Differential Equations
Differential Equations
1/14/14 9:10 PM
11.4 SEPARATION OF VARIABLES We have seen how to sketch solution curves of a differential equation using a slope field and how to calculate approximate numerical solutions. Now we see how to solve certain differential equations analytically, finding an equation for the solution curve. First, we look at a familiar example, the differential equation
whose solution curves are the circles x 2 +y 2 = C .
We can check that these circles are solutions by differentiation; the question now is how they were obtained. The method of separation of variables works by putting all the x -values on one side of the equation and all the y -values on the other, giving y dy = −x dx .
We then integrate each side separately:
This gives the circles we were expecting:
You might worry about whether it is legitimate to separate the dx and the dy . The reason it can be done is explained at the end of this section. Example 1
Using separation of variables, solve the differential equation:
Solution
Separating variables,
and integrating,
gives ln |y | = k x +C for some constant C .
Solving for |y | leads to |y | = e k x +C = e k x e C = A e k x http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-hallet…A0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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Differential Equations
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where A = e C , so A is positive. Thus y = ( ±A )e k x = B e k x
where B = ±A , so B is any nonzero constant. Even though there's no C leading to B = 0, we can have B = 0 because y = 0 is a solution to the differential equation. We lost this solution when we divided through by y at the first step. Thus, the general solution is y = B e k x for any B . The differential equation dy /dx = k y always leads to exponential growth (if k >0) or exponential decay (if k 0 are in Figure 11.39. For k 0
Example 2
For k >0, find and graph solutions of
Solution
The slope field in Figure 11.40 shows the qualitative behavior of the solutions. To find the equation of the solution curves, we separate variables and integrate:
This gives ln |H −20| = −k t +C .
Solving for H leads to: |H −20| = e −k t +C = e −k t e C = A e −k t
or
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Differential Equations
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Again, B = 0 also gives a solution. Graphs for k = 1 and B = −10, 0, 10, with t ≥0, are in Figure 11.40.
Figure 11.40 Slope field and some solution curves for dH /dt = −k (H −20) , with k = 1
This differential equation can be used to represent the temperature, H (t ) , in ∘ C at time t of a cup of water standing in a room at 20 ∘ C . As Figure 11.40 shows, if the initial temperature is 10 ∘ C , the water warms up; if the initial temperature is 30 ∘ C , the water cools down. If the initial temperature is 20 ∘ C , the water remains 20 ∘ C . Example 3
Find and sketch the solution to
Solution
Factoring the right-hand side gives
Separating variables, we get
so ln |P | = 2t −t 2 +C .
Solving for P leads to 2 2 2 |P | = e 2t −t +C = e C e 2t −t = A e 2t −t
with A = e C , so A >0. In addition, A = 0 gives a solution. Thus the general solution to the differential equation is P = B e 2t −t
2
for any B.
To find the value of B , substitute P = 5 and t = 0 into the general solution, giving 2
5 = B e 2·0−0 = B ,
so http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-hallet…A0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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Differential Equations
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2
P = 5e 2t −t .
The graph of this function is in Figure 11.41. Since the solution can be rewritten as P = 5e 1−1+2t −t
2
2 2 = 5e 1e −1+2t −t = (5e )e − (t −1) , 2
the graph has the same shape as the graph of y = e −t , the bell-shaped curve of statistics. Here the maximum, normally at t = 0, is shifted one unit to the right to t = 1.
Figure 11.41 Bell-shaped solution curve
Justification for Separation of Variables Suppose a differential equation can be written in the form
Provided f (y ) ≠0, we write f (y ) = 1/h (y ) , so the right-hand side can be thought of as a fraction,
If we multiply through by h (y ) , we get
Thinking of y as a function of x , so y = y (x ) , and dy /dx = y ′ (x ) , we can rewrite the equation as h (y (x ) ) ·y ′ (x ) = g (x ).
Now integrate both sides with respect to x :
∫ h (y (x ) ) ·y ′ (x ) dx = ∫ g (x ) dx . The form of the integral on the left suggests that we use the substitution y = y (x ) . Since dy = y ′ (x ) dx , we get
∫ h (y ) dy = ∫ g (x ) dx . If we can find antiderivatives of h and g , then this gives the equation of the solution curve. Note that transforming the original differential equation,
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Differential Equations
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into
∫ h (y ) dy = ∫ g (x ) dx looks as though we have treated dy /dx as a fraction, cross-multiplied, and then integrated. Although that's not exactly what we have done, you may find this a helpful way of remembering the method. In fact, the dy /dx notation was introduced by Leibniz to allow shortcuts like this (more specifically, to make the chain rule look like cancellation).
Exercises and Problems for Section 11.4 Exercises 1. Determine which of the following differential equations are separable. Do not solve the equations. a. b. c. d. e. f. g. h. i. j. k. l.
y′ = y y ′ = x +y y ′ = xy y ′ = sin (x +y ) y ′ −x y = 0 y ′ = y /x y ′ = ln (x y ) y ′ = ( sinx ) ( cosy ) y ′ = ( sinx ) ( cosx y ) y ′ = x /y y ′ = 2x y ′ = (x +y ) / (x +2y )
In Exercises 2-28, use separation of variables to find the solutions to the differential equations subject to the given initial conditions. 2. 3.
4. 5.
when t = 0
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Differential Equations
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6. 7.
8. 9.
10. 11.
12. 13.
14.
15.
16.
17.
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18. 19.
20. 21.
22. 23.
24.
25.
26. 27.
28.
Problems 29. (a) Solve the differential equation http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-hallet…A0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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Differential Equations
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Write the solution y as an explicit function of x .
(b) Find the particular solution for each initial condition below and graph the three solutions on the same coordinate plane.
30. (a) Solve the differential equation
Write the solution P as an explicit function of t . (b) Find the particular solution for each initial condition below and graph the three solutions on the same coordinate plane.
31. (a) Find the general solution to the differential equation modeling how a person learns:
(b) Plot the slope field of this differential equation and sketch solutions with y (0) = 25 and y (0) = 110. (c) For each of the initial conditions in part b, find the particular solution and add to your sketch. (d) Which of these two particular solutions could represent how a person learns? 32. A circular oil spill grows at a rate given by the differential equation dr /dt = k /r , where r represents the radius of the spill in feet, and time is measured in hours. If the radius of the spill is 400 feet 16 hours after the spill begins, what is the value of k ? Include units in your answer. 33. Figure 11.42 shows the slope field for dy /dx = y 2.
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Differential Equations
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Figure 11.42
(a) Sketch the solutions that pass through the points (i) (0, 1) (ii) (0, −1) (iii) (0, 0)
(b) In words, describe the end behavior of the solution curves in part a.
(c) Find a formula for the general solution.
(d) Show that all solution curves except for y = 0 have both a horizontal and a vertical asymptote. Solve the differential equations in Problems 34-43. Assume a , b , and k are nonzero constants. 34. 35.
36. 37.
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Differential Equations
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38. 39.
40. 41.
42. 43.
Solve the differential equations in Problems 44-47. Assume x , y , t >0. 44.
45.
46. 47.
48. Figure 11.43 shows the slope field for the equation
(a) Sketch the solutions that pass through (0, 0) . (b) What can you say about the end behavior of the solution curve in part a? http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle…0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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Differential Equations
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(c) For each of the following regions, find a formula for the general solution (i) −1 ≤ y ≤ 1 (ii) y ≤ −1 (iii) y ≥1. (d) Show that each solution curve has two vertical asymptotes. (e) How far apart are the two asymptotes of a solution curve?
Figure 11.43
49. (a) Sketch the slope field for y ′ = x /y . (b) Sketch several solution curves. (c) Solve the differential equation analytically.
50. (a) Sketch the slope field for y ′ = −y /x . (b) Sketch several solution curves. (c) Solve the differential equation analytically. 51. Compare the slope field for y ′ = x /y , Problem 49, with that for y ′ = −y /x , Problem 50. Show that the solution curves of Problem 49 intersect the solution curves of Problem 50 at right angles.
Strengthen Your Understanding In Problems 52-54, explain what is wrong with the statement. 52. Separating variables in dy /dx = x +y gives −y dy = x dx . 53. The solution to dP /dt = 0.2t is P = B e 0.2t .
54. Separating variables in dy /dx = e x +y gives −e y dy = e x dx . In Problems 55-58, give an example of: http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle…0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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Differential Equations
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55. A differential equation that is not separable. 56. An expression for f (x ) such that the differential equation dy /dx = f (x ) +x y − cosx is separable. 57. A differential equation all of whose solutions form the family of functions f (x ) = x 2 +C .
58. A differential equation all of whose solutions form the family of hyperbolas x 2 −y 2 = C . Are the statements in Problems 59-62 true or false? Give an explanation for your answer. 59. For all constants k , the equation y ′ +k y = 0 has exponential functions as solutions. 60. The differential equation dy /dx = x +y can be solved by separation of variables. 61. The differential equation dy /dx −x y = x can be solved by separation of variables.
62. The only solution to the differential equation dy /dx = 3y 2/3 passing through the point (0, 0) is y = x 3.
Copyright © 2013 John Wiley & Sons, Inc. All rights reserved.
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11.4 SEPARATION OF VARIABLES We have seen how to sketch solution curves of a differential equation using a slope field and how to calculate approximate numerical solutions. Now we see how to solve certain differential equations analytically, finding an equation for the solution curve. First, we look at a familiar example, the differential equation
whose solution curves are the circles x 2 +y 2 = C .
We can check that these circles are solutions by differentiation; the question now is how they were obtained. The method of separation of variables works by putting all the x -values on one side of the equation and all the y -values on the other, giving y dy = −x dx .
We then integrate each side separately:
This gives the circles we were expecting:
You might worry about whether it is legitimate to separate the dx and the dy . The reason it can be done is explained at the end of this section. Example 1
Using separation of variables, solve the differential equation:
Solution
Separating variables,
and integrating,
gives ln |y | = k x +C for some constant C .
Solving for |y | leads to |y | = e k x +C = e k x e C = A e k x http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-hallet…A0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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Differential Equations
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where A = e C , so A is positive. Thus y = ( ±A )e k x = B e k x
where B = ±A , so B is any nonzero constant. Even though there's no C leading to B = 0, we can have B = 0 because y = 0 is a solution to the differential equation. We lost this solution when we divided through by y at the first step. Thus, the general solution is y = B e k x for any B . The differential equation dy /dx = k y always leads to exponential growth (if k >0) or exponential decay (if k 0 are in Figure 11.39. For k 0
Example 2
For k >0, find and graph solutions of
Solution
The slope field in Figure 11.40 shows the qualitative behavior of the solutions. To find the equation of the solution curves, we separate variables and integrate:
This gives ln |H −20| = −k t +C .
Solving for H leads to: |H −20| = e −k t +C = e −k t e C = A e −k t
or
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Differential Equations
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Again, B = 0 also gives a solution. Graphs for k = 1 and B = −10, 0, 10, with t ≥0, are in Figure 11.40.
Figure 11.40 Slope field and some solution curves for dH /dt = −k (H −20) , with k = 1
This differential equation can be used to represent the temperature, H (t ) , in ∘ C at time t of a cup of water standing in a room at 20 ∘ C . As Figure 11.40 shows, if the initial temperature is 10 ∘ C , the water warms up; if the initial temperature is 30 ∘ C , the water cools down. If the initial temperature is 20 ∘ C , the water remains 20 ∘ C . Example 3
Find and sketch the solution to
Solution
Factoring the right-hand side gives
Separating variables, we get
so ln |P | = 2t −t 2 +C .
Solving for P leads to 2 2 2 |P | = e 2t −t +C = e C e 2t −t = A e 2t −t
with A = e C , so A >0. In addition, A = 0 gives a solution. Thus the general solution to the differential equation is P = B e 2t −t
2
for any B.
To find the value of B , substitute P = 5 and t = 0 into the general solution, giving 2
5 = B e 2·0−0 = B ,
so http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-hallet…A0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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Differential Equations
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2
P = 5e 2t −t .
The graph of this function is in Figure 11.41. Since the solution can be rewritten as P = 5e 1−1+2t −t
2
2 2 = 5e 1e −1+2t −t = (5e )e − (t −1) , 2
the graph has the same shape as the graph of y = e −t , the bell-shaped curve of statistics. Here the maximum, normally at t = 0, is shifted one unit to the right to t = 1.
Figure 11.41 Bell-shaped solution curve
Justification for Separation of Variables Suppose a differential equation can be written in the form
Provided f (y ) ≠0, we write f (y ) = 1/h (y ) , so the right-hand side can be thought of as a fraction,
If we multiply through by h (y ) , we get
Thinking of y as a function of x , so y = y (x ) , and dy /dx = y ′ (x ) , we can rewrite the equation as h (y (x ) ) ·y ′ (x ) = g (x ).
Now integrate both sides with respect to x :
∫ h (y (x ) ) ·y ′ (x ) dx = ∫ g (x ) dx . The form of the integral on the left suggests that we use the substitution y = y (x ) . Since dy = y ′ (x ) dx , we get
∫ h (y ) dy = ∫ g (x ) dx . If we can find antiderivatives of h and g , then this gives the equation of the solution curve. Note that transforming the original differential equation,
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Differential Equations
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into
∫ h (y ) dy = ∫ g (x ) dx looks as though we have treated dy /dx as a fraction, cross-multiplied, and then integrated. Although that's not exactly what we have done, you may find this a helpful way of remembering the method. In fact, the dy /dx notation was introduced by Leibniz to allow shortcuts like this (more specifically, to make the chain rule look like cancellation).
Exercises and Problems for Section 11.4 Exercises 1. Determine which of the following differential equations are separable. Do not solve the equations. a. b. c. d. e. f. g. h. i. j. k. l.
y′ = y y ′ = x +y y ′ = xy y ′ = sin (x +y ) y ′ −x y = 0 y ′ = y /x y ′ = ln (x y ) y ′ = ( sinx ) ( cosy ) y ′ = ( sinx ) ( cosx y ) y ′ = x /y y ′ = 2x y ′ = (x +y ) / (x +2y )
In Exercises 2-28, use separation of variables to find the solutions to the differential equations subject to the given initial conditions. 2. 3.
4. 5.
when t = 0
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6. 7.
8. 9.
10. 11.
12. 13.
14.
15.
16.
17.
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18. 19.
20. 21.
22. 23.
24.
25.
26. 27.
28.
Problems 29. (a) Solve the differential equation http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-hallet…A0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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Write the solution y as an explicit function of x .
(b) Find the particular solution for each initial condition below and graph the three solutions on the same coordinate plane.
30. (a) Solve the differential equation
Write the solution P as an explicit function of t . (b) Find the particular solution for each initial condition below and graph the three solutions on the same coordinate plane.
31. (a) Find the general solution to the differential equation modeling how a person learns:
(b) Plot the slope field of this differential equation and sketch solutions with y (0) = 25 and y (0) = 110. (c) For each of the initial conditions in part b, find the particular solution and add to your sketch. (d) Which of these two particular solutions could represent how a person learns? 32. A circular oil spill grows at a rate given by the differential equation dr /dt = k /r , where r represents the radius of the spill in feet, and time is measured in hours. If the radius of the spill is 400 feet 16 hours after the spill begins, what is the value of k ? Include units in your answer. 33. Figure 11.42 shows the slope field for dy /dx = y 2.
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Figure 11.42
(a) Sketch the solutions that pass through the points (i) (0, 1) (ii) (0, −1) (iii) (0, 0)
(b) In words, describe the end behavior of the solution curves in part a.
(c) Find a formula for the general solution.
(d) Show that all solution curves except for y = 0 have both a horizontal and a vertical asymptote. Solve the differential equations in Problems 34-43. Assume a , b , and k are nonzero constants. 34. 35.
36. 37.
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38. 39.
40. 41.
42. 43.
Solve the differential equations in Problems 44-47. Assume x , y , t >0. 44.
45.
46. 47.
48. Figure 11.43 shows the slope field for the equation
(a) Sketch the solutions that pass through (0, 0) . (b) What can you say about the end behavior of the solution curve in part a? http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle…0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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(c) For each of the following regions, find a formula for the general solution (i) −1 ≤ y ≤ 1 (ii) y ≤ −1 (iii) y ≥1. (d) Show that each solution curve has two vertical asymptotes. (e) How far apart are the two asymptotes of a solution curve?
Figure 11.43
49. (a) Sketch the slope field for y ′ = x /y . (b) Sketch several solution curves. (c) Solve the differential equation analytically.
50. (a) Sketch the slope field for y ′ = −y /x . (b) Sketch several solution curves. (c) Solve the differential equation analytically. 51. Compare the slope field for y ′ = x /y , Problem 49, with that for y ′ = −y /x , Problem 50. Show that the solution curves of Problem 49 intersect the solution curves of Problem 50 at right angles.
Strengthen Your Understanding In Problems 52-54, explain what is wrong with the statement. 52. Separating variables in dy /dx = x +y gives −y dy = x dx . 53. The solution to dP /dt = 0.2t is P = B e 0.2t .
54. Separating variables in dy /dx = e x +y gives −e y dy = e x dx . In Problems 55-58, give an example of: http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle…0NzA4ODg2NDNjMTEtc2VjLTAwMTQueGZvcm0.enc?course=crs7160&id=ref
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55. A differential equation that is not separable. 56. An expression for f (x ) such that the differential equation dy /dx = f (x ) +x y − cosx is separable. 57. A differential equation all of whose solutions form the family of functions f (x ) = x 2 +C .
58. A differential equation all of whose solutions form the family of hyperbolas x 2 −y 2 = C . Are the statements in Problems 59-62 true or false? Give an explanation for your answer. 59. For all constants k , the equation y ′ +k y = 0 has exponential functions as solutions. 60. The differential equation dy /dx = x +y can be solved by separation of variables. 61. The differential equation dy /dx −x y = x can be solved by separation of variables.
62. The only solution to the differential equation dy /dx = 3y 2/3 passing through the point (0, 0) is y = x 3.
Copyright © 2013 John Wiley & Sons, Inc. All rights reserved.
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