Chapter 2 Second order differential equations
1 Basic concepts
1.1 Linear ODEs
A secondorder linear ODE is of the form
(1)
Note: If the equation has, say as the first term, then divide by to get to the standard linear form (1) with as the first term.
For example, has a standard linear form .
Question: What would the standard linear form for a third order differential equation be? How about a higher order differential equation?
1.2 Linear homogeneous ODEs
If ,
(2)
Example 1.2.
(a) is a nonhomgeneous linear ODE.
(b) is a homogeneous linear ODE. The standard form is .
(c) is a nonlinear ODE.
2 Homogeneous Linear ODEs
2.1 Superposition Principle
If , are two solutions of a linear homogeneous ODE, then any linear combination of them is too.
That is,
is also a solution of that ODE.
Example 2.1.
The functions and are solutions of the homogeneous linear ODE for all . We can verify this by differentiation and substitution. We multiply by any constant, for instance, , and by, say, , and take the sum of the results, claiming that it is a solution. Indeed, differentiation and substitution gives
Note: By careful that the superposition theorem holds for homogeneous linear ODEs only but does not hold for nonhomogeneous linear or nonlinear ODEs.
Exercise: PROVE that any linear combination of solutions of a linear homogenous ODE is also a solution. Does your proof work for a nonhomogenous ODE? Or for a nonlinear ODE?
2.2 Linear independence
Two functions and are said to be linearly independent on an interval where they are defined if
Then if or , we can divide and see that and are proportional,
In contrast, and are said to be linearly dependent on if the above equation also holds for some constants and not both zero.
Thus, in the case of linear independence the two functions are not proportional.
Example 2.2.
(a) and are linearly independent since only true when and are both zero. Or we can say their quotient is (or ).
(b) and are linearly dependent since can be true when and .
Exercise 2.2.
Are the following functions linearly independent on the given interval?
(a)
(b)
(More Exercises: Problem Set 2 question 1)
2.3 General solution/Basis/Particular solution
A general solution of (2) has the form
 The function is on an open interval .
 The functions and are linearly independent on .
 The constants and are arbitrary.
 The functions , are called a basis of solutions of (2) on .
 Choosing , gives particular solutions.
2.4 Initial value problem
An initial value problem consists of (2), along with two initial conditions (since this is a second order ODE), which may be of the form
 The given values and are at the same given on
 These are used to determine and in a general solution
Example 2.4.
Consider and ,
Assume we know that and are solutions of this equation, the general solution for this is given by:
From the 1st initial condition of , ; From the 2nd initial condition of , .
Therefore, we have , , and we have a particular solution .
Exercise: What if, instead of knowing y(0) and y'(0), we knew y(0)=4 and y(1)=e+3/e  how would you find the particular solution? How about if we knew y(0)=4 and y'(1) = e3/e?
3 Homogeneous linear ODEs with constant coefficients
Consider secondorder homogeneous linear ODEs whose coefficients and are constant,
(6)
3.1 Solving homogeneous linear ODEs with constant coefficients
Guess the solution to be
then substitute it into (6), we get
Hence if is a solution of the characteristic equation
(7)
From algebra we further know that the quadratic equation (7) may have three kinds of root, depending on the sign of the discriminant
This leads to table below,
Case

Root of (7)

Basis of (6)

General Solution of (6)


Distinct real ,

,



Real double root

,



Complex conjugate ,

,


Note: For the case , we get only one root. To obtain a second independent solution, we use the method of reduction of order (see advanced topics at the bottom of this worksheet).
Note: For the case of , the roots of (7) and thus the solutions of (6), at first seem to be complexvalued. However, we can obtain a basis of real solutions , where .
Note: Euler's formula
Example 3.1.
Consider the ODE
(a) When (in this case we choose ), , the basis will be , , the general solution is .
(b) When , , the basis will be , , the general solution is .
(c) When (in this case we choose , , the basis will be , , the general solution is .
Exercise 3.1.
(a) Verify the solution of Example 3.1 (b) and (c).
(b) Solve the ODE
(More Exercises: Problem Set 2 Question 3(a) 21, 22, 23)
5 Modeling: Free Oscillations (MassSpring System)
This modeling problem is about Simple Harmonic Motion (SHM). Its model is a homogeneous linear ODE.
We measure the displacement of the body from the 'equilibrium point', when the system is in static equilibrium (at the origin ), with the downward direction regarded as the positive direction, and upward negative.
5.1 Undamped System
If the effect of damping on the system is negligible, we have the equation
(10)
where is the mass of the body and is the spring constant.
By the method we discussed above, we obtain the general solution
(11)
Since it has the period , the frequency of the oscillation is , which is also called the natural freqency of the system.
The general solution can be changed to
(12)
by using . This is called a phaseshifted cosine with amplitude and phase angle .
Figure 1: Harmoic oscillations
Figure 1 shows typical forms of the general solution, all corresponding to some positive initial displacement and different initial velocity .
Example 5.1.
Suppose an iron ball has mass , and a spring has a spring constant . Find the natural frequency when the massspring system execute. What will its motion be if we pull down the weight an additional and let it start with zero initial velocity?
Solution. The natural frequency . From (12) and and , the motion is
5.2 Damping System
When the effect of damping is not negligible, we need to add a damping force and the equation becomes
(13)
where is called the damping constant and is always positive (What would the physical significance of it being negative be?)
Consider the characteristic equation of (13),
(14)
Then we obtain the roots
(15)
As before, there are three cases for the discriminant . Those cases correspond to three types of motion, depending on the amount of damping (high, medium, or low).
Case I.


Distinct real roots ,

(Overdamping)

Case II.


A real double root

(Critical damping)

Case III.


Complex conjugate roots

(Underdamping)

Note: About Case III, , and .
Figure 2: Typical motion in Case I overdamped case


Figure 3: Case II Critical damping

5Figure 4: Damped oscillation in Case III

Example 5.2.
How does the motion in Example 5.1 change if we change the damping constant to one of the following three values, with and as before?
Solution.
Consider the ODE
For Case (I), , the ODE becomes . The discriminant of the characteristic equation is greater than zero. So, it has two roots 9 and 1. The general solution is
By consider the initial conditions, the solution is
Exercise 5.1.
Solve Cases (II) and (III) of the example above.
6 Linear Dependence and Independence of Solutions  Wronskian
Let the ODE have continuous coefficients and on an open interval . Then two solutions and of this equation on are linearly dependent on if and only if their "Wronskian"
is 0 at some in . Furthermore, if at an in , then on ; hence if there is an in at which is not 0, then , are linearly independent on .
Note: The claim is true for that and are analytic functions (infinitely differentiable).
Example 6.1.
A general solution of on any interval is .
Consider and , the Wronskian
Exercise 6.1.
Show linear independence by Wronskian: ,
(More Exercises: Problem Set 2 Question 4)
7 Nonhomogeneous ODEs
The nonhomogeneous linear ODE
(16)
where .
7.1 General solution and particular solution
A general solution of the nonhomogeneous ODE (16) on the open interval is a solution of the form
(17)
 The function is a general solution of the homogeneous ODE on .
 The function is any solution of (16) containing no arbitrary constants.
 Choosing , gives particular solutions.
8 Solving the nonhomogeneous ODE
To solve the nonhomogemeous ODE (16) or an initial value problem for (16), we need to perform the steps below.
Step 1: Find the general solution for the corresponding homogeneous ODE (r(x)=0).
Step 2: Find , as described below. Then we have a general solution (containing arbitrary constants and ).
Step 3: Finally, use the initial conditions to find , , which gives the particular solution.
To find , we use the below method:
8.1 Method of Undetermined Coefficients
The method of undetermined coefficients is suitable for linear ODEs with constant coefficients, say and .
(18)
 Guess a form of that is ``similar to" , but with unknown coefficients.
 For example if , guess .
 Determine the coefficients by substituting that and its derivatives into the ODE
8.1.1 Choice Rules for the Method of Undetermined Coefficients
(a) Basic Rule: If in (18) is one of the functions in the first column in Table, choose in the same line and determine its undetermined coefficients by substituting and its derivatives into (18). (See Example 8.1.1)
(b) Modification Rule: If a term in your choice for happens to be a solution of the homogeneous ODE corresponding to (18), multiply your choice of by (or by if this solution corresponds to a double root of the characteristic equation of the homogeneous ODE). (See Example 8.1.2)
(c) Sum Rule: If is a sum of functions in the first column of the Table below, choose for the sum of the functions in the corresponding lines of the second column. (See Example 8.1.3)
Table
Example 8.1.1.
Application of the basic rule (a) to solve the initial value problem
Solution.
Step 1. Find the general solution of the homogeneous ODE . We get
Step 2. Finding the particular solution .
We first try , . By substitution, . Comparing the coefficient of each power of ( and ), we get and .
But this is a contradiction!
The second line in the Table suggests instead the choice
Comparing the coefficients of , , on both sides, we have , , .
Hence . This gives , and
Step 3: Solution of the initial value problem.
By using the two initial conditions, we get the answer
Example 8.1.2.
Application of the Modification Rule (b)
Solve
Solution.
Step 1. Find .
Consider the homogeneous ODE ,
Ex 8.1.1.
First, we find the general solution of .
As before, this is
Step 2. Finding .
First consider . But this function is a solution of the homogeneous ODE. According to the Modification Rule, we choose
We substitute these expressions into the given ODE and factor out . This yields
Comparing the coefficients of , , and 1, we get . This gives the solution . Hence the given ODE has the general solution
Example 8.1.3.
Application of the Sum Rule (c)
Solve
Solution.
Step 1. Find the general solution of the corresponding homogeneous ODE. (Try it yourself !!!) The general solution should be
Step 2. Find .
We write , where corresponds to the exponential term and to the sum of the sum of the other two terms, i.e., the trigonometric function.
We set .
Ex 8.1.2.
Try to find in this case.
By substitution, we have .
We now set , as in the Table,
Ex 8.1.3.
Again we try to find .
Hence . Together,
Exercise 8.1.
Find a general solution of .
(More Exercises: Problem Set 2 Question 5)
9 Modeling: Free Oscillations. Resonance
Recall the springmass system model in Section 5. We now extend it by including an external force, say , on the right. Then we have
(20)
The resulting motion is called a forced motion with forcing function , which is also known as the input or driving force to the system, and the solution to be obtained is called the output or the response of the system to the driving force.
Of special interest are periodic external forces, and we shall consider a driving force of the form
Then (20) becomes
(21)
By using the method of undetermined coefficients,
(22)
By substitution, we get
Since , then and we obtain
We thus obtain the general solution of (21)
(23)
9.1 Undamped forced oscillations. Resonance
In this case, . Since , become
(24)
Finally, the solution becomes
(25)
We see that it is a superposition of two harmonic oscillations of the natural frequency.
Resonance: In (24) the maximum amplitude of is
(26)
The term is called the resonance factor. If , then and tend to infinity. If , resonance occurs. From (26) we see that is the ratio of the amplitudes of the particular solution and of the input . In case of resonance (21) becomes
(27)
Then (24) is no longer valid, and from the Modification Rule we find the particular solution is
(28)
By substituting this into (27) we get
(29)
Because of the factor , the amplitude of the vibration become larger and larger. Practically speaking, systems with very little damping may undergo large vibrations that can destroy the system.
Figure 5: Particular solution in resonance
Beats: Another interesting type of oscillation is obtained if is close to . Take, for example, the particular solution
(30)
By using , we can write this as
(31)
Since is close to , the difference is small. Hence the period of the last sine function is large, and we obtain an oscillation of the type shown in Figure 6, the dashed curve resulting from the first sine factor.
Figure 6: Beats
9.2 Damped forced oscillations
In this case,  the damping term cannot be negligible.
After a sufficiently long time the output of a damped vibrating system under a purely sinusoidal driving force will practically be a harmonic oscillation whose frequency is that of the input. Then approaches zero. Hence the "transient solution" approaches the "steadystate solution" .
In this case, the amplitude may have a maximum for some depending on the damping constant . This is called practical resonance.
To study the amplitude of as a function of ,
(32)
is called the amplitude of and is the phase angle or phase log. these quantities are
(33)
To find the maximum of , we set the derivative of equal to zero and solve for it to obtain
(34)
By reshuffling terms we have
The right side of this equation becomes negative if , so that then (34) has no real solution and decreases monotonically as increases. If is smaller, , then (34) has a real solution , where
(35)
Substituting (36) into (33) gives
(36)
We see that is always finite when . Furthermore, since the expression
in the denominator of (36) decreases monotonically to zero as goes to zero, the maximum amplitude (36) increases monotonically to infinity, in agreement with result in section 9.1.


Figure 7: Amplication (ratio of the amplitudes of output and input) as a function of for , , hence , and various values of the damping constant

Figure 8: Phase angle , which is less than when , and greater than for

Example 9.1.
Consider the massspring system with , , and . The general solution to the ODE without external force is
When we add the external force to start the oscillation, the steadystate solution is
Exercise 9.1.
Find the transient motion of the massspring system modeled by the given ODE
ADVANCED TOPICS
2.5 Finding a basis if one solution is known
We can find the second linearly independent solution by the method of reduction of order. Let be a known solution of (2) on . Guess the second linearly independent solution is of the form ,
(3)
Ex 2.5.1.
Prove (3).
Substitute , and into (2), we have
(4)
Ex 2.5.2.
Prove (4).
Substituting and , (4) becomes
Separation of variables and integration gives
(5)
Ex 2.5.3.
Prove (5).
Since , we have , thus
Example 2.5.
Consider the differential equation . One solution of it is . We divide the ODE by to get
Then we need to find . Because of (5),
By partial fractions,
Therefore,
It is obvious that and are linearly independent.
Exercise 2.5.
Find of when .
(More Exercises: Problem Set 2 Question 2(a))
4 EulerCauchy equations
EulerCauchy equations are ODEs of the form
(8)
4.1 Solving EulerCauchy equations
To find the basis of (8), we substitute
and its derivatives into (8). Then we have the characteristic equation
(9)
Similar to the case of linear ODEs with constant coefficients, there are three different cases.
Case

Root of (9)

Basis of (8)

General Solution of (8)


Distinct real ,

,



Real double root

,



Complex conjugate ,

,


Note: For the case of , the case of complex roots is of minor practical importance.
Example 4.1.
(a) Consider the ODE . The general solution is (by the table above, first row).
(b) Consider the other ODE . The general solution for all positive is since (by the table above, second row).
Exercise 4.1.
Find a general solution of
(More Exercises: Problem Set 2 Question 3(c))
8.2 Solution by Variation of Parameters
We can get the particular solution on in the form
(19)
where , form a basis of solutions of the corresponding homogeneous ODE
on , and is the Wronskian of , ,
Note: BE CAREFUL: The ODE should be written in the standard form, i.e., if the ODE starts with as in , it should be divided first by to get . Then , and .
Note: Even though this method seems to be very general and always give the answer via a formula, in practice, the integrations above may often cause difficulties, and so may the computation of , if the ODE has nonconstant coefficients. Use the previous methods if they work and you have a choice  they are simpler!
Example 8.2.
Solve the nonhomogeneous ODE
Solution. First, written in standard form, the ODE become
A basis of solutions of the homogeneous ODE an any interval is , (Example 2.1). This gives the Wronskian
From (19), choosing zero constants of integration,
From and the general solution of the homogeneous ODE we obtain the answer
Exercise 8.2.
Solve .
(More Exercises: Problem Set 2 Question 6)
Case II
Figure 2: Typicalmotion in
Case I overdamped case
Example 5.2. How does the motion in Example 5.1 change if we change the damping constant c to one of
the following three values, with y(0) = 0:16 and y0
(0) = 0 as before?
(I) c = 100 kg=sec; (II) c = 60 kg=sec; (III) c = 10 kg=sec:
Solution.
Consider the ODE
10y00
+ cy0
+ 90y = 0:
For Case (I), c = 100, the ODE become 10y00
+ 100y0
+ 90y = 0. The discriminant of the characteristic
equation is greater than zero. So, it has two roots 9 and 1. The general solution is
y = c1e 9t
+ c2e t
:
By consider the initial conditions, the solution is
y = 0:02e 9t
+ 0:18e t
:Example 5.2. How does the motion in Example 5.1 change if we change the damping constant c to one of
the following three values, with y(0) = 0:16 and y0
(0) = 0 as before?
(I) c = 100 kg=sec; (II) c = 60 kg=sec; (III) c = 10 kg=sec:
Solution.
Consider the ODE
10y00
+ cy0
+ 90y = 0:
For Case (I), c = 100, the ODE become 10y00
+ 100y0
+ 90y = 0. The discriminant of the characteristic
equation is greater than zero. So, it has two roots 9 and 1. The general solution is
y = c1e 9t
+ c2e t
:
By consider the initial conditions, the solution is
y = 0:02e 9t
+ 0:18e t
:yyyy
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