Practice Problems for Test 3 - Solution
Last changed 4-16-2000, 5:30 p.m.
Corrections Section appended on April 18, 10 a.m.

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1

Plot (a) does not correspond to the diff. eq. y'=a - x - y.

Reason: From the diff. eq. we see that as x and y get larger, the slope becomes smaller and eventually negative. But plot (a) shows slopes getting larger (steeper) as x and y get larger. So plot (a) cannot be the slope field of the differential equation.

2

3

We may get the equilibrium by solving 2.5*y - 7.5 = 0 for y, which produces y=3. The plot confirms our result. Also, the plot shows that the equilibrium value y=3 is unstable, as solutions that are close to it do separate further from it as x is increased.

4

The equilibrium solutions are and , as one can see by solving for k in

We know that k is positive; To analyze stability we consider the two equilibrium values:

a) y=k . Consider the solution through (0,y1), where y1 is close to k but slightly smaller.

The slope at the point (0,y1) is , which is less than 0.

If now we consider y2 close to k but larger, then the slope at the point (0,y2) is , which is greater than 0.

Thus y=k is an unstable equilibrium solution.

b) y= - k . Consider the solution through (0,y1), where y1 is close to - k but slightly smaller.

The slope at the point (0,y1) is , which is bigger than 0.

If now we consider y2 close to k but larger, then the slope at the point (0,y2) is , which is smaller than 0.

Thus y=k is a stable equilibrium solution.

The slope field plot looks like this:

5

a)

b)

6

a)

b)

FIRST METHOD. The equilibrium solution is Q = 25, as one can see by solving 5 - 0.2*Q=0 for Q.

The slope field plot shows that it is a stable equilibrium. It is also the only one.

After a long time we may expect Q to be quite close to 25.

SECOND METHOD. Solve the differential equation to obtain . When t is large, the

exponential term in the rhs becomes very small, and we see that Q approximates 25.

7a

7b

r = - ln | -0.5 x2 + x + 0.5 |

7c

y = (6 - 2 e8 x ) / (3+e8 x )

7d

A solution in implicit form is 0.5 y2 - y cos(y) + sin(y) = 0.5 x2 ;no explicit solution is possible.

8

a) The population is growing the fastest when dP/dt is the largest possible. Since dP/dt = 0.001*(800 P - P^2) ,

the vertex of the parabola y = 0.001*(800 P - P^2) gives what we want. The vertex is at P = 400. So the population

grows the fastest when it is 400 million.

b) The logistic model we are considering has L = 800, and in our particular example, the initial population is 25 million.

Then population will never surpass 800 million and thus will never reach 900 million, according to the model.

10

a) Considering "up" as the positive direction, we have , that is, .

b) which with M=1 and g=9.8 becomes

c) solving for v yields .

11

a) dQ/dt = k Q

b) Solve to get Q = A*e^(k*t) . Note that 0.9*A = Q(10). Hence

, that is , . Solving for k we get k = .

Also Q(0) = A = 3.0. Then we have the formula

We wish to know t when Q = 5. Solve for t to get t = -48.51, that is, it was 48.5 years before the present time

that the mass was 5 Kg.

12

a) the particle will be in the second quadrant when x < 0 and y > 0, that is,

4 t - t^2 < 0 and t - 5 > 0.

Now 4 t - t^2 < 0 when t < 0 or 4 < t, and t-5 > 0 when t > 5.

Then when t > 5 both requirements are met, that is, the particle is in the second quadrant.

b) dx/dt = 4 - 2 * t and dy/dt = 1, so the speed at time t is

s= , and the speed at time t=4 is

c) The distance traveled is =

Using simpson's rule with n = 30 gives distance = 9.2935

13

y = -0.006 P + 0.7 and the points are shown in the figure

The differential equation is (1/P) * (dP/dt) = -0.006 P + 0.7.

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CORRECTIONS (April 18, 10 a.m.)