KATHERINE HUANG-OUYANG Assignment 9 due 11/06/2013 at 06:00am PST
Math-1B-Lect-2-Fa13
4. (2 pts) Find a positive value of k for which y = cos(kt) satisfies d2y + y = 0. dt 2 k= SOLUTION:
1. (2 pts) Let A and k be positive constants. Which of the given functions is a solution to dy dt = k(y + A)? • A. y = A−1 +CeAkt • B. y = −A +Cekt • C. y = A +Ce−kt • D. y = A−1 +Ce−Akt • E. y = −A +Ce−kt • F. y = A +Cekt
SOLUTION 2 If y = cos(kt), then ddt 2y = −k2 cos(kt), and so we need k such that d2y + y = −k2 cos(2t) + 1 cos(2t) = 0. dt 2 Thus k = 1.
SOLUTION:
is ar stu ed d vi y re aC s o ou urc rs e eH w er as o. co m
SOLUTION We can find the correct answer by plugging the answers into the differential equation to see which satisfies the equation. Here, dtd (−A +Cekt ) = kCekt , so that, plugging in, we have
Answer(s) submitted: • 1
(correct)
kCekt = k(−A +Cekt ) − A,
Correct Answers: • 1
a true statement, so that the correct answer is y = −A +Cekt . Answer(s) submitted: • B
(correct) Correct Answers:
5. (2 pts) Find all values of k for which the function y = sin(kt) satisfies the differential equation y00 + 13y = 0. Separate your answers by commas. Answer(s) submitted:
• sqrt(13), -sqrt(13), 0
• B
(correct)
Correct Answers:
2. (2 pts) Find the value of k for which the constant function dx x(t) = k is a solution of the differential equation 2t 5 − 6x + dt 9 = 0. Answer(s) submitted: • 9/6
(correct) Correct Answers:
6. (2 pts) Verify that every member of the family of funcln x +C tions y = is a solution of the differential equation x 2 0 x y + xy = 1. Answer the following questions.
1. Find a solution of the differential equation that satisfies the initial condition y(1) = 10. Answer: y = 2. Find a solution of the differential equation that satisfies the initial condition y(10) = 1. Answer: y =
Th
• 1.5
• 3.60555127546399, -3.60555127546399, 0
3. (2 pts) For what values of r does the function y = erx satisfy the differential equation y00 − 7y0 + 12y = 0? Answer (separate by commas): r = • 3, 4
Answer(s) submitted:
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Answer(s) submitted:
• (lnx + 10)/x • (lnx + (10-ln10))/x
(correct)
(correct)
Correct Answers:
Correct Answers: • [ln(x)+10]/x • [ln(x)+7.69741]/x
• 3, 4 1
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or 7. (2 pts) It is easy to check that for any value of c, the function c y = x2 + 2 x is solution of equation
Cnxn − 2Cxn = C(n − 2)xn = 0. Thus, if C = 0 we have a solution, y = 0; and if C 6= 0 then we must have n = 2, C may have any value (other than zero), and the solution is y = Cx2 . (b) If y = 30 for x = 5, we cannot have C = 0. Thus, by part (a), we get n = 2. The solution to the differential equation is y = Cx2 . Thus, if y = 30 when x = 5,
xy0 + 2y = 4x2 , (x > 0). Find the value of c for which the solution satisfies the initial condition y(3) = 10. c= Answer(s) submitted: • 9
30 = C(5)2 ,
(correct) Correct Answers: • 9
30 30 2 so that C = (5) 2 , and y = (5)2 · x .
Answer(s) submitted:
8. (3 pts) The solution of a certain differential equation is of the form y(t) = a exp(1t) + b exp(3t), where a and b are constants. The solution has initial conditions y(0) = 3 and y0 (0) = 5. Find the solution by using the initial conditions to get linear equations for a and b.
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y(t) =
• Cxˆ2 • 0 • (6/5)xˆ2
(correct)
Correct Answers:
• C*xˆ2 • 0*xˆn • 30*(x/5)ˆ(2)
Answer(s) submitted: • 2exp(1 t) + 1exp(3t)
(correct)
Correct Answers: • 2 *(eˆ(1 *t)) + 1 *(eˆ(3 *t))
10. (2 pts) A function y(t) satisfies the differential equation dy = y4 + 7y3 − 8y2 . dt
9. (2 pts) (a) Consider the differential equation dy x − 2y = 0. dx
1. For what values of y is y increasing? Answer (in interval notation): 2. For what values of y is y decreasing? Answer (in interval notation): 3. What are the constant solutions of the equation? Answer (separate by commas): y =
Find a general solution to this differential equation that has the form y = Cxn . y= . Find a second solution y = Cxn that might not be a general solution and which may have a different value of n than your first solution. y= . (b) If the solution additionally satisfies y = 30 when x = 5, what is the solution? y= SOLUTION:
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Answer(s) submitted:
• (-inf, -8) U (1, inf) • (-8, 0) U (0, 1) • 0, -8, 1
sh
(correct)
SOLUTION (a) If y = Cxn is a solution to the given differential equation, then we must have d (Cxn ) x − 2(Cxn ) = 0, dx so x(Cnxn−1 ) − 2(Cxn ) = 0,
Correct Answers: • (-infinity,-8) U (1,infinity) • (-8,0) U (0,1) • 0, 1, -8
2
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A. Set y equal to zero and look at how the derivative behaves along the x-axis. B. Do the same for the y-axis by setting x equal to 0 C. Consider the curve in the plane defined by setting y0 = 0 – this should correspond to the points in the picture where the slope is zero. D. Setting y0 equal to a constant other than zero gives the curve of points where the slope is that constant. These are called isoclines, and can be used to construct the direction field picture by hand.
11. (2 pts) Assuming that P ≥ 0, a population is modeled by the differential equation dP P = 1.8P 1 − . dt 4900 1. For what values of P is the population increasing? Answer (in interval notation): 2. For what values of P is the population decreasing? Answer (in interval notation): 3. What are the equilibrium solutions? Answer (separate by commas): P =
1. y0 = 2 sin(3x) + 1 + y (2x + y) 2. y0 = − (2y) 3. y0 = 2y − 2 y 4. y0 = + 3 cos(2x) x
Answer(s) submitted: • (00, 4900) • (4900, inf) • 0, 4900 Correct Answers:
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(correct) • (0,4900) • (4900,infinity) • 0, 4900
12. (2 pts)
Newton’s law of cooling says that the rate of cooling of an object is proportional to the difference between the temperature of the object and that of its surroundings (provided the difference is not too large). If T = T (t) represents the temperature of a (warm) object at time t, A represents the ambient (cool) temperature, and k is a negative constant of proportionality, which equation(s) accurately characterize Newton’s law?
A
B
C
D
• A. dT dt = k(T − A) • B. dT dt = kT (T − A) dT • C. dt = kT (1 − T /A) • D. dT dt = k(A − T ) • E. All of the above • F. None of the above Answer(s) submitted:
(correct)
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• A Correct Answers: • A
Answer(s) submitted: • A • D • C • B
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13. (2 pts) Match the following equations with their direction field. Clicking on each picture will give you an enlarged view. While you can probably solve this problem by guessing, it is useful to try to predict characteristics of the direction field and then match them to the picture. Here are some handy characteristics to start with – you will develop more as you practice.
(correct) Correct Answers: • A • D • C • B 3
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(correct) 14. (2 pts) Match the following equations with their direction field. Clicking on each picture will give you an enlarged view. While you can probably solve this problem by guessing, it is useful to try to predict characteristics of the direction field and then match them to the picture. Here are some handy characteristics to start with – you will develop more as you practice.
Correct Answers: • B • C • A • D
15. (2 pts) Match the direction fields labeled A through D with the differential equation below.
1. y0 = e−x + 2y 2. y0 = 2y + x2 e2x 3. y0 = −2 + x − y (2x + y) 4. y0 = − (2y)
A
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A. Set y equal to zero and look at how the derivative behaves along the x-axis. B. Do the same for the y-axis by setting x equal to 0 C. Consider the curve in the plane defined by setting y0 = 0 – this should correspond to the points in the picture where the slope is zero. D. Setting y0 equal to a constant other than zero gives the curve of points where the slope is that constant. These are called isoclines, and can be used to construct the direction field picture by hand.
A
B
C
D
1. y0 = 2 cos x + sin y 2. y0 = cos x sin y 3. y0 = 3 cos x cos y 4. y0 = 1 − cos x cos y
B
Note: You can click on the graphs to enlarge the images.
C Answer(s) submitted:
sh
Th
Answer(s) submitted: • A • B • D • C
(correct)
D
Correct Answers: • A • B • D • C
• B • C • A • D 4
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16. (2 pts) Consider the two slope fields shown, in figures 1 and 2 below.
figure 1
figure 2 1. Find the equilibrium solutions for the differential equation. Answer (separate by commas): y = 2. If the initial condition is y(0) = c, for what values of c is lim y(t) finite? t→∞ Answer (as an interval):
On a print-out of these slope fields, sketch for each three solution curves to the differential equations that generated them. Then complete the following statements: For the slope field in figure 1, a solution passing through the point (3,-2) has a ? slope. For the slope field in figure 1, a solution passing through the point (-3,-1) has a ? slope. For the slope field in figure 2, a solution passing through the point (2,1) has a ? slope. For the slope field in figure 2, a solution passing through the point (0,3) has a ? slope. SOLUTION:
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Note: You can click on the graph to enlarge the image. Answer(s) submitted: • 0, 3, 6 • [0, 6]
(correct)
Correct Answers: • 0, 3, 6 • [0,6]
SOLUTION There are many possible solution curves that could be drawn. For the slope field in figure 1, a solution passing through the point (3,-2) has a positive slope. For the slope field in figure 1, a solution passing through the point (-3,-1) has a negative slope. For the slope field in figure 2, a solution passing through the point (2,1) has a positive slope. For the slope field in figure 2, a solution passing through the point (0,3) has a negative slope. Answer(s) submitted:
On a print out of this slope field, sketch the solutions that pass through the points (i) (0,0); (ii) (-3,1); and (iii) (-1,0). From your sketch, what is the equation of the solution to the differential equation that passes through (-1,0)? (Verify that your solution is correct by substituting it into the differential equation.) y= SOLUTION:
Correct Answers: • positive • negative • positive • negative
sh
(correct)
Th
• positive • negative • positive • negative
18. (2 pts) The slope field for the equation y0 = x +y is shown below
17. (2 pts) Consider the direction field below for a differential equation y0 = y(y − 3)(y − 6). Use the graph and equation to answer the following questions.
SOLUTION 5
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