infty$, the population becomes extinct. $p(0) = \frac{3}{2}$ and p(0) = 5 grow and decay respectively to the carrying capacity of the environment p = 3.