Practice Problems for Differential Equations - Solution

1 Substitution of y = x2 into the differential equation gives
2 a x = 3 x +
x, which may be rewritten as x = 3 x. Clearly the only value of "a" that makes this equation valid for all values of x is = 3.

2
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.

3

 1 0.104 1.104 0.119 1.223 0.1383 1.3617

4

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.

5

a) N' = k N.

b) N = 5 e0.1682 t

6

a) T' = k (T - 22)

b) T = 22 - 27 e -0.2703 t

c) 9.63 hrs

7

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 = 3.466 min.

8a

8b r = - ln | -0.5 x2 + x + 0.5 |

8c y = 8+ex

8d A solution in implicit form is 0.5 y2 - y cos(y) + sin(y) = 0.5 x2 ;no explicit solution is possible.

9

a) The population is growing the fastest when dP/dt is the largest possible. Since dP/dt = 0.8 P - 0.001 P2, the vertex of the parabola y = 0.8 P - 0.001 P2 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 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.

10

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

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

12

The equation 20 y - 2 y2=0 yields y=0, y=10, so the carrying capacity is L=10.
By substituting y=5 into the right hand side of y' = 20 y - 2 y2 we get that the population is changing initially at the rate of 50 mill/yr.
By substituting y'=22 into the left hand side of y' = 20 y - 2 y2 we get
22= 20 y - 2 y2 . This equation has two positive solutions given approximately by y = 1.258 and y = 8.742

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