8 Differential Equations

8.1

Use separation of variables to solve the following differential equations with given initialconditions.

(a) dydt

= −2ty, y(0) = 10

(b) dydt

= y(1 − y), y(0) = 0.5, (Hint: 1y(y−1)

= 1y−1

− 1y).

Solution

(a) dyy

= −2tdt ⇒ ln |y| = −t2 + C0 ⇒ y = Ce−t2 . y(0) = 10 = C ⇒ y(t) = 10e−t2 .

(b) dyy(y−1)

= −dt ⇒ ∫

( 1y−1

− 1y)dy = − ∫ dt ⇒ ln

∣

∣

∣

∣

y−1y

∣

∣

∣

∣

= −t + C0 ⇒ y−1y

= Ce−t. y(0) = 0.5

⇒ C = −1. Thus, y − 1 = −e−ty, which gives y(t) = 11+e−t .

8.2

Consider the differential equation and initial condition dydt

= 1 − y, y(0) = 0.5.

(a) Solve for y as a function of t using separation of variables.

(b) Use the initial condition to find the value of C.

Solution

(a) The initial condition y(0) = 0.5 guarantees that 1−y > 0, and therefore we can separatevariables:

∫

dy

1 − y=∫

dt = t + C =⇒ − ln |1 − y| = t + C

To find C we set t = 0 and y = 0.5:

− ln(0.5) = C =⇒ C = ln 2.

Solving for y we get:

ln |1 − y| = −t − ln 2 =⇒ 1 − y = e−te− ln 2 =1

2e−t =⇒ y = 1 − 1

2e−t.

Note: We should verify our calculations by doing the check, that is we should check that thefunction

y = 1 − 1

2e−t

satisfies the differential equationdy

dt= 1 − y

and the initial condition y(0) = 0.5.

(b) C = 0.5.

1

8.3

Consider the differential equation and initial condition

dy

dt= 1 + y2, y(0) = 0.5.

Repeat the process described in problem 8.2 and find the solution, i.e. determine the functiony(t) that satisfies this differential equation.

Solution

Since 1 + y2 6= 0 we can separate variables:

∫

dy

1 + y2=∫

dt = t + C,

that istan−1 y = t + C.

Putting t = 0 and y = 0.5 gives

C = tan−1 1

2≈ 0.464

and then solving for y we get

y = tan(

t + tan−1 1

2

)

= tan(t + 0.464).

Again we should do the check.

8.4

Consider the differential equation and initial condition

dy

dt= 1 − y2, y(0) = 0.5.

Solve using separation of variables (Hint: use partial fractions.)

Solution

Separating variables gives∫

dy

1 − y2=∫

dt = t + C.

Using partial fractions we get

∫

dy

1 − y2=

1

2

∫

(

1

1 − y+

1

1 + y

)

dy =1

2(− ln |1 − y| + ln |1 + y|)

2

and therefore1

2ln

∣

∣

∣

∣

∣

1 + y

1 − y

∣

∣

∣

∣

∣

= t + C.

Putting t = 0 and y = 0.5 gives

C =1

2ln 3.

Solving for y we get

1 + y

1 − y= e2t+ln 3 = 3e2t =⇒ 1 + y = 3e2t − 3e2ty

y(1 + 3e2t) = 3e2t − 1 =⇒ y =3e2t − 1

3e2t + 1.

A point worth noting is that the initial condition implies that

1 + y

1 − y> 0.

Again we should do the check.

8.5

Solve the following differential equation and initial condition, assuming that k > 0 is aconstant:

dy

dt= ky2/3 y(0) = y0

Solution

1

y2/3dy = k dt

∫

y−2/3 dy = kt + C

y1/3

1/3= kt + C

y1/3 =1

3kt + C ′

where we have renamed the constant (C ′ = 13C). We note that the constant can be obtained

from the initial conditions. That is, plugging t = 0 into the above, we get

y1/30 = C ′

3

Continuing to solve for the function, we get

y = (1

3kt + C ′)3

Thus

y(t) = (1

3kt + y

1/30 )3.

8.6

A certain cylindrical water tank has a hole in the bottom, out of which water flows. Theheight of water in the tank, h(t), can be described by the differential equation dh

dt= −k

√h

where k is a positive constant. If the height of the water is initially h0, determine how muchtime elapses before the tank is empty.

Solution

h(0) = h0

dh

dt= −k

√h

∫ h(T )

h0

dh√h

= −k∫ T

0dt

2√

h

∣

∣

∣

∣

h(T )

h0

= −kT

2(√

h(T ) −√

h0) = −kT

√

h(T ) =−k

2T +

√

h0

h(T ) = (√

h0 −k

2T )2

the tank is empty when h(T ) = 0, i.e.

√

h0 −k

2T = 0

T =2√

h0

k

4

8.7

The position of a particle is described by

dx

dt=

√1 − x2 0 ≤ t ≤ π

2

(a)Find the position as a function of time given that the partical starts at x = 0 initially.

(b)Where is the particle when t = π2?

(c)At which position(s) is the particle moving the fastest? The slowest?

Solution

(a)∫

dx√1 − x2

=∫

dt

sin−1x = t + C

x = sin(t + c)

at t = 0, x = 0, so C = 0 Therefore, x = sint

(b)at t = π2

x = sinπ2

= 1

(c)

v =dx

dt=

√1 − x2

Therefore v is greatest when x = 0 (then v = 1), and v is smallest when x = +1 (thenv = 0). (We do not consider x = −1 because it is not in the domain.)

8.8

In an experiment involving yeast cells, it was determined that mortality of the cells increasedat a linear rate, i.e. that at time t after the beginning of the experiment the mortality ratewas

m(t) = m0 + rt.

The experiment was started with No cells at time t = 0. Let N(t) represent the populationsize of the cells at time t.

(a) If no “birth” occurs, then

dN(t)/dt = −m(t)N(t).

Solve this differential equation by separation of variables, i.e find N(t) as a function of time.

5

(b) Suppose cells are “born” at a constant rate b, so that the differential equation is

dN(t)/dt = −m(t)N(t) + bN(t).

Determine how this affects the population size at time t.

Solution

(a) We have to solve the differential equation

dN(t)/dt = −(m0 + rt)N(t).

Separation of variables yields

∫

dN

N=∫

−(m0 + rt)dt,

henceln(N) = −m0t − rt2/2 + c.

Taking exponentials, this yields

N(t) = Ce−m0t−rt2/2

where the constant C must be determined from the initial condition N(0) = N0, henceC = N0, i.e.

N(t) = N0e−m0t−rt2/2.

(b) The only difference to the previous question is that we add a constant b to the constant−m0 in (a):

dN(t)

dt= −(m0 + rt + b)N(t)

∫

dN

N=

∫

−(m0 + b + rt)dt

ln(N) = −m0t − bt − rt2

2+ C

N(t) = N0e−(m0+b)t−rt2/2

8.9

Muscle cells are known to be powered by filaments of the protein actin which slide past oneanother. The filaments are moved by cross-bridges of myosin, that act like little motors,which attach, pull the filaments, and then detach. Let n be the fraction of myosin cross-bridges that are attached at time t. A model for cross-bridge attachment is:

dn

dt= k1(1 − n) − k2n, n(0) = n0

6

where k1 > 0, k2 > 0 are constants.

(a) Solve this differential equation and determine the fraction of attached cross-bridges n(t)as a function of time t.

(b) What value does n(t) approach after a long time?

(c) Suppose k1 = 1.0 and k2 = 0.2. Starting from n(0) = 0, how long does it take for 50% ofthe cross-bridges to become attached ?

Solution

(a) We have to solve the differential equation

dn

dt= k1(1 − n) − k2n = k1 − k1n − k2n = k1 − n(k1 + k2)

Separation of variables yields

∫

dn

k1 − n(k1 + k2)=∫

dt

hence −1

k1 + k2ln |k1 − n(k1 + k2)| = t + c

and thusk1 − n(k1 + k2) = Ce−(k1+k2)t. (1)

Thus, the general solution is

n(t) =−Ce−(k1+k2)t + k1

k1 + k2.

The constant C must be determined by the initial condition n(0) = n0 using equation (1):

C = k1 − n0(k1 + k2).

(b) As t → ∞, we get n(t) → k1

k1+k2

, because the exponential term approaches 0.(c) Plugging in k1 = 1.0 and k2 = 0.2, we have

n(t) =−Ce−1.2t + 1

1.2.

From (a) we find that C = 1 − 0(1.2) = 1. We have to solve for n(t) = 0.5, hence

0.5 = (−e−1.2t + 1)/1.2

which yieldse−1.2t = 0.4 =⇒ t = − ln(0.4)/1.2 ≈ 0.7636.

7

8.10

The velocity of an object falling under the effect of gravity with air resistance is given by:dvdt

= f(v) = g − kv, v(0) = vo, where g > 0 is acceleration due to gravity and k > 0 is africtional coefficient (both constant).

(a) Sketch the function f(v) as a function of v. Identify a value of v for which no changeoccurs, i.e., for which f(v) = 0. This is called a steady state value of the velocity or a fixed

point. Interpret what this value represents.

(b) Explain what happens if the initial velocity is larger or smaller than this steady statevalue. Will the velocity increase or decrease ? (Recall that the sign of the derivative dv/dttells us whether v(t) is an increasing or decreasing function of t.

(c) Use separation of variables to find the function v(t).

Solution

(a) See the sketch below. The steady state value is v = gk. At this value of v there is no

change in the velocity (that is dvdt

= 0) and so v(t) = gk

for all time t that the differentialequation remains valid.

dv/dt = g − kv

v

Stable steady state = g/k

g/k

Figure 1: Plot for problem 8.10

(b) If v(0) > gk

thendv

dt< 0

for all time t that the differential equation remains valid, and so v(t) decreases to the terminalvelocity g

k. On the other hand, if v(0) < g

k, then

dv

dt> 0

8

for all time t that the differential equation remains valid, and so v(t) increases to the terminalvelocity g

k.

(c) Separating variables gives

∫ dv

g − kv=∫

dt = t + C :

that is

−1

kln |g − kv| = t + C

ln |g − kv| = −kt − kC

g − kv = C ′e−kt

Putting t = 0 and v(0) = v0 givesC ′ = g − kv0

and then solving for v we get:

g − (g − kv0)e−kt = kv

v(t) =g

k−(

g

k− v0

)

e−kt.

8.11

A model for the velocity of a sky diver (slightly different from the model we have alreadystudied) is

dv

dt= 9 − v2

(a) What is this skydiver’s ”terminal velocity”; that is, near what velocity will the sky divereventually stabilize?

(b) Starting from rest, how long will it take to reach half of this velocity?

Solution

(a) Velocity will stop changing and reach a ”terminal” value when

dv

dt= 0

i.e.0 = 9 − v2

v = 3.

We reject v = −3 because it is in the wrong direction.

9

(b)dv

dt= 9 − v2

dv

9 − v2= dt

∫ 1.5

0

dv

(3 − v)(3 + v)=∫ T

0dt = T

(time to reach velocity v = 32) Partial Fractions:

1

(3 + v)(3 − v)=

a

(3 + v)+

B

3 − v

1 = A(3 − v) + B(3 + v)

A = B =1

6

T =1

6

∫ 1.5

0

1

(3 + v)+

1

(3 − v)dv =

1

6(ln |3 + v| − ln |3 − v|)

∣

∣

∣

∣

1.5

0=

1

6ln |4.5

1.5|

T =1

6ln 3 = 0.183 sec

8.12 Newton’s Law of Cooling

Consider the differential equation for Newton’s Law of Cooling

dT

dt= −k(T − E)

and the initial condition: T (0) = T0. Solve this differential equation by the method ofseparation of variables i.e. find T (t). Interpret your result.

Solution

dT

dt= −k(T − E), solve for T (t).

Separation of variables:

dT

T − E= −k dt

Integration:

∫

dT

T − E= −

∫

k dt

ln |T − E| = −kt + C0

10

If T > E, we get:

T − E = Ce−kt

T = E + Ce−kt

Using the initial condition, T (0) = T0, we find that T0 = E + C ⇒ C = T0 − E. This givesus Newton’s Law of Cooling, T (t) = E + (T0 − E)e−kt.However, if T < E, we get:

−T + E = Ce−kt

T = E − Ce−kt

In this case, with the initial condition we get T0 = E −C, or C = E −T0. Plugging this intothe equation gives:

T = E − (E − T0)e−kt =⇒ T = E + (T0 − E)e−kt.

Again, we end up with the familiar equation for Newton’s Law of Cooling.Interpret result: For t → ∞, T (t) → E + very small quantity ≈ E.Thus, the coffee temperature approaches the value of the environment as time goes on. Ifinitially it is greater than the environment, i.e. if T0 > E, then it approaches equilibriumfrom above. If T0 < E, then the temperature approaches equilibrium from below.

8.13

Let V (t) be the volume of a spherical cell that is expanding by absorbing water from itssurface area. Suppose that the rate of increase of volume is simply proportional to thesurface area of the sphere, and that, initially, the volume is V0. Find a differential equationthat describes the way that V changes. Use the connection between surface area and volumein a sphere to rewrite your differential equation in terms of the volume alone. You shouldget an equation in the form

dV

dt= kV 2/3,

where k is some constant. Solve the differential equation to show how the volume changesas a function of the time. (Hint: recall that for a sphere, V = (4/3)πr3, S = 4πr2.)

Solution

The volume changes at a rate proportional to the surface area, i.e.

dV

dt= αS.

V = (4/3)πr3 ⇒ r = (3V/4π)1/3

S = 4πr2 = 4π[(3V/4π)1/3]2 = βV 2/3

11

where β is a constantβ = 4π(3/4π)2/3

ThusdV

dt= αβV 2/3 = kV 2/3,

where k is just a constant. This equation is in the same form as the one solved in problem8.5, and the method of solution will be similarThus

V (t) = (1

3kt + V

1/30 )3.

Qualitative Methods

8.14

In problems 8.2, 8.3, and 8.4 above, your method was to calculate a formula for the unknownfunction y(t). Now you will use qualitative methods for these same problems instead: Sketcha graph of the expression dy/dt versus y in each case. (E.g., in problem 8.2 you will besketching the straight line f(y) = 1 − y). In each case, determine the values of y at whichdy/dt = 0 (steady states, or critical points).

Solution

See the sketches in Figure 2.

8.15

Find the fixed points (steady states, critical points) of the following differential equationsand classify their stability.

(a)dy

dt= y2 − 4 (b)

dy

dt= 4 − y2

(c)dy

dt= y(y − 1)(y + 1) (d)

dy

dt= cos(2πy)

Solution

(a) f(y) = y2 − 4. f(y) = 0 gives us two steady states: y = −2 and y = 2. Since f ′(y) = 2y,f ′(−2) = −4 < 0 ⇒ y = −2 is stable. f ′(2) = 4 > 0 ⇒ y = 2 is unstable.

(b) f(y) = 4− y2. f(y) = 0 gives us two steady states: y = −2 and y = 2. But f ′(y) = −2y.The stability is the reverse of (a). y = −2 is unstable, y = 2 is stable.

12

dy/dt = 1− y

y

Stable steady state = 1

1

Stable steady state: NONE

dy/dt = 1+y2

y

Stable steady state: ± 1

−1 1

dy/dt = 1−y2

y

Figure 2: Plots for problem 8.14

(c) f(y) = y(y − 1)(y + 1). The steady states are: y = −1, 0, 1. f ′(y) = (y − 1)(y + 1) +y(y + 1) + y(y − 1). Thus, f ′(−1) = 2, f ′(0) = −1, f ′(1) = 2. Therefore, y = −1, 1 areunstable, y = 0 is stable.

(d) f(y) = cos(2πy). For 0 < y < 1, there are 2 steady states: y = 1/4, 3/4. Sincef ′(y) = −2π sin(2πy), f ′(1/4) = −2π and f ′(3/4) = 2π. Thus, y = 1/4 is stable andy = 3/4 is unstable.

8.16

For each part of problem 8.15, sketch the qualitative behavior, i.e. the flows along the yaxis.

Solution

See Figure 3.

13

y

dy/dt

−2 2

dy/dt

y−2 2

y

dy/dt

−1 0 1 y−1.25 −.75 −.25 .25 .751.25

Figure 3: Plots for problem 8.16

8.17

Determine the stability type of the equilibrium solution (fixed point) in the sky-diving equa-tion

dv

dt= g − kv.

Interpret your result in the context of the terminal velocity of the sky-diver.

Solution

The only steady state is v = gk

and it is stable. This follows from the Linear StabilityCondition since f ′(v) = −k < 0, where f(v) = g − kv. This means that any object in freefall will eventually reach the terminal velocity v = g

k.

8.18

Newton’s Law of Cooling states that the temperature of an object T (t) changes at a rate thatdepends on the difference between that temperature and the temperature of the environment,E: dT

dt= −k(T − E), T (0) = T0. Assume that E, k are positive constants.

(a) Sketch the expression dTdt

= −k(T − E) as a function of T . Identify points on yoursketch that correspond to value(s) of the temperature that would not change (i.e. at whichdT/dt = 0).

14

(b) Use your sketch to identify ranges of the temperature T for which dT/dt is negative orpositive, i.e. for which the temperature is increasing or decreasing. Use arrows on the T axisto indicate the “flow”.

(c) Interpret your sketch. Give a verbal description of what Newton’s Law of Cooling issaying about the temperature T (t) of the body as the time t increases.

Solution

(a) See the sketch in Figure 4. The steady state temperature is just the ambient temperature,that is, T = E.

dT/dt = −k (T−E)

T

Stable steady state = E

E

Figure 4: Plot for problem 8.18

(b) From the differential equation dTdt

= −k(T − E) it is clear that if T > E then dTdt

isnegative and if T < E then dT

dtis positive. See the sketch in Figure 4 for the “direction of

flow”.

(c) Newton’s law of cooling is saying that objects either warm up to the room temperature(if T (0) < E) or cool down to it (if T (0) > E) , and this happens exponentially. In all casesT (t) → E as t → ∞.

8.19

Find a differential equation that has exactly three steady states, at y = 0, 2, 4 and for whichonly y = 2 is an unstable steady state. How would you change this to an equation for whichy = 2 is stable and the other two states are unstable?

Solution

The differential equation f(x) = dydt

= ry(y − 2)(y − 4) has exactly 3 steady states aty = 0, 2, 4. Now we need to determine if the constant r should be positive or negative inorder that y = 2 be unstable. The derivative of this function is:

f ′(y) = r(y − 2)(y − 4) + ry(y − 4) + ry(y − 2).

15

Thus f ′(2) = ry(y−4) = −4r. For y = 2 to be an unstable steady state, f ′(2) > 0 ⇒ −4r >0 and r < 0. Therefore, the differential equation is dy

dt= −ry(y− 2)(y− 4) (r > 0 constant).

If we change the sign of the right hand side of the equation, i.e. dydt

= ry(y− 2)(y− 4), y = 2becomes stable while the other two become unstable.

8.20

Consider the differential equation

dy

dt= y2 − r,

where r is some positive constant. Find the fixed points of this equation and determine theirstability. What happens as r is decreased ? What happens when r = 0? The Linear StabilityCondition fails (WHY?). But would you describe the fixed point as stable or unstable inthat case?

Solution

The fixed points (steady states) of the differential equationdy

dt= y2 − r are y = ±√

r. From

the graph we see that −√r is stable; whereas

√r is unstable. As r → 0+ the steady states

±√r coalesce and both become 0. The linear stability condition fails because the derivative

of f(y) = y2 at y = 0 is 0 i.e. the slope is neither positive nor negative. On the other hand,y = 0 is an unstable steady state because the trajectories point away from it for y > 0.

8.21

You are a resource ecologist, hired to manage the population of fish in some lakes. Recordsindicate that in some lakes, the population of fish stabilizes at a very low level, and that inother lakes, the population stabilizes at a high level. The outcome seems to depend on thestarting density of fish in the lake. You are asked to model the density of the fish populationy(t) in a lake by a differential equation:

dy

dt= f(y)

(a)You are asked to predict whether any major changes in the behavior of the fish populationwould occur if the lake was continuously stocked (i.e. new fish were continuously broughtfrom a hatchery and added to the lake). How would the model change? Use a diagramand explain clearly what you expect would happen, emphasizing any abrupt transitions(bifurcations) that might occur.

(b)Repeat part (b) but assume instead that the lake is used for fishing, and that fish arecontinuously harvested at some constant rate (and no new fish are added). What happensin this case?

16

Solution

(a)The equation would then be dydt

= f(y)+ s where s is a positive constant ”stocking rate”.Expect the transition shown in figure 5 .

increase s

Figure 5:

The fish population should always stabilize at the high level beyond some thus hold valueof the stocking rates.

(b) The model would change to dydt

= f(y) − h where h is constant harvesting rate. Weexpect the transition shown in figure 6 .

increase in h

Figure 6:

The fish population may be driven to extinction.

8.22

(a) Consider the differential equation dydt

= sin(y). By plotting dy/dt versus y, identifyvalues of y which would not change (steady states), and regions for which y is increasing or

17

dy/dt = sin(y)

y

Figure 7: Plot for problem 8.22 (a)

decreasing. Use this diagram to summarize all the possible outcomes (behaviors of y(t)) forvarious initial values of y.

(b) Now consider a related equation, dydt

= sin(y)+A where A is some positive constant. Howdoes the behavior that you described in (a) change as the constant A is increased gradually?For what value(s) of A does the behavior change dramatically ? Such changes are calledbifurcations.

Solution

(a) The steady states are those values of y such that sin y = 0, thus y = nπ, where n isany integer. See the plot in Figure 7. The regions where y(t) is increasing are given by2nπ ≤ y ≤ (2n + 1)π, (i.e between an even and an odd multiple of π) and the regions wherey(t) is decreasing are given by (2n−1)π ≤ y ≤ 2nπ, where n is any integer (i.e between an oddand an even multiple of π). If y(0) = nπ then y(t) = nπ for all t, if 2nπ < y(0) ≤ (2n + 1)πthen y(t) → (2n + 1)π as t → ∞, and if (2n − 1)π ≤ y(0) < 2nπ then y(t) → (2n − 1)π.

(b) The steady state solutions ofdt

dt= sin y + A are those values of y, if any, such that

sin y + A = 0. If |A| < 1 then there are infinitely many such y values and the behavior ofthe solutions is as in part (a). See Figure 8. If A > 1 (resp. A < 1) the behavior is quitedifferent, in that all solutions tend to +∞ (resp. −∞). If A = 1 (resp. A = −1) then steadystates occur at intervals of 2π, and all solutions tend to increase (resp. decrease) to the nextsteady state. Thus, bifurcation occurs for A = ±1. (But, note that the question only askedfor positive A values.)

18

dy/dt = sin(y) + A

y

typical picutre if −1 < A < 1

dy/dt = sin(y) + A

y

A = 1A = −1 similar

dy/dt = sin(y) + A

yA > 1 (A < −1 similar)Stable steady state: NONE

Figure 8: Plots for problem 8.22 (b)

19

Orthogonal Trajectories

8.23

Find the curve passing through the point (1,2) and orthogonal to the family of curves x2y = k.

Solution

We find the slope of the curve x2y = k at the point (x,y) by implicit differentiation: 2xy +x2 dy

dx= 0 or dy

dx= −2y

x. We are interested in moving in a perpendicular direction; that is, in

the direction such that the slope is the negative reciprocal of − 2yx

, i.e.:

dy

dx=

x

2y

This is a separable equation and we solve it in the usual way:

∫

2ydy

dx=∫

2ydy =∫

xdx

This means that

y2 =x2

2+ C.

The constant C can be determined by requiring that this curve pass through (1,2). Thismeans that the curve is

y2 =x2

2+

7

2.

8.24

Sketch the family of curves

x2

9+ y2 = k2

for k constant. Find the orthogonal trajectories by setting up and solving the appropriatedifferential equation. Sketch these on the same picture.

Solution

See Figure 9 for the sketch. The tangent line at a point (x, y), x 6= 0, on the curve x2

9+y2 =

k2 has slope − x9y

, and therefore the differential equation for the orthogonal trajectories isdydx

= 9yx. Assume y 6= 0. Then, separating variables leads to

∫ dyy

= 9∫ dx

x, that is y = Kx9,

where K is a constant. Thus, the orthogonal trajectories are x = 0 and the power curvesy = Kx9, where K is a constant. Note: K = 0 gives y = 0.

20

(1/9) x^2 + y^2 = k^2

3k−3k

k

−k

Figure 9: Plot for problem 8.24

8.25

Find the family of curves that are orthogonal to the parabola

y = Cx2

Important remark about a possible pitfall: C is a constant only for a given parabola, not forthe whole family. For this reason, we need to eliminate it from the general expression for theslope of the parabola (dy/dx = 2Cx). This can be done by plugging in C = y/x2 in placeof the constant C, (i.e getting dy/dx = 2y/x, a relationship that holds for all the parabola).Proceed from this point to find the orthogonal curves.

Solution

dy/dx = 2Cx = 2y/x. Thus, the slope of the orthogonal trajectories is dy/dx = −x/(2y).Solve this separable equation:

∫

2y dy = − ∫ x dx we obtain: y2 = −x2/2+C, i.e. x2/2+y2 =C. As the constant, C, takes on values, this generates a family of ellipses.

8.26

Sketch the family of curves

xy = k

21

for k constant. Find the orthogonal trajectories by setting up and solving the appropriatedifferential equation. Sketch these on the same picture.

Solution

xy = C

Find slope dydx

: x dydx

+ y = 0 ⇒ dydx

= − yx.

Orthogonal trajectories:

dy

dx=

x

y⇒ y dy = x dx

⇒∫

y dy =∫

x dx

⇒ y2

2=

x2

2+ const

⇒ x2 − y2 = K

Applications to Biology

8.27

According to Klaassen and Lindstrom (1996) J theor Biol 183:29-34, the “fuel load” (nectar)carried by a hummingbird, F (t) depends on the rate of intake (from flowers) and the rate ofconsumption due to metabolism. They assume that intake takes place at a constant rate α.They also assume that consumption increases when the bird is heavier (carrying more fuel).Suppose that fuel is consumed at a rate proportional to the amount of fuel being carried(with proportionality constant β).

(a) What features are being neglected or simplified in this model?

(b) Write down the differential equation model for F (t).

(c) Find F (t) as a function of time t.

(d) Klaassen and Lindstrom determined that for a hummingbird, α = 0.48 gm fuel /day andβ = 0.09 /day. Determine the steady state level of fuel carried by the bird.

Solution

(a) This model is simplified because it assumes that the rate of intake and the rate ofconsumption are constant. It also does not take into account the fact that energy must beexpended at each flower for the hummingbird to get any fuel.

(b)

dF/dt = α − βF

22

(c) Separation of variables yields∫ dF

α−βF=∫

dt, and hence −1β

ln |α− βF | = t +C, which, as

in (9b), yields α − βF (t) = ce−βt, hence F (t) = (α − ce−βt)/β.

(d) The equilibrium level is reached when dF/dt = 0. This happens when α − βF = 0, i.e.when F = α/β = 0.48/0.09 ≈ 5.33. This is the steady state level of fuel carried (i.e., whenthe hummingbird carries that amount of fuel, then intake and consumption balance eachother).

8.28

A biochemical reaction S → P is catalysed by an enzyme E. The speed of the reactiondepends on the concentration of the substrate S. It is found that the substrate concentration,x changes at the rate

dx

dt= rx − Kx

k + x

where r, K, k are positive constants. (The second term is often called “Michaelis Mentenkinetics” in chemistry.) How many steady state concentration values are there? What arethese values? Explain the behaviour of the solutions x(t) for very large and for very smallvalues of r. You may use sketches of the flow along the x axis.

Solution

To find the steady states we set dxdt

= 0 ⇒ rx − Kxk+x

= 0 ⇒ x(rk+rx−K)k+x

= 0. There are two

s.s.: x = 0, Kr− k. For very small r, K

r− k is likely the positive s.s. that is unstable (i.e.

k < Kr), x(t) decreases and approaches x = 0 as t increases if x(0) < K

r− k. For very large

r, Kr− k is likely the negative s.s. (i.e. k > K

r) which has no meaning for the biochemical

reaction. In this case, x = 0 can become unstable. In case of a large r and very small k,Kr− k can remain positive and unstable. See Figure 10.

8.29 Harvesting 1

Consider the Logistic equation with harvesting,

dN

dt= rN

(

1 − N

K

)

− H

(a) Explain the distinction between this term and another possible term that might havebeen used, −µy to represent a mortality or removal rate.

(b) Show that by defining a new variable, y = N/K, the equation in part can get rewrittenin the form:

dy

dt= ry(1 − y) − h

23

K/r−k0 x

dx/dt

Small r, K/r−k positive and unstable

x

dx/dt

K/r−k 0

Large r, K/r−k negative

K/r−k0 x

dx/dt

Large r, very small k, K/r−k positive and unstable

Figure 10: Plots for problem 8.28

What is the new constant h?

(c) Find the steady states of this equation (all values of y for which dy/dt = 0) and describethe stability properties of each one. There may be several cases to consider, and you areasked to list the possible outcomes and how they depend on parameters.

(d) Use a sketch of dy/dt versus y to explain what happens to the population for very low,intermediate, and high starting population sizes.

(e) What harvesting level is reasonable in this model? What happens if the harvesting isincreased to “unreasonable” levels? At what value of h does this transition take place?

Solution

Logistic equation with harvesting:

dN

dt= rN

(

1 − N

K

)

− H

where K is the carrying capacity and N is the population size.

24

(a) The harvesting term H is a constant removal rate, i.e. it does not depend on the popu-lation size. A term like −µy, where y = N/K, does depend on the population (linearly).

(b) If y =N

Kthen dy

dt= 1

KdNdt

= rNK

(

1 − NK

)

− HK

= ry(1 − y) − h, where h = HK

.

(c) The steady states are those values of y such that ry(1 − y) − h = 0. We rearrangethe equation such that we can use the quadratic formula: ry2 − ry + h = 0, and get y =

r ±√

r2 − 4rh

2r=

1 ±√

1 − 4h/r

2

. Three cases now arise: 4h/r > 1, 4h/r = 1 or

4h/r < 1. In the first case there are no steady state solutions anddy

dtis always negative.

(The level of harvesting is so great the population dies out.) In the second case there is onlyone steady state solution, namely y = 1/2, and it is unstable. For the last case there are

2 steady state solutions, namely y =

1 −√

1 − 4h/r

2

and y =

1 +√

1 − 4h/r

2

. The

smaller one is unstable and the larger is stable.

(d) y = p+ is stable while y = p− is unstable. For y(0) < p−, the population will becomeextinct. For any y(0) > p−, the population will settle to the steady state size y = p+. SeeFigure 11.

p− p

+y

dy/dt

Figure 11: Plot for problem 8.29

(e) A reasonable harvesting level is one that satisfies 4hr

< 1, or h < r4. If harvesting increases

to unreasonable levels (h > r4) then the population dies off. The transition takes place at

h = r4. This corresponds to H = rK

4.

8.30 Harvesting 2

(a) The Logistic Equation with harvesting discussed in a problem above has some unrealisticbehaviour. Explain why its behaviour for y = 0 is not biologically realistic.

(b) As a “fix” for this difficulty, we might consider a “corrected” equation of the form

dy

dt= ry(1 − y) − h

ay

a + y

25

where now the harvesting rate is h aya+y

. Explain what feature of this new equation makes it

more convincing. Explain this new dependence of harvesting on the population level. (i.e.what is being assumed about harvesting when the population is very low? when it is veryhigh?) It will be helpful to sketch how the harvesting depends on the population level.

(c) Determine the steady states of this new model equation by finding values of y for whichdy/dt = 0. Describe the stability properties of each one. There may be several cases toconsider, and you are asked to list the possible outcomes and how they depend on theparameters.

(d) It is slightly awkward to try to graph dy/dt versus y since the expression on the RHS(right hand side) of the equation is now a bit nasty. However, it is easy to plot each of theterms in the equation, i.e. f1(y) = ry(1 − y), f2(y) = hay/(a + y) on the same graph andask where one term is larger than the other. (This would depend on the values of r andh. You might want to pick values for these for your plot. Then ask yourself how your plotchanges if these values were larger or smaller.) By so doing, you can identify both steadystates (intersections) and regions where dy/dt is positive or negative. Use this idea to find aqualitative description of the behaviour of y in this modified model.

Solution

Fixing the harvesting term:

dy

dt= ry(1 − y) − h

ay

a + y

(a) In problem 8.29, the harvesting term was H, a constant. For y = 0 in this case, dydt

=−h(= −H

K).This says that the population will continue to decrease even when it is extinct.

(b) The new harvesting term is h aya+y

. It depends on the population size and unlike in

problem 8.29, when y = 0, dydt

= 0 with this new harvesting term.Sketch of harvesting versus y (in Figure 12):Low population: y small ⇒ h ay

a+y≈ hy (almost linear).

High population: y → ∞ ⇒ h aya+y

≈ ha (approaches some limiting value). Even if there isa higher and higher population, you can only harvest it so fast!

(c) dydt

= 0 ⇒ ry(1 − y) = haya+y

⇒ y = 0, and r(1 − y) = haa+y

.

⇒ r(1 − y)(a + y) = ha ⇒ ry2 + (ra − r)y + ha − ra = 0.

y =r − ra ±

√

(ra − r)2 − 4r(ha − ra)

2r

There are three steady-state values, at

y = 0,r − ra −

√

(ra − r)2 − 4r(ha − ra)

2r,

r − ra +√

(ra − r)2 − 4r(ha − ra)

2r.

26

a+yh ay

y

y=ha

Figure 12: Plot of modified harvesting term for problem 8.30

a+yh ay

yQP

ry(1−y)

Figure 13: Plot for problem 8.30 (d)

There are now three cases: (1) (ra−r)2−4r(ha−ra) > 0. In this case, there are 3 steady state

values at y = 0,r − ra −

√

(ra − r)2 − 4r(ha − ra)

2r,

r − ra +√

(ra − r)2 − 4r(ha − ra)

2r.

The middle one is unstable, while y = 0,r − ra +

√

(ra − r)2 − 4r(ha − ra)

2rare both stable.

(2) (ra − r)2 − 4r(ha − ra) = 0. In this case there are 2 steady states at y = 0 (unstable)

and y =r − ra

2r=

1 − a

2(stable). (3) (ra − r)2 − 4r(ha − ra) < 0. In this case there is only

one steady state at y = 0, which is stable.

(d) See Figure 13.Intersections: ry(1 − y) = h ay

a+y⇒ y = 0, and y = P, y = Q.

Ay2 + By + C = 0 ⇒ y =r − ra ±

√

(ra − r)2 − 4r(ha − ra)

2r(= P, Q) .

27

dydt

yP Q

Figure 14: Plot for problem 8.30 (d): Case (i)

a+yh ayry(1−y)

P = Q 1 y

dydt

P = Q y

Figure 15: Plot for problem 8.30 (d): Case (ii)

Cases: for various combinations of r, a, and h, we may have(i) both intersections at y = P , y = Q (distinct roots)(ii) one intersection at y = P = Q(iii) neither intersection (no real roots)Note: y = 0 is always an intersection point (i.e. extinction is always a steady state here).

Case (i): See Figure 14. It occurs when (ra − r)2 − 4r(ha − ra) > 0 [B2 − 4AC > 0].

dy

dt= ry(1 − y) − h

ay

a + y

If y0 < P , then population goes extinct.If y0 > P , then population approaches equilibrium size at y = Q.

Case (ii): See Figure 15. It occurs when (ra − r)2 − 4r(ha − ra) = 0 [B2 − 4AC = 0].Population approaches equilibrium size at y = P (= Q).

28

a+yh ay

y1

ry(1−y)

dydt

y

Figure 16: Plot for problem 8.30 (d): Case (iii)

Case (iii): See Figure 16. It occurs when (ra − r)2 − 4r(ha − ra) < 0 [B2 − 4AC < 0].Population goes extinct!

8.31 Mouse Population Type 1

A population of mice reproduces at a per capita rate r and is preyed upon by a cat. Thepredation rate is proportional to the density of the mice. Explain the following simple modelfor the population of mice:

dy

dt= ry − py

Determine what happens to the mice. Show that the outcome depends on the relative valuesof the parameters r and p.

Solution

This model assumes that the population size of the mice increases at a rate r, and decreasesonly due to the predation rate p of the cat. Both rates are assumed proportional to thecurrent population size. From the graphs of dy

dtvs. y, we see that there are 3 cases: (i) if

r > p, the population size will increase exponentially (ii) if r = p, the population size willremain constant and (iii) if r < p, the population size will decrease to extinction.

8.32 Mouse population Type 2

Consider a mouse and cat model in which the cat (predator) has a so-called “Type II”response, i.e. in which the differential equation is

dy

dt= ry − Pmax

y

a + y.

29

Explain the differences in this version of the model. What is being assumed about thepredator? One steady state of this equation is y=0. Find the other steady state. (Does theother steady state always exist? What has to be true for this steady state to be biologicallymeaningful?) Use the Linear Stability Condition to show that this second steady state isunstable whenever it exists. Interpret your findings.

Solution

The predation term Pmaxy

a+yin this problem is similar to the modified harvesting term in

problem 8.30 (see Figure 12). When the population of mice is small, the cat’s predation rateincreases nearly linearly with population density (i.e. it is good at catching a small numberof mice). However, as the mice population grows and approaches infinity, the predation ratelevels off - the cat cannot keep catching more and more mice; there is a limit to its predation.The second steady state can be found by setting f(y) = dy

dt= 0:

ry(a + y) − Pmaxy

a + y= 0.

That is, ar + ry − Pmax = 0 and y =Pmax − ar

r. For this steady state to be biologically

meaningful, Pmax > ar.To find stability we compute f ′(y):

f ′(y) = r − Pmax

(

a + y − y

(a + y)2

)

= r − aPmax

(a + y)2

Linear Stability Condition:

f ′(

Pmax − ar

r

)

= r − Pmaxar2

(ar + Pmax − ar)2= r − ar2

Pmax

= r(

1 − ar

Pmax

)

Since Pmax > ar for this steady state to exist, 1 − ar

Pmax> 0. Therefore f ′

(

Pmax−arr

)

is

always positive, and this steady state is unstable.This means that whenever the mice population size is less than Pmax−ar

r, the population will

decrease to extinction, while if it is larger than Pmax−arr

, the population will continue growingin size.If the second steady state does not exist (i.e. if Pmax < ar) then y = 0 is the only steadystate, and it is unstable. The mice population will continue growing, no matter what itsoriginal size.

8.33 Mouse population Type 3

Consider a population of rodents (density y(t)) that grow at an exponential rate (dy/dt = ry),and suppose that a cat is brought in to control the population. The cat is a predator witha “Holling Type III” response, i.e. it is not very good at catching a small number of mice,

30

it gets better at finding prey when there are quite a few of them, but if there are too many,it can’t keep catching more and more of them. A model for the rodent population withpredation of this type might be:

dy

dt= ry − P

y2

1 + y2

Explain this model. (You may want to use graphical arguments). Use qualitative methods todetermine the predictions for the outcome of predation on the prey. Pay particular attentionto the way that these predictions depend on the parameter r.

Solution

There are 3 steady states: y = 0, P2r±√

( P2r

)2 − 1 if P2r

> 1. In this case only the middle one:

y = P2r

−√

( P2r

)2 − 1 is stable and is less than 1. When P2r

= 1, y = 1 is the steady state

resulting from the collapsing of the two nonzero steady states. It is unstable. When P2r

< 1,y = 0 becomes the only steady state which is unstable. y goes to infinity as time evolves.See Figure 17.

dy/dt

y0

P/2r>1

dy/dt

y

P/2r=1

dy/dt

y

P/2r<1

Figure 17: Plots for problem 8.33

31