Math 132: Calculus for Physical and Life Sciences 2 Solutions for Problem Set 7 Due Friday, April 4, 2008, at the beginning of class.

General Directions: You must show all work for credit on these problems.

1. All parts of this question deal with the differential equation y′ = x + y.

(a) Verify that y = −x − 1 is a solution of this differential equation. Solution: With y = −x − 1, we have y′ = \frac{dy}{dx} = −1. On the right side of the equation x + y = x + (−x − 1) = −1. Therefore y = −x − 1 solves this differential equation.

(b) Sketch the direction field of this equation by hand, showing the slopes at all points with −3 ≤ x ≤ 3, −3 ≤ y ≤ −3 with integer coordinates. Solution: (See plot with part (c) below).

(c) On your direction field graph, sketch approximate graphs of the solutions of this differential equation satisfying the initial conditions y(0) = −2, y(0) = 0, y(0) = 1. Solution: Here is a Maple plot showing this direction field and the solutions.

Note that more direction field segments are plotted here than the problem asked for, because Maple is using a default grid that is finer than the grid of points with integer coordinates.

(d) Use Euler’s method with n = 5, Δx = .2 to approximate the solution of the equation with the initial condition y(0) = 1 for 0 ≤ x ≤ 1.

Solution: We have

x_{0} = 0  y_{0} = 1 given from initial condition
x_{1} = .2  y_{1} = y_{0} + (x_{0} + y_{0})Δx = 1 + 1(.2) = 1.2
x_{2} = .4  y_{2} = y_{1} + (x_{1} + y_{1})Δx = 1.2 + (1.4)(.2) = 1.48
x_{3} = .6  y_{3} = y_{2} + (x_{2} + y_{2})Δx = 1.48 + (1.88)(.2) = 1.856
x_{4} = .8  y_{4} = y_{3} + (x_{3} + y_{3})Δx = 1.856 + (2.456)(.2) = 2.3472
x_{5} = 1.0  y_{5} = y_{4} + (x_{4} + y_{4})Δx = 2.3472 + (3.1472)(.2) = 2.97664

2. All parts of this question deal with the differential equation y′ = 3 − y.

(a) Sketch the direction field of this equation by hand, showing the slopes at all points with −1 ≤ x ≤ 5, 0 ≤ y ≤ 5 with integer coordinates. Solution: See plot with part (b) below. Note again that Maple is using a finer default grid to draw the direction field.

(b) On your direction field graph, sketch approximate graphs of the solutions of this differential equation satisfying the initial conditions y(0) = 0, y(0) = 3, and y(0) = 4. What is the long term behavior of each? Solution:

From the graph, we see that lim_{x→+∞} y(x) = 3 for all three solutions.

(c) Use Euler’s method with n = 5, Δx = .2 to approximate the solution of the equation with the initial condition y(0) = 1 for 0 ≤ x ≤ 1.

x_{0} = 0  y_{0} = 1 given from initial condition
x_{1} = .2  y_{1} = y_{0} + (3 − y_{0})Δx = 1.4
x_{2} = .4  y_{2} = y_{1} + (3 − y_{1})Δx = 1.72
x_{3} = .6  y_{3} = y_{2} + (3 − y_{2})Δx = 1.976
x_{4} = .8  y_{4} = y_{3} + (3 − y_{3})Δx = 2.1808
x_{5} = 1.0  y_{5} = y_{4} + (3 − y_{4})Δx = 2.34464

(d) This is a separable equation. Derive the general solution by separating variables and integrating. Solution: Separating variables and integrating yields:

\frac{dy}{3 − y} = dx
∫ \frac{dy}{3 − y} = ∫ dx
− ln |3 − y| = x + c
3 − y = be^{−x} (where b = ±e^{−c})
y = 3 − be^{−x}.

(e) What is the exact formula for the solution of the differential equation with the initial condition y(0) = 1? Solution: Using the answer from part (d), we solve for b from the initial condition:

1 = 3 − be^{0} ⇒ b = 2.

So the solution with this initial condition is y = 3 − 2e^{−x}.

3. Solve each of the following differential equations, finding an explicit formula for y as a function of x if possible (you can leave the solution in implicit form if not). In some cases you are given initial conditions; use them to determine the arbitrary constant in the general solution.

(a) y′ = \frac{tan(y)}{x^{2} + 4x + 3}, find the general solution.

Solution: Separate variables and integrate, using partial fractions on the right and tan(y) = \frac{sin(y)}{cos(y)} on the left:

\frac{1}{tan(y)}dy = \frac{1}{x^{2} + 4x + 3} dx
∫ \frac{cos(y)}{sin(y)} dy = ∫ \frac{1/2}{x + 1} + \frac{−1/2}{x + 3} dx
ln | sin(y)| = \frac{1}{2} ln |x + 1| − \frac{1}{2} ln |x + 3| + c
= ln \sqrt{|\frac{x + 1}{x + 3}|} + c (by properties of ln)

sin(y) = b · \sqrt{|\frac{x + 1}{x + 3}|} (where b = ±e^{c})

y = sin^{−1} (b · \sqrt{|\frac{x + 1}{x + 3}|}) .

(b) y′ = \frac{(x − 1)y^{5}}{x^{2}(2y^{3} − y)}, find the general solution.

Solution: Separate variables, simplify, and integrate by the power rule,

\frac{2y^{3} − y}{y^{5}} dy = \frac{x − 1}{x^{2}} dx
2y^{−2} − y^{−4} dy = x^{−1} − x^{−2} dx
∫ 2y^{−2} − y^{−4} dy = ∫ x^{−1} − x^{−2} dx
−2y^{−1} + \frac{1}{3}y^{−3} = ln |x| + x^{−1} + c.

(This is one where the solution is defined implicitly by the equation.)

(c) y′ = 2xy with y(0) = 4. Solution: Separate variables, and integrate,

\frac{1}{y} dy = 2x dx
ln |y| = x^{2} + c
y = be^{x^{2}} (where b = ±e^{c}).

From the initial condition 4 = be^{0}, so b = 4. The solution is y = 4e^{x^{2}}.

(d) y′ = x(y^{2} + 1) with y(0) = 1.

Solution: Separate variables, and integrate,

\frac{1}{y^{2} + 1} dy = x dx
tan^{−1}(y) = \frac{x^{2}}{2} + c
y = tan (\frac{x^{2}}{2} + c) .

From the initial condition 1 = tan(0 + c), so c = \frac{π}{4} gives one solution, and

y = tan (\frac{x^{2}}{2} + \frac{π}{4} ). (Note that in this case the periodicity of the tangent function

means that other values of c with tan(c) = 1 will define equivalent functions.)

(e) y′ = 2xy^{2} + 3x^{2}y^{2} with y(1) = −1. Solution: To see that this equation is separable, we factor the right hand side: y′ = (2x + 3x^{2})y^{2} which is of the form g(x)h(y) as required. Then we separate variables and integrate as before.

∫ y^{−2} dy = ∫ 2x + 3x^{2} dx
−\frac{1}{y} = x^{2} + x^{3} + c
y = \frac{−1}{x^{2} + x^{3} + c}.

From the initial condition −1 = \frac{−1}{2 + c} so c = −1. The solution is

y = \frac{−1}{x^{2} + x^{3} − 1}.

(f) y′ = 3y(1 − y/20) with y(0) = 5. (Note: this is a logistic equation.) Solution: By the general form of the solutions of logistic equations derived in class, we have M = 20 and k = 3, so

y = \frac{20}{1 + ae^{−3x}}

for some constant a. To determine a, we use the initial condition: 5 = \frac{20}{1 + ae^{0}} , so 1 + a = 4 and a = 3. The solution is

y = \frac{20}{1 + 3e^{−3x}} .

4. (Radiocarbon dating.) C^{14} is a radioactive isotope of carbon with half-life of approximately 5700 years. It occurs naturally and is incorporated in living tissues by normal metabolic processes. Assume that any C^{14} present when the tissue dies decays exponentially from that time on. Carbon extracted from an ancient skull contains only one-sixth as much C^{14} as carbon extracted from present-day bone. How old is the skull?

Solution: Since the C^{14} is decaying exponentially, writing Q for the amount, we have Q′ = kQ for some negative constant k. The solutions of this differential equation are

Q(t) = Q(0)e^{kt}.

Since the half-life is 5700 years, whatever Q(0) was,

Q(5700) = Q(0)e^{5700k} = \frac{1}{2}Q(0).

Therefore solving the equation e^{5700k} = \frac{1}{2} for k, we have k = \frac{ln(1/2)}{5700} ≐ −0.1216 × 10^{−3}. Now thinking of Q(0) as the amount of C^{14} that was in the skull at the time the animal it came from died, in the time elapsed since death to the present, the amount present

has decayed to \frac{1}{6}Q(0). Hence we want to solve the equation

\frac{1}{6}Q(0) = Q(0)e^{(−.1216×10^{−3})t}

for t to find the age:

t = \frac{ln(1/6)}{−.1216 × 10^{−3}} ≐ 14375 years.

5. (Newton’s Law of Heating/Cooling.) A pitcher of buttermilk initially at 25^{∘} C is cooled by setting it out on the front porch on a cold day when the exterior temperature is a constant 0^{∘} C. Suppose that the buttermilk has reached 15^{∘} C after 20 minutes.

(a) What will the temperature be after 25 minutes?

(b) When will the temperature reach 5^{∘} C? Solution: For both parts, we use the differential equation T′ = k(T − A) from Newton’s Law. The solutions are T = A + be^{kt}. Now in the situation of the problem, A = 0 (the ambient temperature on the porch). We have T (0) = 25, so 25 = 0 + be^{0} which shows b = 25. In addition from T (20) = 15, 15 = 25e^{20k} so k = \frac{ln(15/25)}{20} = ln(.6)/20 ≐ −.02554. So the temperature at all later times is

T (t) = 25e^{(−0.02554)t}.

(a) After 25 minutes T (25) = 25e^{(−0.02554)(25)} ≐ 13.2^{∘} C.

(b) The temperature reaches 5^{∘} C when 5 = 25e^{(−0.02554)t}, so t = \frac{ln(.2)}{−0.02554} ≐ 63 minutes.

6. The population of fish in a lake is attacked by a microscopic water-borne parasite at t = 0, and as a result the population declines at a rate proportional to the square root of the population from that time on.

(a) Express this statement about the rate of growth of the population P as a differential equation. (There should be a constant of proportionality, say −k, in your equation.) Solution: The differential equation is

P′ = \frac{dP}{dt} = −k\sqrt{P} .

(We write the constant of proportionality as −k here because we know that since P is decreasing over time, P′ < 0. The value of k will be positive with this choice.)

(b) Use separation of variables to find the general solution of your differential equation. Solution:

∫ \frac{dP}{\sqrt{P}} = ∫ −k dt
2P^{1/2} = −kt + c
P = (\frac{−k}{2} t + \frac{c}{2})^{2} .

(Note that \frac{c}{2} is just another arbitrary constant, though.)

(c) At t = 0 there were 900 fish in the lake; 441 were left after 6 weeks. When did the fish population disappear entirely? Solution: We take t as the time in weeks for simplicity. (The only differences if we used a different time unit would be in the values of k and c.) From the answer to part (b), 900 = (0 + \frac{c}{2})^{2} so \frac{c}{2} = 30. Thus P = (\frac{−k}{2}t + 30)^{2}. When t = 6, 441 = (−3k + 30)^{2}, so 21 = −3k + 30 and k = 3. The population as a function of time is

P = (\frac{−3}{2}t + 30)^{2} .

The population reaches P = 0 when \frac{−3}{2}t + 30 = 0, so t = 20 weeks.