MA282 - Spring 2012
Study Guide for EXAM #2
EXAM #2 will be given on Friday, March 23. It will include material from Sections 2.8-2.9, 3.13.5, 3.8 and 3.11. This sheet is designed to help you organize your studying, not to be
exhaustive. You have to study all materials; textbook, notebook, homework and quizzes.
Topics
1st order equations (with initial conditions)

Nonlinear Models - Logistic Equations

Modeling with a system of 1st order equations
– Mixture(linear) & Predator–Prey Model(nonlinear)
Higher order equations (with initial conditions)

Theory of Linear Equations
– Linear dependence/independence, Fundamental Set of Solutions, …

Reduction of Order

Homogeneous Equations with Constant Coefficients

Non-homogeneous Equations with Constant Coefficients
– Method of Undetermined Coefficients

Non-homogeneous Equations with Constant Coefficients
- Variation of Parameters

Linear Models
– Free Undamped Motion, Free Damped Motion

Nonlinear Models
– Simple pendulum and Linearization
Review of Analytic Methods
1st order Equations
Separable Equations
To solve a separable equation of the form ∶
𝑑𝑦
= 𝑔(𝑥)ℎ(𝑦)
𝑑𝑥
Step1. Write DE in separable form:
1
𝑑𝑦 = 𝑔(𝑥)𝑑𝑥
ℎ(𝑦)
Step2. Integrate both sides:
∫
1
𝑑𝑦 = ∫ 𝑔(𝑥) 𝑑𝑥
ℎ(𝑦)
Separable Equations: Things to be Remembered
To get all solutions, you should FIRST check the singular solutions.
Method of Variation of Parameters (Integrating Factor)
Idea: Find the solution 𝑦𝑐 of associated homogeneous equation. Assume 𝑦𝑝 = 𝑢(𝑥)𝑦1 (𝑥)
where 𝑦1 (𝑥) is a solution of associated homo equation. Then the general solution is 𝑦 = 𝑦𝑐 + 𝑦𝑝 .
Here is the easy way.
To find a solution of equation
𝑑𝑦
+ 𝑃(𝑥)𝑦 = 𝑓(𝑥), do the following
𝑑𝑥
Step1. Calculate the integrating factor 𝑚(𝑥) = 𝑒 ∫ 𝑃(𝑥)𝑑𝑥
Step2. Multiply both sides of the DE by 𝑚(𝑥)
𝑚(𝑥)
𝑑𝑦
+ 𝑚(𝑥)𝑃(𝑥)𝑦 = 𝑚(𝑥)𝑓(𝑥)
𝑑𝑥
Step3. Write the LHS of the resulting equation in Step2 as
𝑑𝑦
+ 𝑚(𝑥)𝑃(𝑥)𝑦 = [𝑚(𝑥)𝑦(𝑥)]′
𝑑𝑥
Step4. Integrate the equation = [𝑚(𝑥)𝑦(𝑥)]′ = 𝑚(𝑥)𝑓(𝑥) to get
𝑚(𝑥)
𝑚(𝑥)𝑦(𝑥) = ∫ 𝑚(𝑥)𝑓(𝑥) 𝑑𝑥 + 𝐶
Step5. Solve for 𝑦(𝑥)
𝑦(𝑥) =
1
𝐶
∫ 𝑚(𝑥)𝑓(𝑥) 𝑑𝑥 +
𝑚(𝑥)
𝑚(𝑥)
Variation of Parameters: Things to be Remembered
The recipe only works for the linear equations in the standard form.
2nd order Equations
Reduction of Order
If 𝑦1 is a solution of DE
𝑦 ′′ + 𝑃(𝑥)𝑦 ′ + 𝑄(𝑥)𝑦 = 0
we substitute 𝑦2 (𝑥) = 𝑢(𝑥)𝑦1 (𝑥) and reduce the order of the equation by setting 𝑤 = 𝑢′ .
Then a second linearly independent solution 𝑦2 can be obtained by
𝑒 ∫ 𝑃(𝑥)𝑑𝑥
𝑦2 = 𝑦1 ∫
𝑑𝑥
𝑦12
Reduction of Order: Things to be Remembered
The formula in reduction of order only works for the equations in the standard form.
Homogeneous 2nd order Equations with Constant Coefficients
To solve the equation 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐 = 0, do the following
Step1. Find the corresponding auxiliary equation:
𝑎𝑚2 + 𝑏𝑚 + 𝑐 = 0
Step2. Solve this auxiliary equation:
𝑚1 , 𝑚2 =
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
Step3. Get the general solution:
Case1. 𝑚1 ≠ 𝑚2 : 𝑦 = 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑒 𝑚2 𝑥
Case2. 𝑚1 = 𝑚2 : 𝑦 = 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑥𝑒 𝑚1 𝑥
Case3. 𝑚1 , 𝑚2 = 𝛼 ± 𝑖𝛽: 𝑦 = 𝑐1 𝑒 𝛼𝑥 cos 𝛽𝑥 + 𝑐2 𝑒 𝛼𝑥 sin 𝛽𝑥
Remark Analogous for higher order Equations with Constant Coefficients
Method of Undetermined Coefficients for Nonhomo Equations with Constant Coefficients
Consider an equation 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐 = 𝑔(𝑥) with 𝑔(𝑥) of either form 𝑃𝑙 (𝑥)𝑒 𝑚𝑥 𝑐𝑜𝑠(𝑘𝑥) or
𝑃𝑙 (𝑥)𝑒 𝑚𝑥 𝑠𝑖𝑛(𝑘𝑥), where 𝑃𝑙 (𝑥) is a polynomial in 𝑥 of degree 𝑙 . To find a particular solution 𝑦𝑝 ,
do the following
Step1. Find the complementary function 𝑦𝑐
Step1. Take the trial function 𝑦𝑝
𝑦𝑝 (𝑥) = (𝐴0 + 𝐴1 𝑥 + 𝐴2 𝑥 2 + · · · + 𝐴𝑙 𝑥 𝑙 )𝑒 𝑚𝑥 cos(𝑘𝑥)
+ (𝐵0 + 𝐵1 𝑥 + 𝐵2 𝑥 2 + · · · + 𝐵𝑙 𝑥 𝑙 )𝑒 𝑚𝑥 cos(𝑘𝑥)
Step3. Check the duplicate terms
Step4. Eliminate the duplicate part by multiplying by 𝑥 𝑠 , where 𝑠 > 0 is the smallest such
integer.
Method of Variation of Parameters for Nonhomo DEs with Constant Coefficients
To find a particular solution of the nonhomogeneous linear DE 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐 = 𝑔(𝑥)
do the following
Step1. Find the complementary function 𝑦𝑐 = 𝑐1 𝑦1 + 𝑐2 𝑦2
Step2. Assume 𝑦𝑝 = 𝑢1 𝑦1 + 𝑢2 𝑦2
Step3. Write down the system of equations:
𝑢1′ 𝑦1 + 𝑢2′ 𝑦2 = 0,
𝑢1′ 𝑦1 ′ + 𝑢2′ 𝑦2 ′ = 𝑔(𝑥)
Step4. Solve for 𝑢1 ′, 𝑢2 ′:
𝑢1′ = −
𝑦2 𝑔(𝑥)
,
𝑊(𝑦1 , 𝑦2 )
𝑢2′ =
𝑦1 𝑔(𝑥)
𝑊(𝑦1 , 𝑦2 )
Step5. Obtain 𝑢1 , 𝑢2 :
𝑢1 = − ∫
𝑦2 𝑔(𝑥)
𝑦1 𝑔(𝑥)
𝑑𝑥 , 𝑢2 = ∫
𝑑𝑥
𝑊(𝑦1 , 𝑦2 )
𝑊(𝑦1 , 𝑦2 )
Variation of Parameters: Things to be Remembered
The formula in variation of parameters only works for the equations in the standard form.