Chapter 9

Determine whether the differential equation is linear. \( y' + x \sqrt y = x^2 \)

A population grows according to the given logistic equation, where \( t \) is measured in weeks. (a) What is the carrying capacity? What is the value of \( k? \) (b) Write the solution of the equation. (c) What is the population after 10 weeks? \( \frac {dP}{dt} = 0.04P (1 - \frac {P}{1200}), P(0) = 60 \)

Solve the differential equation. \( \frac {dy}{dx} = 3x^2y^2 \)

For each predator-prey system, determine which of the variables, \( x \) or \( y, \) represents the prey population and which represents the predator population. Is the growth of the prey restricted just by the predators or by other factors as well? Do the predators feed only on the prey or do they have additional food sources? Explain. (a) \( \frac {dx}{dt} = -0.05x + 0.0001xy \) \( \frac {dy}{dt} = 0.1y - 0.005xy \) (b) \( \frac {dx}{dt} = 0.2x - 0.0002x^2 - 0.006xy \) \( \frac {dy}{dt} = -0.015y + 0.00008xy \)

Each system of differential equation is a model for two species that either compete for the same resources or cooperate for mutual benefit (flowering plants and insect pollinators, for instance). Decide whether each system describe competition or cooperation and explain why it is a reasonable model. (Ask yourself what effect an increase in one species has on the growth rate of the other.) (a) \( \frac {dx}{dt} = 0.12x - 0.0006x^2 + 0.00001xy \) \( \frac {dy}{dt} = 0.08x + 0.00004xy \) (b) \( \frac {dx}{dt} = 0.15x - 0.0002x^2 - 0.0006xy \) \( \frac {dy}{dt} = 0.2y - 0.00008y^2 - 0.0002xy \)

Determine whether the differential equation is linear. \( y' - x = y \tan x \)

A population grows according to the given logistic equation, where \( t \) is measured in weeks. (a) What is the carrying capacity? What is the value of \( k? \) (b) Write the solution of the equation. (c) What is the population after 10 weeks? \( \frac {dP}{dt} = 0.02P - 0.0004P^2, P(0) = 40 \)

Solve the differential equation. \( \frac {dy}{dx} = x \sqrt y \)

The system of differential equations \( \frac {dx}{dt} = 0.5x - 0.004x^2 - 0.001xy \) \( \frac {dy}{dt} = 0.4y - 0.001y^2 - 0.002xy \) is a model for the populations of two species. (a) Does the model describe cooperation, or competition, or a predator-prey relationship? (b) Find the equilibrium solutions and explain their significance.

Determine whether the differential equation is linear. \( ue^t = t + \sqrt t \frac {du}{dt} \)

Solve the differential equation. \( xyy' = x^2 + 1\)

(a) For what values of \( r \) does the function \( y = e^{rx} \) satisfy the differential equation \( 2y^{"} + y^{'} - y = 0? \) (b) if \( r_1 \) and \( r_2 \) are the values of \( r \) that you found in part (a), show that every member of the family of functions \( y = ae^{r_1{x}} + be^{r_2{x}} \) is also a solution.

Lynx eat snowshoe hares and snowshoe hares eat woody plants like willows. Suppose that, in the absence of hares, the willow population will grow exponentially and the lynx population will decay exponentially. In the absence of lynx and willow, the hare population will decay exponentially. If \( L(t), H(t), \) and \( W(t) \) represent the populations of these three species at time \( t, \) write a system of differential equations as a model for their dynamics. If the constants in your equation are all positive, explain why you have used plus or minus signs.

Determine whether the differential equation is linear. \( \frac {dR}{dt} + t \cos R = e^{-t} \)

Suppose that a population grows according to a logistic model with carrying capacity 6000 and \( k = 0.0015 \) per year. (a) Write the logistic differential equation for these data. (b) Draw a direction field (either by hand or with a computer algebra system). What does it tell you about the solution curves? (c) Use the direction field to sketch the solution curves for initial populations of 1000, 2000, 4000, and 8000. What can you say about the concavity of these curves? What is the significance of the inflection points? (d) Program a calculator or computer to use Euler's method with step size \( h = 1 \) to estimate the population after 50 years if the initial population is 1000. (e) If the initial populations is 1000, write a formula for the population after \( t \) years . Use it to find the population after 50 years and compare with your estimate in part (d). (f) Graph the solution in part (e) and compare with the solution curve you sketched in part (c).

Solve the differential equation. \( y' + xe^y = 0 \)

(a) For what values of \( k \) does the function \( y = \cos kt \) satisfy the differential equation \( 4y^{"} = - 25y? \) (b) For those values of \( k. \) verify that every member of the family functions \( y = A \sin kt + B \cos kt \) is also a solution.

Solve the differential equation. \( y' + y = 1 \)

The Pacific halibut fishery has been modeled by the differential equation \( \frac {dy}{dt} = ky (1 - \frac {y}{M}) \) where \( y(t) \) is the biomass (the total mass of the members of the population) in kilograms at time \( t \) (measured in years), the carrying capacity is estimated to be \( M = 8 \times 10^7 \) kg, and \( k = 0.71 \) per year. (a) If \( y(0) = 2 \times 10^7 \) kg, find the biomass a year later. (b) How long will it take for the biomass to reach \( 4 \times 10^7 \) kg?

Solve the differential equation. \( (e^y - 1)y' = 2 + \cos x \)

Which of the following functions are solutions of the differential equation \( y^{"} + y = \sin x ? \) (a) \( y = \sin x \) (b) \( y = \cos x \) (c) \( y = \frac {1}{2} x \sin x \) (d) \( y = - \frac{1}{2} x \cos x \)

Solve the differential equation. \( y' - y = e^x \)

Suppose a population \( P(t) \) satisfies \( \frac {dP}{dt} = 0.4 P - 0.001P^2 P(0) = 50 \) where \( t \) us measured in years. (a) What is the carrying capacity? (b) What is \( P'(0)? \) (c) When will the population reach \( 50% \) of the carrying capacity?

Solve the differential equation. \( \frac {du}{dt} = \frac {1 + t^4}{ut^2 + u^4t^2} \)

Solve the differential equation. \( y' = x - y \)

Solve the differential equation. \( \frac {d \theta}{dt} = \frac {t \sec \theta}{\theta e^{t^2}} \)

(a) What can you say about a solution of the equation \( y{'} = - y^{2} \) just by looking at the differential equation? (b) Verify that all members of the family \( y = 1/(x + C) \) are solutions of the equation in part (a). (c) Can you think of a solution of the differential equation \( y^{'} = - y^{2} \) that is not a member of the family in part (b)? (d) Find a solution of the initial-value problem. \( y^{'} = - y^2 \) \( y (0) = 0.5 \)

Solve the differential equation. \( \frac {dH}{dR} = \frac {RH^2 \sqrt{1 + R^2}}{\ln H} \)

(a) What can you say about the graph of a solution of the equation \( y^{'} = xy^3 \) when \( x \) is close to \( 0? \) What if \( x \) is large? (b) Verify that all members of the family \( y = (c - x^2) ^{{-}{1/2}} \) are solutions of the differential equation \( y^{'} = xy^3. \) (c) Graph several members of the family of solutions on a common screen. Do the graphs confirm what you predicted in part (a)? (d) Find a solution of the initial-value problem \( y^{'} = xy^3 \) \( y(0) = 2 \)

Solve the differential equation. \( 4x^3y + x^4y' = \sin^3x \)

Solve the differential equation. \( xy' + y = \sqrt x \)

The population of the world was about 6.1 billion in 2000. Birth rates around that time ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Let's assume that the carrying capacity for world population is 20 billion. (a) Write the logistic differential equation for these data. (Because the initial population is small compared to the carrying capacity, you can take \( k \) to be an estimate of the initial relative growth rate.) (b) Use the logistic model to estimate the world population in the year 2010 and compare with the actual population of 6.9 billion. (c) Use the logistic model to predict the world population in the years 2100 and 2500.

Solve the differential equation. \( \frac {dp}{dt} = t^2 - p + t^2 - 1 \)

A population is modeled by the differential equation \( \frac {dP}{dt} = 1.2 P \left(1 - \frac{P}{4200} \right) \) (a) For what values of \( P \) is the population increasing? (b) For what values of \( P \) is the population decreasing? (c) What are the equilibrium solutions?

Solve the differential equation. \( 2xy' + y = 2 \sqrt x \)

(a) Assume that the carrying capacity for the US population is 800 million. Use it and the fact that the population was 282 million in 2000 to formulate a logistic model for the US population. (b) Determine the value of \( k \) in your model by using the fact that the population in 2010 was 309 million. (c) Use your model to predict the US population in the years 2100 and 2200. (d) Use your model to predict the year in which the US population will exceed 500 million.

Solve the differential equation. \( \frac {dz}{dt} + e^{t+z} = 0 \)

Sketch a direction field for the differential equation. Then use it to sketch three solution curves. $$ y^{\prime}=x-y+1 $$

The Fitzhugh-Nagumo model for the electrical impulse in a neuron states that, in the absence of relaxation effects, the electrical potential in a neuron \( v(t) \) obeys the differential equation \( \frac {dv}{dt} = - v [v^2 - (1 + a) v + a] \) where \( a \) is a positive constant such that \( 0 < a < 1\. \) (a) For what values of \( v \) is \( v \) unchanging (that is, \( dv/dt = 0)?. \) (b) For what values of \( v \) is \( v \) increasing? (c) For what values of \( v \) is \( v \) decreasing?

Solve the differential equation. \( xy' - 2y = x^2, x > 0 \)

One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction \( y \) of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by \( y. \) (b) Solve the differential equation. (c) A small town has 1000 inhabitants. At \( 8 AM, \) 80 people have heard a rumor. By noon half the town has heard it. At what time will \( 90% \) of the population have heard the rumor?

Find the solution of the differential equation that satisfies the given initial condition. \( \frac {dy}{dx} = xe^y, y(0) = 0 \)

Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. \( y' = y - 2x, (1,0) \)

Solve the differential equation. \( y' + 2xy = 1 \)

Biologist stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 10,000. The number of fish tripled on the first year. (a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after \( t \) years. (b) How long will it take for the population to increase to 5000?

Find the solution of the differential equation that satisfies the given initial condition. \( \frac {dy}{dx} = \frac {x \sin x}{y}, y(0) = -1 \)

Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. \( y' = 1 - xy, (0,0) \)

Solve the differential equation. \( t^2 \frac {dy}{dt} + 3ty = \sqrt {1 + t^2}, t > 0 \)

Find the solution of the differential equation that satisfies the given initial condition. \( \frac {du}{dt} = \frac {2t + \sec^2t}{2u}, u(0) = 0 \)

Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. \( y' = y + xy, (0,0) \)