**Practice
Problems for Test 3 - Solution
**

**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.

1 | 0.104 | |

1.104 | 0.119 | |

1.223 | 0.1383 | |

1.3617 |

We may get the equilibrium point by solving 2.5*y - 7.5 = 0 for y, which produces y=3. The slope field plot shown above 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 y=3 as x is increased.

a) N' = k N.

b) N = 5 e^{0.1682 t}

**5**

a) T' = k (T - 22)

b) T = 22 - 27 e ^{-0.811 t}

**6**

Use the relation RATE OF CHANGE OF Q
= R_in - R_out. We get,

Q' = 2 - (Q/10)*2, that is, Q' = 2 - 0.2 Q. Solve this equation
to get

Q = 10 - 10 e^{-0.2 t}. Finally set Q = 5 and solve for
t to get t = 1.735 min.

**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.

a) The population is growing the fastest
when dP/dt is the largest possible. Since dP/dt = **0.8
P - 0.001 P ^{2}**, the vertex
of the parabola

b) The logistic model we are considering
has carrying capacity **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 logistic model properties.

**9
**a) Considering "up"
as the positive direction, we have
, that is, dv/dt = -9.8.

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

c) solving for v yields .The
initial velocity is 30 m/s, so

30 = -9.8/k + A, that is, A = 30 + 9.8/k.

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.

**11**

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

**12**

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

**13**

The equation **20 y - 2 y ^{2}=0
**yields

By substituting y'=22 into the left hand side of

**14**

a) x = 0 + -3 t , y = 2 + -3 t, where 0 < = t < = 1.

b) x = cos t, y = sin t , where pi < = t < = 2 pi

b) x = t, y = t ^{ 2 }, where
1 < = t < = 2