1. Differential equations
Chapter 15
Differential equations
15.1 Basic concepts, separable and homogeneous equations
First-order linear equations
Exact equations
Strategy for solving first-order equations
12.1 Basic concepts,separable and homogeneous equations
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2. 15.5 Second-Order Linear Equations
A second-order linear differential equation has the form
(1)
where P, Q, R, and G are continuous functions.
If G(x) = 0 for all x, such equations are called second-order homogeneous linear equations. (This use of the word homogeneous has nothing to do with the meaning given in Section 15.1.)
(2)

3. If for some x, Equation 1 is nonhomogeneous.
Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions and of such an equation, then the linear combination is also a solution.
(3)Theorem If and are both solutions of the linear equation (2) and and are any constants, then the function is also a solution of Equation 2.

4. Proof Since and are solutions of Equation 2, we have
Therefore
Thus is a solution of Equation 2.

5. Let x and y are two variables, if neither x nor y is a constant multiple of the other, we say x and y are two linearly independent variables. For instance, the function
and are linearly dependent, but
and are linearly independent.

6. The second theorem says that the general solution of a homogeneous linear equation is a linear combination of two linearly independent solutions.
(4)Theorem If and are linearly independent solutions of Equation 2 , then the general solution is given by
where and are arbitrary constants.

7. In general, it is not easy to discover particular solutions to a second-order linear equation. But it is always possible to do so if the coefficient functions P, Q and R are constant functions, that is, if the differential equation has the form
(5)
It is not hard to think of some likely candidates for particular solutions of Equation 5. For example, the exponential function y because its derivatives are constants multiple of itself: Substitute these expression into Equation 5

8. Notice is never 0 so is a solution of Equation 5 if r is a root of the equation
(6)
which is called the auxiliary equation(or characteristic equation) of Equation 5. Using the quadratic formula, the root and of the auxiliary equation can be found:
(7)
We distinguish three cases according to the sign of the discriminant

9. Case 1
In this case the roots and of the auxiliary equation are real and distinct, so and are two linearly independent solutions of Equation 5.
(8) If the roots and of the auxiliary equation ` are real and unequal, then the general solution of is
Example 1 Solve the equation
Example 2 Solve the equation

10. Case 2
In this case ; that is, the root of the auxiliary equation are real and equal. Denote r as the common value of and , we have
(9)
We know that is one solution of Equation 5. We now verify that is also a solution:

11. Since and are linearly independent solutions, Theorem 4 provides us with the general solution:
(10) If the auxiliary equation has only one real root r, then the general solution of ` is
Example 3 Solve the equation

12. Case 3
In this case the roots and of the auxiliary equation are complex numbers, we can write
Using Euler’s equation
we write the solution of the differential equation as

14. We summarize the discussion as follows:
(11) If the roots of auxiliary equation are the complex numbers , then the general solution of is
Example 4 Solve the equation

15. Initial-value and boundary-value problems
An initial-value problem for the second-order Equation 1 or 2 consists of finding a solution y of the differential equation that also satisfies initial conditions of the form
where and are given constants. If P, Q, R, and G are continuous on an interval and there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem.

16. A boundary-value problem for Equation 1 consists of finding a solution of the differential equation that also satisfies boundary conditions of the form
In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution.

17. Example 5 Solve the initial-value problem
Example 7 Solve the boundary-value problem

18. 15.1 Basic concepts, separable and homogeneous equations
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19. Basic concepts In Sections 8， 10 Two kinds of equations Def:
We have known the concept of the differential equations.
10 Two kinds of equations
Def:
An ordinary differential equation:
involves an unknown function of a single variable and some of its derivatives.
Such as
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20. A partial differential equation:
involves an unknown function of two or more variables and some of its partial derivatives.
Such as

21. 20 The order of the equation
The order of a differential equation is the order of the highest derivative that appears in the equation.
Is an ordinary differential equation of order 1
Is a third-order differential equation
Is a second-order partial differential equation
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22. Separable equations
Note: We study only ordinary differential equations mainly
In general,a first-order differential equation has the form :
where F is some function of the two variables x and y
10 Define the separable equations
When F can be factored as a function of x times a function
then
Is called a separable equation
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23. We integrate both sides：
Form (1)
Also can be written as:
(2)
We integrate both sides：
This defines y implicitly as a function of x.
Sometimes ,we can solve for y in terms of x ，
thus , y=f(x) are called general solutions of the equation
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24. We say this is an initial problem。
20 An initial-value problem
The separable equation
Has a solution satisfies an initial condition of the form
(3)
We say this is an initial problem。
In general, there is a unique solution to the initial problem given by equations (2),(3)
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25. Solution (a) Example 1 Write it as
(a) Solve the differential equation
(b) Solve the initial-value problem
Solution (a)
Write it as
This is the general solution, involves an arbitrary constant C
Solution (b)
y(0) =1 tell us x=o,y=1 So
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26. Solve the differential equation
Example 2
Solve the differential equation
Solution:
Write it as
We integrate both sides So
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27. Solve the differential equation
Example 3
Solve the differential equation
Solution：
Write it as
We integrate both sides So
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28. Homogeneous equations A first-order differential equation
If F(x,y) can be written as
The form (4)
is called homogeneous equations
For instance:
Turn it as :
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29. This is a separable differential equation ,
We see
make the change of variable
then
thus or
This is a separable differential equation ,
it’s general solution
So, the original differential equation has the general solution
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30. Solve the differential equation
Example 5
Solve the differential equation
Solution:
write it as
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31. Solve the homogeneous differential equation
Example 6
Solve the homogeneous differential equation
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32. 15.1Homework
P 955
1.3.5.

33. 15.2 First-order linear equations
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34. The general solutions of the first-order linear equations
(1)
It’s solution is
(2)
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35. When Q(x)=0, form (1) become
(3)
This is a separate equations ,solve it
Thus
(4)
Is the solution of the equation form (3),
C is an arbitrary constant
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36. The solution is written as
Equation
(3)
The solution is written as
Let us replace the constant C in it by arbitrary
We look for the solution of the equation of the form
(5)
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37. Differentiating Equation (5),we get
Put it into (5),
thus: The solution of the form (1) is
(6)
Another method see page 997
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38. Solve the differential equation
Example 1
Solve the differential equation
Solution:
this is a linear equation
By the general solution form (6)
First step:
The second step:
Last step:
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39. Find the solution of the initial-value problem
Example 2
Find the solution of the initial-value problem
Solution :
We must first put the differential equation into standard form:
First:
Second:
Last:
Since y(1)=2,
We have
Therefore,
the solution to the initial-value problem is
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40. find the solution of the initial-value problem
Example 3
find the solution of the initial-value problem
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41. A Bernoulli differential equation
Observe that ,if n=0 or 1,it is linear.
the substitution
transforms the Bernoulli equation into the linear equation
Example1
Solve the differential equation
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42. 15.2 homework

43. 15.3 Exact equations

44. Definition of the exact
Suppose the equation
(1)
Then y=f(x) satisfies a first-order differential equation obtained by using the Chain Rule to differentiate both sides of equation with respect to x:
(2)
A differential equation of the form of Equation (2)is called exact。
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45. A first –order differential equation of the form
Definition：
A first –order differential equation of the form
(3)
is called exact if there is a function f(x,y) such that
If f(x,y) is known,thus the solution is given implicitly by
(4)
We may be able to solve Equation 4 for y as an explicit
function of x.
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46. Theorem
We have the following convenient method for the exactness of a differential equation.
Theorem:
Suppose
and
have continuous partial derivatives on a simply –connected domain.
The the differential equation
is exact if and only if
(5)
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47. Find solutions of exact equation
Example 1 Solve the differential equation
Solution
Here
and
have continuous partial derivatives on
Also
So the differential equation is exact by theorem .
Thus there exists a function f such that
(6)
(7)
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48. To determine f we first integrate 6 with respect to x :
(8)
Now we differentiate 8 with respect to y:
(9)
Comparing 7 and 9,we see that
and so
We do not need the arbitrary constant here
Thus So
is the solution.
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49. Solve the differential equation
Example 2
Solve the differential equation
Solution
Here
and
have continuous partial derivatives on
Also
So the differential equation is exact by theorem .
thus there exists a function f such that
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50. To determine f we first integrate
with respect to x :
Now we differentiate it with respect to y:
Comparing them,we see
that
then
We do not need the arbitrary constant here
Thus So
is the solution .
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51. Integrating factors The differential equation is not exact.
If the differential equation
is not exact,
we look for an integrating factor I(x,y)
Such that ,after multiplication by I(x,y),
the resulting equation
(10 )
is exact.
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52. Equation (10) is exact if That is or (11)
In general,it is harder to solve this partial differential equation than the original differential equation.
But it is sometimes possible to find I that is a function of x or y alone.
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53. Suppose I is a function of x alone.
Then
So equation 11 becomes
(12) if
is a function of x alone ,
Then Equation 12 is a first –order linear (and separable ) ordinary differential equation.
It can be solved for I(x)
Then Equation 10 is exact and can be solved In Example 1 or 2.
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54. Solve the differential equation
Example 3
Solve the differential equation
Solution :
Here
since
the given equation is not exact .
But
is a function of x alone.
So by Equation 12 there is an integrating factor I that satisfies
we get
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55. Multiplying the original equation by x ,we get
(13)
If we let
Then
So Equation 13 is now exact.thus there is a function f such that
Integrating the first of these equations,we get so
Comparison then gives
,so g is a constant
(which we can take to be 0)
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56. therefore
The solution is
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57. 15.3homework

58. 15.4 Strategy for solving first-order equations

59. In solving first-order differential equations we used the technique for separable equations in Sections 8.1 and the method for linear equations in Section 15.2.(on the text).
We also develops methods for solving homogeneous equations in Section 15.1 (on the text) and exact equations in Section 15.3 (on the text)。
In this section we present a miscellaneous collection of first-order differential equations and part of the problem is to recognize which technique should be used on each equation.
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60. As with the strategy of integration (Section7
As with the strategy of integration (Section7.6) and the strategy of testing series (Section 10.7),the main idea is to classify the equation according to its form.
Here , however , the important thing is not so much the form of the functions involved e as it is the form of the equation itself .
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61. Recall that a separable equation can be written in the form
(1)
that is ,the expression for dy/dx can be factored and a product of a function of x and a function of y.
A linear equation can be put into the form
(2)
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62. A homogeneous equation can be expressed in the form
(3)
An exact equation has the form
(4)
where
If an equation has none of these forms ,we can ,as a last resort , attempt to find and integrating factor and thus equation exact .
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63. In each of these cases ,some preliminary algebra may be required in order to put a given equation into one of the preceding forms .
(This step is analogous to Step l of the strategy for integration: algebraic implification. )
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64. It may happen that a given equation is of more than one type .
For instance, the equation
is separable because dy/dx can be written as
(compare with Equation 1)
It is also linear ,since we can write the equation as
(compare with Equation 2)
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65. Furthermore ,it is also homogeneous because we can write it as
(compare with Equation 2).
In such a case we could solve the equation using any one of the corresponding methods ,although one of the methods might be easier than the others .
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66. In the following examples we identify the type of each equation without working our the details of the solutions .
Example1
Initially ,this equation may not appear to be in any of the forms of Equations 1,2,3,or 4,but observe that we can factor the right side and therefore write the equation as
We now recognize the equation ad being separable and we can solve it using the methods of Section 15.1.
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67. But if we solve for y’, we get
Example 2
The equation is clearly mot separable , nor separable, nor is it linear .
Since
it is not exact .
But if we solve for y’, we get
Which shows that y’ is a function of y/x and the equation is homogeneous (see Equation 3).
(We could have anticipated this because the expressions x2,y2,and 2xy are all of degree 2.)
The change of variable v=y\x converts the equation into a separable equation .
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68. This equation is not separable ,linear ,or homogeneous .
Example3
This equation is not separable ,linear ,or homogeneous .
We suspect is might be exact, so we write it in the form If
then
Therefore ,the equation is indeed exact and can be dolled by the methods of Section 15.3.
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69. 15.4homework

70. If we put the equation in the form
Example4
If we put the equation in the form
We recognize it as having the form of Equation 2.
It is therefore linear and can be solved using the integrating factor

71. 15.6 Nonhomogeneous Linear Equations
In this section we deal with second-order nonhomogeneous linear differential equations with constant coefficient :
(1)
where a, b, c are constants and G is a continuous function.
The related homogeneous equation
(2)
is called the complementary equation and plays an important role in the solution of the original nonhomogeneous equation (1).

72. (3)Theorem The general solution of the nonhomogeneous differential equation (1) can be written as
where is a particular solution of Equation 1 and is the general solution of the complementary Equation 2.

73. Proof We want to verify that if y is any solution of Equation 1, then is a solution of the complementary Equation 2.
Indeed

74. We know from Section 15.5 that the solution of the complementary equation is , where and are linearly independent solutions of Equation 2.
Therefore, Theorem 3 says that we know the general solution of the nonhomogeneous equation as soon as we know a particular solution.
There are two methods for finding a particular solution. The method of undetermined coefficients is straightforward but works only for a restricted class of functions G. The method of variation of parameters works for every function G but is usually more difficult to apply in practice.

75. The method of undetermined coefficients
We first illustrate the method of undetermined coefficients for the equation
Case 1 G(x) is a polynomial
It is reasonable to guess that the particular solution is a polynomial of the same degree as G because if y is a polynomial, then is also a polynomial. We therefore substitute a polynomial (of the same degree as G ) into the differential equation and determine the coefficients.
Example 1 Solve the equation

76. Case 2 G(x) is of the form , where C and k are constants.
We take as a trial solution a function of the same form ` , because the derivatives of are constant multiples of
Example 2 Solve the equation
Case G(x) is either Ccoskx or Csinkx
Because of the rules for differentiating the sine and cosine functions, we take as a trial particular solution a function of the form
Example 3 Solve the equation

77. Case 4 G(x) is a product of functions of the preceding types
We take the trial solution to be a product of functions of the same type. For instance, in solving the differential equation
we would try
Case 5 G(x) is a sum of functions of these types
We use the easily verified principle of superposition, which says if and are solutions of
respectively, then is a solution of

78. Example 4 Solve
Finally we note that the recommended trial solution sometimes turns out to be a solution of the complementary equation and therefore cannot be a solution of the nonhomogeneous equation. In such cases we multiply the recommended trial solution by x (or if necessary) so that no term in is a solution of the complementary equation.
Example 5 Solve

79. Summary of undetermined coefficient
G(x) = First try
Modification: If any term of is a solution of the complementary equation, multiply by x (or if necessary) .

80. The method of variation of parameters
Suppose we have already solved the homogeneous equation and written the solution as
(4)
where and are linearly independent solutions.
Let us replace the constants and in Equation 4 by arbitrary functions and
We look for a particular solution of the nonhomogeneous equation of the form
(5)

81. Differentiating Equation 5, we get
(6)
Since and are arbitrary functions, we can impose two conditions on them. One condition is that is a solution of the differential equation; we can choose the other condition so as to simplify our calculations. In view of the expression in Equation 6, let us impose the condition that
(7)
Then

82. Substituting in the differential equation, we getor
(8)
But and are solutions of the complementary equation, so
and Equation 8 simplifies to
(9)

83. Equation 7 and 9 form a system of two equations in the unknown functions and After solving this system we may be able to integrate to find and then the particular solution is given by Equation 5.
Example 6 Solve the equation

84. 15.8 Series Solutions
So far, the only second-order differential equations that we have been able to solve are linear equations with constant coefficients.
Even a simple-looking equation like
(1)
is not easy to solve. But it is important to be able to solve equations such as Equation 1 since they arise from physical problems and, in particular, in connection with the Schrödinger equation(薛定谔方程) in quantum mechanics.

85. In such a case, we use the method of power series; that is, we look for a solution of the form
The method is to substitute this expression into the differential equation and determine the values of the coefficients
Example 1 Use power series to solve the equation
Solution We assume there is a solution of the form
(2)

86. By Theorem 10.9.2 we can differentiate power series term by term. So
(3)
Rewrite as
In order to compare the expression for y and more easily, we substituting these expressions into the differential equation, we obtain

87. or
(5)
If two power series are equal, then the corresponding coefficients must be equal (from Section 10.10). Therefore, the coefficients of in Equation 5 must be 0:
(6)
Equation 6 is called a recursion relation. If and are known, this equation allows us to determine the remaining coefficients recursively by putting n = 0, 1, 2, 3, …in succession.

89. By now we see the pattern:
For the even coefficients,
For the odd coefficients,
Putting these values back into Equation 2, we write the solution as

90. Notice that there are two arbitrary constants and , which is to be expected.
Note 1: We recognize the series obtained in Example 1 as being the Maclaurin series for cosx and sinx. (See Equation and ) Therefore, we could write the solution as
But we are not usually able to express power series solutions of differential equations in terms of known functions.

91. Example 2 Solve
Solution We assume there is a solution of the form
Then
and
Substituting in the differential equation, we get

92. This is true if the coefficient of is 0:
(7)
We solve this recursion relation by putting n = 0, 1, 2, 3, … successively in Equation 7:

93. In general, the even coefficients are given by
and the odd coefficients are given by

94. The solution is Or
(8)

95. Note 2: In Example 2 we had to assume that the differential equation had a series solution. But now we could verify directly that the function given by Equation 8 is indeed a solution.

96. Note 3: Unlike the situation of Example 1, the power series that arise in the solution of Example 2 do not define elementary functions. The functions
and
are perfectly good functions but they cannot be expressed in terms of familiar functions. We can use these power series expressions for and to compute approximate values of the functions and even to graph them.

97. Note 4: If we are asked to solve the initial-value problem
we would observe from Theorem that
This would simplify the calculations in Example 2, since all of the even coefficients would be 0. The solution to the initial-value problem is