**Practice Problems for Test 3 - Solution
Last changed 4-16-2000, 5:30 p.m.
**

**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 x^{2} + x + 0.5 |

**7c**

y = (6 - 2 e^{8 x} ) / (3+e^{8 x} )

**7d**

A solution in implicit form is
0.5 y^{2} - y cos(y) + sin(y) = 0.5 x^{2}
;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.

Answer 2 1.3613 PLEASE CHANGE TO 1.3617

Answer 4, case b). Thus y=k is ... PLEASE CHANGE TO Thus y=-k is ...

Answer 6. PLEASE CHANGE ALL Q's to M's

Answer 10, part a) ...that is, dv/dt=9.8 CHANGE TO dv/dt = -9.8

Answer 10, part c) INCOMPLETE AS IS; SUBSTITUTE INITIAL CONDITION TO OBTAIN A IN TERMS OF k.