1. ENGG2013 Unit 25 Second-order Linear DE
Apr, 2011.

2. Yesterday First-order DE System of first-order DE Objectives:
Method of separating variable
Method of integrating factor
System of first-order DE
Eigenvalues determine the convergence behaviour near a critical point.
Objectives:
Solve the initial value problem
Given the initial value, find the trajectory
Transient-state analysis of electronic circuits
Understand the system behaviour
Does the system converge?
Stable equilibrium point, unstable equilibrium point.
kshum

3. Linear Second-order DE
Homogeneous
Non-homogeneous
kshum

4. Linear Second-order constant-coeff. DE
Homogeneous
Non-homogeneous
kshum

5. Vibrating spring without damping
x(t): vertical displacement
Hooke’s law: Force = k x
k is the spring constant, k > 0 (the constant k is sometime called the spring modulus.)
Newton’s law: F = m x’’
m is the mass, m > 0
k x
m x’’ x
Second-order, autonomous linear, constant-coefficient and homogeneous
m x’’ + k x = 0
Assumptions:
Spring has negligible weight
No friction
kshum

6. Solutions to undamped spring-mass model
Normalize by m
Direct substitution verifies that and are solutions, e.g.
{\color{blue} \begin{align*} &\sin''\left( \sqrt{\frac{k}{m}} t \right) + \frac{k}{m}\sin\left( \sqrt{\frac{k}{m}} t \right) \\
&= \sqrt{\frac{k}{m}}\cos'\left( \sqrt{\frac{k}{m}} t \right) + \frac{k}{m}\sin\left( \sqrt{\frac{k}{m}} t \right) \\
&= -\frac{k}{m} \sin\left( \sqrt{\frac{k}{m}} t \right) + \frac{k}{m}\sin\left( \sqrt{\frac{k}{m}} t \right) \\
&= 0
\end{align*}
kshum

7. Natural frequency
Let
We know that cos( t) and sin( t) are both solutions.
 is called the natural frequency.
Dimension check: The unit of  is Hz = s-1.
The spring constant k has unit kg s-2.
k/m has unit s-2.
Square root of k/m has unit s-1.
kshum

8. Principle of superposition (aka linearity principle)
For linear and homogenous differential equation, the linear combination of two solutions is also a solution.
 For any real numbers a and b, a cos( t) + b cos( t)
is a solution to x’’+ 2 x=0.
\begin{align*}
&[a\ {\color{red} \cos(\omega t)} + b\ {\color{blue} \sin(\omega t)}]'' + \omega^2 [a\ {\color{red} \cos(\omega t)} + b\ {\color{blue}\sin(\omega t)}] \\
&= a\ {\color{red} \cos''(\omega t)} + b\ {\color{blue}\sin''(\omega t)}+ \omega^2 a\ {\color{red} \cos(\omega t)} + \omega^2 b\ {\color{blue}\sin(\omega t)} \\
&= a [{\color{red} \cos''(\omega t)}+ \omega^2\ {\color{red} \cos(\omega t)} ] +b [{\color{blue}\sin''(\omega t)}+ \omega^2\ {\color{blue}\sin(\omega t)}]\\
&= = 0
\end{align*}
kshum

9. GRAPHICAL METHOD
kshum

10. Graphical illustration of spring-mass model
Define the displacement-velocity vector
Reduction to system of two first-order differential equations.
{\color{blue}\begin{bmatrix} x \\ v \end{bmatrix}' =
\begin{bmatrix}
0 & 1 \\ -\omega^2 & 0
\end{bmatrix}
\begin{bmatrix} x \\ v \end{bmatrix} }
kshum

11. Phase plane for the vibrating spring
Sample solution
The trajectory is a ellipse
kshum

12. Order reduction technique
equivalent
\begin{bmatrix} x \\ x' \end{bmatrix}' =
\begin{bmatrix} 0 & 1 \\
-b & a
\end{bmatrix}
\begin{bmatrix} x \\ x' \end{bmatrix}
Second-order DE with constant coefficients is basically the same as a system of two first-order DE.
kshum

13. Vibrating spring with damping
Vibrating spring in honey
m x’’ = – k x – d x’
honey
Force exerted by the spring
Assumption:
Magnitude of damping force is directly proportional to x’.
d> 0
Damping due to viscosity
Equivalent form
kshum

14. Phase plane for damped spring
Sample solutions
Convergence to the origin
kshum

15. METHOD OF DIAGONALIZATION
kshum

16. Recall: Diagonalization
Definition: an n  n matrix M is called diagonalizable if we can find an invertible matrix S, such that is a diagonal matrix, or equivalently,
Example:
Diagonalization is useful in decoupling a linear system for instance.

17. Solution to vibrating spring with damping
Characteristic equation
Eigenvalues
\begin{bmatrix} 0 & 1 \\
-3 & -1
\end{bmatrix}
\begin{bmatrix} x \\ v \end{bmatrix}

18. Solution to vibrating spring with damping
Eigenvectors have complex components
Concatenate
Two eigenvectors are not scalar multiple of each other,
because the two eigenvalues are distinct . Hence inverse exists.
Diagonalize

19. Solution to vibrating spring with damping
Substitute by the diagonalized matrix
Change of variables
\begin{bmatrix} x \\ y\end{bmatrix}
\leftarrow \begin{bmatrix} 0 & 2 \\ 1& -5 \end{bmatrix}^{-1}
\begin{bmatrix} C_A \\ C_B \end{bmatrix}
An uncoupled system

20. Solution to vibrating spring with damping
Solve the uncoupled system
K1 and K2 are constants
Substitute back
(C1, C1, a and b are constants)

21. Solution to vibrating spring with damping

22. A general strategy for linear system
Decouple (Diagonalization)
Solve each subsystem separately
Piece them together
Solved

23. METHOD OF GUESS-AND-VERIFY
kshum

24. 2nd-order constant-coeff. DE
Homogeneous
b and c are constants.
Non-homogeneous
f(t) is a function of independent variable t.
kshum

25. The homogeneous case
Idea: try a function in the form ekt as a solution.
k is some constant.
Substitute ekt into x’’+bx’+cx=0 and try to solve for the constant k.
Apply the superposition principle: any linear combination of solutions is also a solution.
kshum

26. Examples Solve x’’+3x’+2x=0. Solve x’’–4x’+4x=0. Solve x’’+9x= 0.
General solution: x(t) = c1 e–2t+ c2e–t
General solution: x(t) = c1 e2t+ c2 t e2t
General solution:
x(t) = c1 e3i t+ c2 e-3i t
= d1 sin(t)+ d2 cos(t)
kshum

27. Summary of the three cases
Differential equation
Characteristic equation
Case
Roots
Basis of solutions
General Solution 1
Distinct real  and 
et, et
c1et+c2et 2
Repeated root 
et, tet
c1et+c2tet 3
Complex roots  =r+i, =r–i
e(r+i)t, e(r–i)t
er(c1cos t+c2sin t)
kshum

28. The non-homogeneous case
Property: If x1(t) and x2(t) are two solutions to
then their difference x1(t) – x2(t) is a solution to the homogeneous counterpart
kshum

29. Consequence
Suppose that xp(t) is some solution to x’’+bx’+cx=f(t) (given to you by a genie for example)
 Any solution of x’’+bx’+cx=f(t) can be written as xh(t) +xp(t)
A solution to the homogeneous DE
x’’+bx’+c=0
A particular solution
kshum

30. Method of trial and error (aka as the method of undetermined coefficient)
To solve the non-homogeneous DE x’’+bx’+cx=f(t)
Find a particular solution by trial and error (and experience)
Let the particular solution be xp(t).
Solve the homogeneous version x’’+bx’+cx=0.
Let the homogeneous solution be xh(t).
The general solution is xh(t)+ xp(t).
kshum

31. How to guess a particular solution
f(t)
Choice for xp(t)
K xn
cnxn+cn-1xn-1+…+c1x1+c0
K eat
Ceat
K sin(t)
c1 sin(t)+c2 cos(t)
K cos(t)
K eat sin(t)
eat (c1 sin(t)+c2 cos(t))
K eat cos(t)
K, C, a, , c0, c1, c2,… are constants
kshum

32. Example Solve x’’+3x’+2x= e5t. Try c e5t as a particular solution.
(c e5t )’’+3(c e5t )’+(c e5t )= e5t
25c e5t+15c e5t+5c e5t= e5t
25 c + 15c + 5c=1  c = 1/45
Let xp(t) = e5t/45 as a particular solution.
General solution is c1e–2t+ c2e–t + e5t/45
Homogeneous solution
Particular solution
kshum

33. Summary Graphical method using phase plane. Method of diagonalization
Reduction to two 1st-order linear DE.
Method of diagonalization
Need to reduce the second-order DE to a system of first-order DE.
Time-consuming but theoretically sound.
Method of undetermined coefficients
Find a solution quickly, but not systematic.
Good for calculation in examination.
kshum

34. Final Exam Date: 6th May (Friday) Venue: NA Gym Time: 9:30~11:30
Coverage: Everything in Lecture Notes and Homeworks
Close-book exam
You may bring a calculator, and a handwritten A4-size and double-sided crib sheet.
kshum