Upload
geeess87
View
70
Download
18
Embed Size (px)
DESCRIPTION
Higher order differential Equations
Citation preview
EEEE707: Engineering Analysis
Dr. Eli Saber Department of Electrical and Microelectronic Engineering
Chester F. Carlson Center for Imaging Science Rochester Institute of Technology, Rochester, NY 14623 USA
Chapter 3 Higher Order Differential
Equations
9/30/2014 Dr. Eli Saber 2
Section 3.1 Theory of Linear Equations
9/30/2014 Dr. Eli Saber 3
Theory of Linear Equations
Objective: Investigate Differential Equations of Order 2++
9/30/2014 Dr. Eli Saber 4
Theory of Linear Equations
IVP: Initial Value Problem BVP: Boundary Value Problem • IVP: Solve:
𝑎𝑎𝑛𝑛 𝑥𝑥 𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥
𝑑𝑑𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1 + … + 𝑎𝑎1 𝑥𝑥
𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔(𝑥𝑥)
Subject to:
𝑦𝑦 𝑥𝑥0 = 𝑦𝑦0, 𝑦𝑦′ 𝑥𝑥0 = 𝑦𝑦, … . . ,𝑦𝑦 𝑛𝑛−1 (𝑥𝑥0) = 𝑦𝑦𝑛𝑛−1 i.e. seek a function defined on interval 𝐼𝐼 containing 𝑥𝑥0 that satisfies the D.E. and the 𝑛𝑛 initial conditions
9/30/2014 Dr. Eli Saber 5
Initial Value and Boundary Value Problems
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 6
Initial Value and Boundary Value Problems
Theorem: Existence of a Unique Solution (for 1st order D.E.) Let 𝑅𝑅 be a Rectangular region in the x-y plane defined by 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏; 𝑐𝑐 ≤ 𝑦𝑦 ≤ 𝑑𝑑 that contains the point 𝑥𝑥0,𝑦𝑦0 . If 𝑓𝑓 𝑥𝑥, 𝑦𝑦 & 𝑑𝑑𝑦𝑦/𝑑𝑑𝑥𝑥 are continuous on 𝑅𝑅, then there exists some Interval 𝐼𝐼0: 𝑥𝑥0 − ℎ, 𝑥𝑥0 + ℎ ; ℎ > 0 contained in [𝑎𝑎, 𝑏𝑏] and a unique function 𝑦𝑦 𝑥𝑥 defined on 𝐼𝐼0 that is a solution of the Initial Value Problem.
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 7
Initial Value and Boundary Value Problems
Theorem: Existence of a Unique Solution (for nth order D.E.)
𝑎𝑎𝑛𝑛 𝑥𝑥 𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥
𝑑𝑑𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1 + … + 𝑎𝑎1 𝑥𝑥
𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔(𝑥𝑥)
Let 𝑎𝑎𝑛𝑛 𝑥𝑥 , 𝑎𝑎𝑛𝑛−1 𝑥𝑥 , … , 𝑎𝑎1 𝑥𝑥 ,𝑎𝑎0 𝑥𝑥 & 𝑔𝑔 𝑥𝑥 be continuous on an interval 𝐼𝐼, and let 𝑎𝑎𝑛𝑛 𝑥𝑥 ≠ 0 ∀ 𝑥𝑥 𝜖𝜖𝐼𝐼 If 𝑥𝑥 = 𝑥𝑥0 is any point in 𝐼𝐼, a solution 𝑦𝑦(𝑥𝑥) of the IVP exists on the interval and is unique.
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 8
Initial Value and Boundary Value Problems
E.g. 3𝑦𝑦′′′ + 5𝑦𝑦′′ − 𝑦𝑦′ + 7𝑦𝑦 = 0 𝑦𝑦 1 = 0; 𝑦𝑦′ 1 = 0; 𝑦𝑦′′ 1 = 0 Solution: 𝒚𝒚 = 𝟎𝟎 Since D.E. is linear with constant coefficients, the unique solution theorem is fulfilled. Hence, 𝒚𝒚 = 𝟎𝟎 is the only solution on any interval containing x=1
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 9
Initial Value and Boundary Value Problems
E.g. 𝑦𝑦′′ − 4𝑦𝑦 = 12𝑥𝑥 𝑦𝑦 0 = 4;𝑦𝑦′ 0 = 1 Solution: 𝒚𝒚 = 𝟑𝟑𝒆𝒆𝟐𝟐𝟐𝟐 + 𝒆𝒆−𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟐𝟐 1. D.E. is linear with constant coefficients 2. The coefficients as well as 𝑔𝑔(𝑥𝑥) are continuous 3. 𝑎𝑎2(𝑥𝑥) = 1 ≠ 0 on any interval 𝐼𝐼 containing 𝑥𝑥 = 0
The unique solution theorem is fulfilled. Hence, 𝒚𝒚 = 𝟑𝟑𝒆𝒆𝟐𝟐𝟐𝟐 + 𝒆𝒆−𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟐𝟐 is the unique solution on interval 𝑰𝑰
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 10
Initial Value and Boundary Value Problems
Check: 𝑦𝑦′′ − 4𝑦𝑦 = 12𝑥𝑥; Solution: 𝑦𝑦 = 3𝑒𝑒2𝑥𝑥 + 𝑒𝑒−2𝑥𝑥 − 3𝑥𝑥 Now, 𝑦𝑦′ = 6𝑒𝑒2𝑥𝑥 − 2𝑒𝑒−2𝑥𝑥 − 3 And, 𝑦𝑦′′ = 12𝑒𝑒2𝑥𝑥 + 4𝑒𝑒−2𝑥𝑥 𝑦𝑦′′ − 4𝑦𝑦 = 12𝑒𝑒2𝑥𝑥 + 4𝑒𝑒−2𝑥𝑥 − 4 3𝑒𝑒2𝑥𝑥 + 𝑒𝑒−2𝑥𝑥 − 3𝑥𝑥 = 12𝑒𝑒2𝑥𝑥 + 4𝑒𝑒−2𝑥𝑥 − 12𝑒𝑒2𝑥𝑥 − 4𝑒𝑒−2𝑥𝑥 + 12𝑥𝑥
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 11
Initial Value and Boundary Value Problems
Check: 𝑦𝑦′′ − 4𝑦𝑦 = 12𝑥𝑥;𝑦𝑦 = 3𝑒𝑒2𝑥𝑥 + 𝑒𝑒−2𝑥𝑥 − 3𝑥𝑥 Now, 𝑦𝑦′ = 6𝑒𝑒2𝑥𝑥 − 2𝑒𝑒−2𝑥𝑥 − 3 And, 𝑦𝑦′′ = 12𝑒𝑒2𝑥𝑥 + 4𝑒𝑒−2𝑥𝑥 𝑦𝑦′′ − 4𝑦𝑦 = 12𝑒𝑒2𝑥𝑥 + 4𝑒𝑒−2𝑥𝑥 − 4 3𝑒𝑒2𝑥𝑥 + 𝑒𝑒−2𝑥𝑥 − 3𝑥𝑥 = 12𝑒𝑒2𝑥𝑥 + 4𝑒𝑒−2𝑥𝑥 − 12𝑒𝑒2𝑥𝑥 − 4𝑒𝑒−2𝑥𝑥 + 12𝑥𝑥 = 12𝑥𝑥 Verified.
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 12
Initial Value and Boundary Value Problems
E.g. 𝑥𝑥2𝑦𝑦′′ − 2𝑥𝑥𝑦𝑦′ + 2𝑦𝑦 = 6 𝑦𝑦 0 = 3; 𝑦𝑦′ 0 = 1 Solution: 𝑦𝑦 = 𝑐𝑐𝑥𝑥2 + 𝑥𝑥 + 3 in interval 𝐼𝐼 ≡ (−∞,∞)
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 13
Initial Value and Boundary Value Problems
Check: 𝑦𝑦 = 𝑐𝑐𝑥𝑥2 + 𝑥𝑥 + 3 ⇒ 𝑦𝑦′ = 2𝑐𝑐𝑥𝑥 + 1; 𝑦𝑦′′ = 2𝑐𝑐 𝑥𝑥2𝑦𝑦′′ − 2𝑥𝑥𝑦𝑦′ + 2𝑦𝑦 = 𝑥𝑥2 2𝑐𝑐 − 2𝑥𝑥 2𝑐𝑐𝑥𝑥 + 1 + 2 𝑐𝑐𝑥𝑥2 + 𝑥𝑥 + 3 = 2𝑐𝑐𝑥𝑥2 − 4𝑐𝑐𝑥𝑥2 − 2𝑥𝑥 + 2𝑐𝑐𝑥𝑥2 + 2𝑥𝑥 + 6 = 2𝑐𝑐𝑥𝑥2 − 4𝑐𝑐𝑥𝑥2 − 2𝑥𝑥 + 2𝑐𝑐𝑥𝑥2 + 2𝑥𝑥 + 6 𝑥𝑥2𝑦𝑦′′ − 2𝑥𝑥𝑦𝑦′ + 2𝑦𝑦 = 6
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 14
Initial Value and Boundary Value Problems
IVP Check: 𝑦𝑦 = 𝑐𝑐𝑥𝑥2 + 𝑥𝑥 + 3 • 𝑦𝑦 0 = 3 ⇒ 3 = 𝑐𝑐 0 2 + 0 + 3 ⇒ 3 = 3
• 𝑦𝑦′ 0 = 1
𝑦𝑦 = 𝑐𝑐𝑥𝑥2 + 𝑥𝑥 + 3 ⇒ 𝑦𝑦′ = 2𝑐𝑐𝑥𝑥 + 1 1 = 2𝑐𝑐 0 + 1 ⇒ 1 = 1
• Note:
For 𝑦𝑦 = 𝑐𝑐𝑥𝑥2 + 𝑥𝑥 + 3 , the initial conditions of 𝑦𝑦 0 = 3 & 𝑦𝑦′ 0 = 1 did not provide a unique value for 𝑐𝑐
Hence: 𝑦𝑦 = 𝑐𝑐𝑥𝑥2 + 𝑥𝑥 + 3 is a solution for the D.E. 𝑥𝑥2𝑦𝑦′′ − 2𝑥𝑥𝑦𝑦′ + 2𝑦𝑦 = 6 ∀𝑐𝑐 i.e. there is no unique solution But what w.r.t. unique solution theorem?
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 15
Initial Value and Boundary Value Problems
Let us apply the theorem towards this example.
𝑥𝑥2 𝑦𝑦′′ − 2𝑥𝑥 𝑦𝑦′ + 2 𝑦𝑦 = 6 Note: 𝑎𝑎2 𝑥𝑥 = 𝑥𝑥2 = 0 𝑓𝑓𝑓𝑓𝑓𝑓 𝑥𝑥 = 0 and 𝑥𝑥 ∈ 𝐼𝐼 = (−∞,∞) 𝒂𝒂𝟐𝟐 𝟐𝟐 ≠ 𝟎𝟎∀ 𝟐𝟐 = 𝟐𝟐𝟎𝟎 ∈ 𝑰𝑰 this condition is NOT satisfied
𝑎𝑎2 𝑥𝑥 = 𝑥𝑥2
𝑎𝑎1 𝑥𝑥 = −2𝑥𝑥
𝑎𝑎0 𝑥𝑥 = 2
𝑔𝑔 𝑥𝑥 = 6
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 16
Initial Value and Boundary Value Problems
• BVP (Boundary Value Problem):
𝐷𝐷.𝐸𝐸. ∶ 𝑎𝑎2 𝑥𝑥𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
+ 𝑎𝑎1 𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔 𝑥𝑥
With 𝑦𝑦 𝑎𝑎 = 𝑦𝑦0 & 𝑦𝑦 𝑏𝑏 = 𝑦𝑦1 Other boundary value conditions could be: 𝑦𝑦𝑦 𝑎𝑎 = 𝑦𝑦0 & 𝑦𝑦 𝑏𝑏 = 𝑦𝑦1 𝑦𝑦 𝑎𝑎 = 𝑦𝑦0 & 𝑦𝑦𝑦 𝑏𝑏 = 𝑦𝑦1 𝑦𝑦𝑦 𝑎𝑎 = 𝑦𝑦0 & 𝑦𝑦𝑦 𝑏𝑏 = 𝑦𝑦1 General Boundary Conditions: 𝑨𝑨𝟏𝟏𝒚𝒚 𝒂𝒂 + 𝑩𝑩𝟏𝟏𝒚𝒚′ 𝒂𝒂 = 𝑪𝑪𝟏𝟏 𝑨𝑨𝟐𝟐𝒚𝒚 𝒃𝒃 + 𝑩𝑩𝟐𝟐𝒚𝒚′ 𝒃𝒃 = 𝑪𝑪𝟐𝟐
Boundary conditions
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 17
Initial Value and Boundary Value Problems
Note: Even when the conditions for Unique Solution theorem are met, a BVP may have: 1) Many solutions 2) Unique Solution 3) No solution
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 18
Initial Value and Boundary Value Problems
E.g. 𝑑𝑑2𝑥𝑥𝑑𝑑𝑡𝑡2
+ 16𝑥𝑥 = 0
Solution: 𝑥𝑥 = 𝑐𝑐1 cos4𝑡𝑡 + 𝑐𝑐2 sin 4𝑡𝑡 Check: 𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡
= −4𝑐𝑐1 sin4𝑡𝑡 + 4𝑐𝑐2 cos4𝑡𝑡
𝑑𝑑2𝑥𝑥𝑑𝑑𝑡𝑡2
= −16𝑐𝑐1 cos4𝑡𝑡 − 16𝑐𝑐2 sin 4𝑡𝑡
𝑑𝑑2𝑥𝑥𝑑𝑑𝑡𝑡2
+ 16𝑥𝑥 = −16𝑐𝑐1 cos4𝑡𝑡 − 16𝑐𝑐2 sin 4𝑡𝑡 + 16 𝑐𝑐1 cos4𝑡𝑡 + 𝑐𝑐2 sin4𝑡𝑡 = 0
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 19
Initial Value and Boundary Value Problems
Now, let us consider these different sets of BV Conditions: 𝑥𝑥 = 𝑐𝑐1 cos4𝑡𝑡 + 𝑐𝑐2 sin 4𝑡𝑡
1) 𝑥𝑥 0 = 0; 𝑥𝑥𝜋𝜋2
= 0
• 𝑥𝑥 0 = 0 ⇒ 0 = 𝑐𝑐1 1 + 𝑐𝑐2 0 ⇒ 𝒄𝒄𝟏𝟏 = 𝟎𝟎
• 𝑥𝑥 𝜋𝜋2
= 0 ⇒ 0 = 𝑐𝑐1 cos(2𝜋𝜋) + 𝑐𝑐2 sin(2𝜋𝜋) ⇒ 0 = 𝑐𝑐1 1 + 𝑐𝑐2 0
– But 𝑐𝑐1 = 0
– That means, 𝑐𝑐2 0 = 0
– Implies 𝑐𝑐2 can be anything
• Infinite solutions since 𝑐𝑐2 can be anything
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 20
Initial Value and Boundary Value Problems
2) 𝑥𝑥 0 = 0; 𝑥𝑥𝜋𝜋8
= 0
• 𝑥𝑥 0 = 0 ⇒ 0 = 𝑐𝑐1 1 + 𝑐𝑐2 0 ⇒ 𝒄𝒄𝟏𝟏 = 𝟎𝟎
• 𝑥𝑥 𝜋𝜋8
= 0 ⇒ 0 = 𝑐𝑐1 cos 𝜋𝜋2
+ 𝑐𝑐2 sin 𝜋𝜋2⇒ 0 = 𝑐𝑐1 0 + 𝑐𝑐2 1
– But 𝑐𝑐1 = 0
– That means, 𝑐𝑐2 1 = 0
– Implies 𝒄𝒄𝟐𝟐 = 𝟎𝟎
– 𝟐𝟐 = 𝟎𝟎 is the solution of this new boundary problem
• Unique solution ≡ 𝟐𝟐 = 𝟎𝟎
𝑥𝑥 = 𝑐𝑐1 cos4𝑡𝑡 + 𝑐𝑐2 sin4𝑡𝑡
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 21
Initial Value and Boundary Value Problems
3) 𝑥𝑥 0 = 0;𝑥𝑥𝜋𝜋2
= 1
• 𝑥𝑥 0 = 0 ⇒ 0 = 𝑐𝑐1 1 + 𝑐𝑐2 0 ⇒ 𝒄𝒄𝟏𝟏 = 𝟎𝟎
• 𝑥𝑥 𝜋𝜋2
= 1 ⇒ 1 = 𝑐𝑐1 cos 4𝜋𝜋 ∙ 𝜋𝜋2
+ 𝑐𝑐2 sin 4𝜋𝜋 ∙ 𝜋𝜋2⇒ 1 = 𝑐𝑐1 cos(2𝜋𝜋) + 𝑐𝑐2 sin(2𝜋𝜋)
– 1 = 0 1 + 𝑐𝑐2 0
– That means, 𝑐𝑐2 0 = 1
– Implies 𝒄𝒄𝟐𝟐 = 𝟏𝟏𝟎𝟎
= 𝑵𝑵.𝑫𝑫.
– Not possible to find 𝑐𝑐2
• No solution for BVP
𝑥𝑥 = 𝑐𝑐1 cos4𝑡𝑡 + 𝑐𝑐2 sin4𝑡𝑡
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 22
Differential Operators “D”
E.g. 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 = 𝐷𝐷𝑦𝑦
𝑑𝑑𝑦𝑦2
𝑑𝑑𝑥𝑥2 =𝑑𝑑𝑑𝑑𝑥𝑥
𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 = 𝐷𝐷 𝐷𝐷𝑦𝑦 = 𝐷𝐷2𝑦𝑦
i.e. 𝑑𝑑𝑑𝑑𝑥𝑥
cos4𝑥𝑥 = −4 sin4𝑥𝑥 ⇒ 𝐷𝐷 cos4𝑥𝑥 = −4 sin 4𝑥𝑥
⇒ 𝐼𝐼𝑛𝑛 𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑓𝑓𝑎𝑎𝑔𝑔: 𝒅𝒅𝒏𝒏𝒚𝒚𝒅𝒅𝟐𝟐𝒏𝒏 = 𝑫𝑫𝒏𝒏𝒚𝒚
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 23
Differential Equations
Note: • 𝐷𝐷 𝑐𝑐 𝑓𝑓 𝑥𝑥 = 𝑐𝑐 𝐷𝐷𝑓𝑓(𝑥𝑥)
• 𝐷𝐷 𝑓𝑓 𝑥𝑥 + 𝑔𝑔 𝑥𝑥 = 𝐷𝐷𝑓𝑓 𝑥𝑥 + 𝐷𝐷𝑔𝑔(𝑥𝑥)
• 𝐷𝐷 𝛼𝛼 𝑓𝑓 𝑥𝑥 + 𝛽𝛽 𝑔𝑔 𝑥𝑥 = 𝛼𝛼 𝐷𝐷 𝑓𝑓 𝑥𝑥 + 𝛽𝛽 𝐷𝐷 𝑔𝑔 𝑥𝑥
– 𝛼𝛼,𝛽𝛽 are constants
Linear
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 24
Differential Equations
Let 𝑦𝑦′′ + 5𝑦𝑦′ + 6𝑦𝑦 = 5𝑥𝑥 − 3
This can be written was 𝑑𝑑2𝑦𝑦
𝑑𝑑𝑥𝑥2+ 5 𝑑𝑑𝑦𝑦
𝑑𝑑𝑥𝑥+ 6𝑦𝑦 = 5𝑥𝑥 − 3
Which can also be written as: 𝐷𝐷2𝑦𝑦 + 5𝐷𝐷𝑦𝑦 + 6𝑦𝑦 = 5𝑥𝑥 − 3
Similarly, 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑑𝑑𝑛𝑛𝑦𝑦
𝑑𝑑𝑥𝑥𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑑𝑑
𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1
+ … + 𝑎𝑎1 𝑥𝑥 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 0
can be written as 𝐿𝐿 𝑦𝑦 = 0
And, 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑑𝑑𝑛𝑛𝑦𝑦
𝑑𝑑𝑥𝑥𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑑𝑑
𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1
+ … + 𝑎𝑎1 𝑥𝑥 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔(𝑥𝑥)
can be written as 𝐿𝐿 𝑦𝑦 = 𝑔𝑔(𝑥𝑥)
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 25
Differential Equations Definition: 𝑛𝑛𝑡𝑡𝑡order differential operator is:
𝐿𝐿 = 𝑎𝑎𝑛𝑛 𝑥𝑥 𝐷𝐷𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝐷𝐷𝑛𝑛−1 + ⋯+ 𝑎𝑎1 𝑥𝑥 𝐷𝐷 + 𝑎𝑎0(𝑥𝑥)
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 26
Superposition Principle
Theorem: Superposition Principle – Homogeneous Equations Let 𝑦𝑦1, 𝑦𝑦2, … ,𝑦𝑦𝑘𝑘 be solutions of the Homogeneous 𝑛𝑛𝑡𝑡𝑡 order differential equation on an interval 𝐼𝐼. Then the linear combination
𝑦𝑦 = 𝑐𝑐1𝑦𝑦1 𝑥𝑥 + 𝑐𝑐2𝑦𝑦2 𝑥𝑥 + ⋯+ 𝑐𝑐𝑘𝑘𝑦𝑦𝑘𝑘 𝑥𝑥 ,where 𝑐𝑐1, 𝑐𝑐2, … , 𝑐𝑐𝑘𝑘 as are arbitrary constants, is also a solution on 𝐼𝐼 Corollaries: • A constant multiple 𝑦𝑦 = 𝑐𝑐1𝑦𝑦1(𝑥𝑥) of the solution 𝑦𝑦1(𝑥𝑥) of a homogeneous
linear differential equation is also a solution
• A homogeneous linear differential equation always possesses the trivial solution 𝑦𝑦 = 0
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 27
Superposition Principle
E.g. 𝑥𝑥3 𝑑𝑑3𝑦𝑦
𝑑𝑑𝑥𝑥3− 2𝑥𝑥 𝑑𝑑𝑦𝑦
𝑑𝑑𝑥𝑥+ 4𝑦𝑦 = 0
And 𝑦𝑦1 = 𝑥𝑥2 & 𝑦𝑦2 = 𝑥𝑥2 ln 𝑥𝑥 are both solutions Check: First solution: 𝑦𝑦1 = 𝑥𝑥2
𝑦𝑦 = 𝑥𝑥2 ⇒𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 = 2𝑥𝑥 ⇒
𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 = 2 &
𝑑𝑑3𝑦𝑦𝑑𝑑𝑥𝑥3 = 0
Implies, 𝑥𝑥3 0 − 2𝑥𝑥 2𝑥𝑥 + 4 𝑥𝑥2 = −4𝑥𝑥2 + 4𝑥𝑥2 = 0
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 28
Superposition Principle
E.g. 𝑥𝑥3 𝑑𝑑3𝑦𝑦
𝑑𝑑𝑥𝑥3− 2𝑥𝑥 𝑑𝑑𝑦𝑦
𝑑𝑑𝑥𝑥+ 4𝑦𝑦 = 0
And 𝑦𝑦1 = 𝑥𝑥2 & 𝑦𝑦2 = 𝑥𝑥2 ln 𝑥𝑥 are both solutions Check: Second solution: 𝑦𝑦1 = 𝑥𝑥2 ln 𝑥𝑥
𝑦𝑦 = 𝑥𝑥2 ln 𝑥𝑥 ⇒𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
= 2𝑥𝑥 ln 𝑥𝑥 + 𝑥𝑥21𝑥𝑥
= 2𝑥𝑥 ln 𝑥𝑥 + 𝑥𝑥
𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
= 2 ln 𝑥𝑥 +𝑥𝑥𝑥𝑥
+ 1 = 2 ln 𝑥𝑥 + 3 ⇒𝑑𝑑3𝑦𝑦𝑑𝑑𝑥𝑥3
=2𝑥𝑥
Implies, 𝑥𝑥3 2𝑥𝑥− 2𝑥𝑥 2𝑥𝑥 ln 𝑥𝑥 + 𝑥𝑥 + 4 𝑥𝑥2 ln 𝑥𝑥 = 2𝑥𝑥2 − 4𝑥𝑥2 ln 𝑥𝑥 − 2𝑥𝑥2 + 4𝑥𝑥2 ln 𝑥𝑥 = 0
By superposition 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟐𝟐𝟐𝟐𝟐𝟐 𝐥𝐥𝐥𝐥 𝟐𝟐 is also a solution
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 29
Linear (Dependence & Independence)
Definition: • A set of functions 𝑓𝑓1 𝑥𝑥 ,𝑓𝑓2 𝑥𝑥 , … ,𝑓𝑓𝑛𝑛(𝑥𝑥) is said to be
linearly dependent on an Interval 𝑰𝑰 if there exists constants 𝒄𝒄𝟏𝟏, 𝒄𝒄𝟐𝟐, … , 𝒄𝒄𝒏𝒏 not all zero such that:
𝒄𝒄𝟏𝟏𝒇𝒇𝟏𝟏 𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒇𝒇𝟐𝟐 𝟐𝟐 + ⋯+ 𝒄𝒄𝒏𝒏𝒇𝒇𝒏𝒏 𝟐𝟐 = 𝟎𝟎 ∀𝟐𝟐 ∈ 𝑰𝑰
• A set of functions is linearly independent on an interval if the only constants for which
𝒄𝒄𝟏𝟏𝒇𝒇𝟏𝟏 𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒇𝒇𝟐𝟐 𝟐𝟐 + ⋯+ 𝒄𝒄𝒏𝒏𝒇𝒇𝒏𝒏 𝟐𝟐 = 𝟎𝟎 ∀𝟐𝟐 ∈ 𝑰𝑰 are 𝒄𝒄𝟏𝟏 = 𝒄𝒄𝟐𝟐 = 𝒄𝒄𝟑𝟑 = ⋯ = 𝒄𝒄𝒏𝒏 = 𝟎𝟎
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 30
Linear (Dependence & Independence)
Definition: • A set of functions 𝑓𝑓1 𝑥𝑥 ,𝑓𝑓2 𝑥𝑥 , … ,𝑓𝑓𝑛𝑛(𝑥𝑥) is said to be
linearly dependent on an Interval 𝑰𝑰 if there exists constants 𝒄𝒄𝟏𝟏, 𝒄𝒄𝟐𝟐, … , 𝒄𝒄𝒏𝒏 not all zero such that:
𝒄𝒄𝟏𝟏𝒇𝒇𝟏𝟏 𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒇𝒇𝟐𝟐 𝟐𝟐 + ⋯+ 𝒄𝒄𝒏𝒏𝒇𝒇𝒏𝒏 𝟐𝟐 = 𝟎𝟎 ∀𝟐𝟐 ∈ 𝑰𝑰
• A set of functions is linearly independent on an interval if the only constants for which
𝒄𝒄𝟏𝟏𝒇𝒇𝟏𝟏 𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒇𝒇𝟐𝟐 𝟐𝟐 + ⋯+ 𝒄𝒄𝒏𝒏𝒇𝒇𝒏𝒏 𝟐𝟐 = 𝟎𝟎 ∀𝟐𝟐 ∈ 𝑰𝑰 are 𝒄𝒄𝟏𝟏 = 𝒄𝒄𝟐𝟐 = 𝒄𝒄𝟑𝟑 = ⋯ = 𝒄𝒄𝒏𝒏 = 𝟎𝟎
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 31
Wronskian
Definition: Suppose each of the functions 𝑓𝑓1 𝑥𝑥 ,𝑓𝑓2 𝑥𝑥 , … ,𝑓𝑓𝑛𝑛(𝑥𝑥) possesses at least 𝑛𝑛 − 1 derivatives Then
𝑊𝑊 𝑓𝑓1,𝑓𝑓2, … ,𝑓𝑓𝑛𝑛 =
𝑓𝑓1 𝑓𝑓2 … ⋯ 𝑓𝑓𝑛𝑛𝑓𝑓1′ 𝑓𝑓2′ … … 𝑓𝑓𝑛𝑛′
⋮𝑓𝑓1
(𝑛𝑛−1) 𝑓𝑓2(𝑛𝑛−1) … 𝑓𝑓𝑛𝑛
(𝑛𝑛−1)
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 32
Wronskian
Criterion for Linearly Independent solutions Let 𝑦𝑦1, 𝑦𝑦2, … ,𝑦𝑦𝑛𝑛 be n-solutions of the homogeneous linear nth order differential equation on an interval 𝐼𝐼. Then the set of solutions is linearly independent on 𝐼𝐼 if and only if
𝑾𝑾 𝒇𝒇𝟏𝟏,𝒇𝒇𝟐𝟐, … ,𝒇𝒇𝒏𝒏 ≠ 𝟎𝟎 ∀𝟐𝟐 ∈ 𝑰𝑰
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 33
Wronskian
E.g. 𝑦𝑦1 = 𝑒𝑒3𝑥𝑥 and 𝑦𝑦2 = 𝑒𝑒−3𝑥𝑥 are both the solutions of the homogeneous linear equation 𝑦𝑦′′ − 9𝑦𝑦 = 0; 𝐼𝐼 = (−∞,∞) Check: 𝑊𝑊 𝑒𝑒3𝑥𝑥, 𝑒𝑒−3𝑥𝑥 = 𝑒𝑒3𝑥𝑥 𝑒𝑒−3𝑥𝑥
3𝑒𝑒3𝑥𝑥 −3𝑒𝑒−3𝑥𝑥
= 𝑒𝑒3𝑥𝑥 −3𝑒𝑒−3𝑥𝑥 − 𝑒𝑒−3𝑥𝑥 3𝑒𝑒3𝑥𝑥 = −3 − 3 = −6 ≠ 0 Thus, 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝟑𝟑𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆−𝟑𝟑𝟐𝟐 is the general solution
𝑾𝑾 𝒇𝒇𝟏𝟏,𝒇𝒇𝟐𝟐, … ,𝒇𝒇𝒏𝒏 ≠ 𝟎𝟎 ∀𝟐𝟐 ∈ 𝑰𝑰
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 34
Wronskian
E.g. 𝑦𝑦′′′ − 6𝑦𝑦′′ + 11𝑦𝑦′ − 6𝑦𝑦 = 0 The functions 𝑦𝑦1 = 𝑒𝑒𝑥𝑥;𝑦𝑦2 = 𝑒𝑒2𝑥𝑥 & 𝑦𝑦3 = 𝑒𝑒3𝑥𝑥 satisfy the D.E. above Check:
𝑊𝑊 𝑒𝑒𝑥𝑥, 𝑒𝑒2𝑥𝑥, 𝑒𝑒3𝑥𝑥 =𝑒𝑒𝑥𝑥 𝑒𝑒2𝑥𝑥 𝑒𝑒3𝑥𝑥𝑒𝑒𝑥𝑥 2𝑒𝑒2𝑥𝑥 3𝑒𝑒3𝑥𝑥𝑒𝑒𝑥𝑥 4𝑒𝑒2𝑥𝑥 9𝑒𝑒3𝑥𝑥
= 𝑒𝑒𝑥𝑥 2𝑒𝑒2𝑥𝑥 3𝑒𝑒3𝑥𝑥
4𝑒𝑒2𝑥𝑥 9𝑒𝑒3𝑥𝑥− 𝑒𝑒2𝑥𝑥 𝑒𝑒𝑥𝑥 3𝑒𝑒3𝑥𝑥
𝑒𝑒𝑥𝑥 9𝑒𝑒3𝑥𝑥+ 𝑒𝑒3𝑥𝑥 𝑒𝑒𝑥𝑥 2𝑒𝑒2𝑥𝑥
𝑒𝑒𝑥𝑥 4𝑒𝑒2𝑥𝑥= 𝑎𝑎𝑓𝑓𝑡𝑡𝑒𝑒𝑓𝑓 𝑠𝑠𝑓𝑓𝑔𝑔𝑠𝑠𝑠𝑠𝑛𝑛𝑔𝑔
= 2𝑒𝑒6𝑥𝑥 ≠ 0 Hence, 𝒆𝒆𝟐𝟐, 𝒆𝒆𝟐𝟐𝟐𝟐,𝒆𝒆𝟑𝟑𝟐𝟐 form a fundamental set & 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟑𝟑𝒆𝒆𝟑𝟑𝟐𝟐 is the general solution
𝑾𝑾 𝒇𝒇𝟏𝟏,𝒇𝒇𝟐𝟐, … ,𝒇𝒇𝒏𝒏 ≠ 𝟎𝟎 ∀𝟐𝟐 ∈ 𝑰𝑰
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 35
Non Homogeneous Equations
𝑎𝑎𝑛𝑛 𝑥𝑥 𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛
+ 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑑𝑑𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1
+ … + 𝑎𝑎1 𝑥𝑥 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔(𝑥𝑥)
,where 𝑔𝑔(𝑥𝑥) ≠ 0 • If 𝑦𝑦𝑝𝑝 (free of arbitrary parameter) satisfies the equation above, 𝑦𝑦𝑝𝑝 is called
particular solution E.g. 𝑦𝑦′′ + 9𝑦𝑦 = 27 Let 𝑦𝑦𝑝𝑝 = 3 ⇒ 𝑦𝑦′′ + 9𝑦𝑦 = 0 + 9 3 = 𝟐𝟐𝟐𝟐 • If 𝑦𝑦1,𝑦𝑦2, … , 𝑦𝑦𝑛𝑛 are solutions of Homogeneous equations and 𝑦𝑦𝑝𝑝 is any particular
solution,
𝑦𝑦 = 𝑐𝑐1𝑦𝑦1 𝑥𝑥 + 𝑐𝑐2𝑦𝑦2 𝑥𝑥 + ⋯+ 𝑐𝑐𝑛𝑛𝑦𝑦𝑛𝑛 𝑥𝑥 + 𝑦𝑦𝑝𝑝 a
General solution Complementary S𝑓𝑓𝑔𝑔𝑜𝑜𝑡𝑡𝑠𝑠𝑓𝑓𝑛𝑛 𝒚𝒚𝒄𝒄
Particular Solution 𝒚𝒚𝒑𝒑
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 36
Non Homogeneous Equations
E.g. 𝑦𝑦′′′ − 6𝑦𝑦′′ + 11𝑦𝑦′ − 6𝑦𝑦 = 3𝑥𝑥 non-homogeneous equation
Let 𝒚𝒚𝒑𝒑 = −𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐− 𝟏𝟏
𝟐𝟐𝟐𝟐. Is it a solution?
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 37
Non Homogeneous Equations
E.g.
𝑦𝑦𝑝𝑝′ = −12
; 𝑦𝑦𝑝𝑝′′ = 0; 𝑦𝑦𝑝𝑝′′′ = 0
⇒ 𝑦𝑦′′′ − 6𝑦𝑦′′ + 11𝑦𝑦′ − 6𝑦𝑦 = 0 − 6 0 + 11 −12
− 6 −1112
−12𝑥𝑥
= −112 +
112 + 3𝑥𝑥 = 𝟑𝟑𝟐𝟐
Verified.
𝑦𝑦′′′ − 6𝑦𝑦′′ + 11𝑦𝑦′ − 6𝑦𝑦 = 3𝑥𝑥 ; 𝑦𝑦𝑝𝑝= −1112− 1
2𝑥𝑥
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 38
Non Homogeneous Equations
Homogeneous Equation: 𝑦𝑦′′′ − 6𝑦𝑦′′ + 11𝑦𝑦′ − 6𝑦𝑦 = 0 Let 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒𝑥𝑥 + 𝑐𝑐2𝑒𝑒2𝑥𝑥 + 𝑐𝑐3𝑒𝑒3𝑥𝑥 be a complimentary solution Hence, the general solution is given by:
𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝑥𝑥 + 𝑐𝑐2𝑒𝑒2𝑥𝑥 + 𝑐𝑐3𝑒𝑒3𝑥𝑥 + (−1112 −
12 𝑥𝑥)
𝑦𝑦′′′ − 6𝑦𝑦′′ + 11𝑦𝑦′ − 6𝑦𝑦 = 3𝑥𝑥 ; 𝑦𝑦𝑝𝑝= −1112− 1
2𝑥𝑥
𝒚𝒚𝒄𝒄 𝒚𝒚𝒑𝒑
Section 3.2 Reduction of Order
9/30/2014 Dr. Eli Saber 39
Reduction of Order
9/30/2014 Dr. Eli Saber 40
Introduction
2nd 𝑓𝑓𝑓𝑓𝑑𝑑𝑒𝑒𝑓𝑓 𝐻𝐻𝑓𝑓𝑚𝑚𝑓𝑓𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑓𝑓𝑜𝑜𝑠𝑠 𝐷𝐷.𝐸𝐸.: 𝑎𝑎2 𝑥𝑥 𝑦𝑦′′ + 𝑎𝑎1 𝑥𝑥 𝑦𝑦′ + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 0 Solution: 𝑦𝑦 = 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2 Where 𝑦𝑦1&𝑦𝑦2 are linearly independent (L.I.) solutions on 𝐼𝐼 Objective: Assume that we know 𝑦𝑦1(𝑥𝑥) solution seek a 2nd solution 𝑦𝑦2(𝑥𝑥) such that 𝑦𝑦1 𝑥𝑥 & 𝑦𝑦2(𝑥𝑥) are independent on 𝐼𝐼
Reduction of Order
9/30/2014 Dr. Eli Saber 41
Introduction
Approach:
• Recall if 𝑦𝑦1 𝑥𝑥 & 𝑦𝑦2(𝑥𝑥) are L.I. => 𝑦𝑦2𝑦𝑦1
is non-constant
𝑦𝑦2𝑦𝑦1
= 𝑜𝑜 𝑥𝑥 𝑓𝑓𝑓𝑓 𝑦𝑦2 𝑥𝑥 = 𝑜𝑜 𝑥𝑥 𝑦𝑦1(𝑥𝑥)
Seek to find 𝑜𝑜(𝑥𝑥) in order to find
𝒚𝒚𝟐𝟐 𝟐𝟐 = 𝒖𝒖 𝟐𝟐 𝒚𝒚𝟏𝟏(𝟐𝟐)
Reduction of Order
9/30/2014 Dr. Eli Saber 42
E.g. Given 𝑑𝑑2𝑦𝑦
𝑑𝑑𝑥𝑥2− 𝑦𝑦 = 0; 𝐼𝐼 = (−∞,∞) and assume that 𝑦𝑦1 = 𝑒𝑒𝑥𝑥 is a solution. Find
a second solution 𝑦𝑦2 Check:
𝑦𝑦 = 𝑒𝑒𝑥𝑥 ⇒𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 = 𝑒𝑒𝑥𝑥 ⇒
𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 = 𝑒𝑒𝑥𝑥
And substituting back in the equation, 𝑑𝑑2𝑦𝑦
𝑑𝑑𝑥𝑥2− 𝑦𝑦 = 𝑒𝑒𝑥𝑥 − 𝑒𝑒𝑥𝑥 = 𝟎𝟎
Reduction of Order
9/30/2014 Dr. Eli Saber 43
Let 𝑦𝑦 𝑥𝑥 = 𝑜𝑜 𝑥𝑥 𝑦𝑦1 𝑥𝑥 = 𝑜𝑜 𝑥𝑥 𝑒𝑒𝑥𝑥
⇒𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
= 𝑜𝑜 𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝑜𝑜′ 𝑥𝑥 𝑒𝑒𝑥𝑥
⇒𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 = 𝑜𝑜 𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝑜𝑜′ 𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝑜𝑜′ 𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝑜𝑜′′ 𝑥𝑥 𝑒𝑒𝑥𝑥
⇒𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 = 𝑜𝑜 𝑥𝑥 𝑒𝑒𝑥𝑥 + 2𝑜𝑜′ 𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝑜𝑜′′ 𝑥𝑥 𝑒𝑒𝑥𝑥
Hence, 𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 − 𝑦𝑦 = 0 ⇒ 𝑜𝑜 𝑥𝑥 𝑒𝑒𝑥𝑥 + 2𝑜𝑜′ 𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝑜𝑜′′ 𝑥𝑥 𝑒𝑒𝑥𝑥 − 𝑜𝑜 𝑥𝑥 𝑒𝑒𝑥𝑥 = 0
Reduction of Order
9/30/2014 Dr. Eli Saber 44
𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
− 𝑦𝑦 = 0 ⇒ 2𝑜𝑜′ 𝑥𝑥 𝑒𝑒𝑥𝑥 + 𝑜𝑜′′ 𝑥𝑥 𝑒𝑒𝑥𝑥 = 0
⇒ 𝑒𝑒𝑥𝑥 𝑜𝑜′′ + 2𝑜𝑜′ = 0 But 𝑒𝑒𝑥𝑥 ≠ 0. ⇒ 𝑜𝑜′′ + 2𝑜𝑜′ = 0 Let 𝑤𝑤 = 𝑜𝑜𝑦 change of variable ⇒ 𝑤𝑤′ + 2𝑤𝑤 = 0 (Linear First Order D.E.) ⇒𝑑𝑑𝑤𝑤𝑑𝑑𝑥𝑥
+ 2𝑤𝑤 = 0
⇒𝑑𝑑𝑤𝑤𝑑𝑑𝑥𝑥
= −2𝑤𝑤
Reduction of Order
9/30/2014 Dr. Eli Saber 45
⇒𝑑𝑑𝑤𝑤𝑑𝑑𝑥𝑥
= −2𝑤𝑤
⇒𝑑𝑑𝑤𝑤𝑤𝑤
= −2 𝑑𝑑𝑥𝑥
⇒ �𝑑𝑑𝑤𝑤𝑤𝑤 = �−2𝑑𝑑𝑥𝑥
⇒ ln 𝑤𝑤 = −2𝑥𝑥 + 𝑐𝑐 ⇒ 𝑤𝑤 = 𝑒𝑒−2𝑥𝑥+𝑐𝑐 = 𝑒𝑒−2𝑥𝑥 𝑒𝑒𝑐𝑐 = 𝑒𝑒−2𝑥𝑥 𝑐𝑐1 ⇒ 𝑤𝑤 = 𝑐𝑐1𝑒𝑒−2𝑥𝑥
Reduction of Order
9/30/2014 Dr. Eli Saber 46
Introduction
𝑤𝑤 = 𝑐𝑐1𝑒𝑒−2𝑥𝑥
But 𝑤𝑤 = 𝑜𝑜′ ⇒ 𝑜𝑜′ = 𝑐𝑐1𝑒𝑒−2𝑥𝑥 ⇒𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥
= 𝑐𝑐1𝑒𝑒−2𝑥𝑥
Hence, ∫𝑑𝑑𝑜𝑜 = ∫ 𝑐𝑐1𝑒𝑒−2𝑥𝑥𝑑𝑑𝑥𝑥
⇒ 𝑜𝑜 = − 12 𝑐𝑐1𝑒𝑒
−2𝑥𝑥 + 𝑐𝑐2
Hence, 𝑦𝑦 = 𝑜𝑜 𝑥𝑥 𝑒𝑒𝑥𝑥 = −𝑐𝑐12𝑒𝑒−2𝑥𝑥 + 𝑐𝑐2 𝑒𝑒𝑥𝑥
⇒ 𝑦𝑦 = −𝑐𝑐12 𝑒𝑒−𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝑥𝑥
Reduction of Order
9/30/2014 Dr. Eli Saber 47
𝑦𝑦 = −𝑐𝑐12 𝑒𝑒−𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝑥𝑥
Let, 𝑐𝑐1 = −2 & 𝑐𝑐2 = 0 ⇒ 𝑦𝑦2 𝑥𝑥 = 𝑒𝑒−𝑥𝑥 Let us check for independence in the two solutions 𝑊𝑊 𝑒𝑒𝑥𝑥, 𝑒𝑒−𝑥𝑥 = 𝑒𝑒𝑥𝑥 𝑒𝑒−𝑥𝑥
𝑒𝑒𝑥𝑥 −𝑒𝑒−𝑥𝑥 = −𝑒𝑒𝑥𝑥𝑒𝑒−𝑥𝑥 − 𝑒𝑒𝑥𝑥𝑒𝑒−𝑥𝑥 = −1 − 1 = −2 ≠ 0
𝑒𝑒𝑥𝑥 & 𝑒𝑒−𝑥𝑥 are independent
General solution: 𝑦𝑦 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥
Wronskian
Reduction of Order
9/30/2014 Dr. Eli Saber 48
Check: 𝑦𝑦 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥
⇒𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
= 𝛼𝛼1𝑒𝑒𝑥𝑥 − 𝛼𝛼2𝑒𝑒−𝑥𝑥
⇒𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥
Hence, 𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 − 𝑦𝑦 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥 − 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥 = 𝟎𝟎
Reduction of Order
9/30/2014 Dr. Eli Saber 49
General case: 𝑎𝑎2 𝑥𝑥 𝑦𝑦′′ + 𝑎𝑎1 𝑥𝑥 𝑦𝑦𝑦 + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 0
𝑑𝑑𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑒𝑒 𝑏𝑏𝑦𝑦 𝑎𝑎2 𝑥𝑥 ⇒ 𝑦𝑦′′ +𝑎𝑎1 𝑥𝑥𝑎𝑎2 𝑥𝑥
𝑦𝑦′ +𝑎𝑎0 𝑥𝑥𝑎𝑎2 𝑥𝑥
𝑦𝑦 = 0
⇒ 𝑦𝑦′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦 = 0 𝑃𝑃 𝑥𝑥 & 𝑄𝑄(𝑥𝑥) are continuous on 𝐼𝐼
Assume 𝑦𝑦1 𝑥𝑥 is a known solution on 𝐼𝐼 and 𝑦𝑦1 𝑥𝑥 ≠ 0∀𝑥𝑥 ∈ 𝐼𝐼
P(x) Q(x)
Reduction of Order
9/30/2014 Dr. Eli Saber 50
Introduction
Let 𝑦𝑦 𝑥𝑥 = 𝑜𝑜 𝑥𝑥 𝑦𝑦1 𝑥𝑥 ⇒ 𝑦𝑦′ 𝑥𝑥 = 𝑜𝑜 𝑥𝑥 𝑦𝑦1′ 𝑥𝑥 + 𝑜𝑜′ 𝑥𝑥 𝑦𝑦1 𝑥𝑥 ⇒ 𝒚𝒚′ = 𝒖𝒖𝒚𝒚𝟏𝟏′ + 𝒖𝒖′𝒚𝒚𝟏𝟏 ⇒ 𝑦𝑦′′ = 𝑜𝑜𝑦𝑦1′′ + 𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′′𝑦𝑦1 ⇒ 𝒚𝒚′′ = 𝒖𝒖𝒚𝒚𝟏𝟏′′ + 𝟐𝟐𝒖𝒖′𝒚𝒚𝟏𝟏′ + 𝒖𝒖′′𝒚𝒚𝟏𝟏 Now, 𝑦𝑦′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦 = 0 Replacing, 𝑜𝑜𝑦𝑦1′′ + 2𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′′𝑦𝑦1 + 𝑃𝑃 𝑥𝑥 𝑜𝑜𝑦𝑦1′ + 𝑜𝑜′𝑦𝑦1 + 𝑄𝑄 𝑥𝑥 𝑜𝑜𝑦𝑦1 = 0
Reduction of Order
9/30/2014 Dr. Eli Saber 51
Rearranging terms, ⇒ 𝑜𝑜 𝑦𝑦1′′ + 𝑃𝑃𝑦𝑦1′ + 𝑄𝑄𝑦𝑦1 + 𝑦𝑦1𝑜𝑜′′ + 2𝑦𝑦1′ + 𝑃𝑃𝑦𝑦1 𝑜𝑜′ = 0 ⇒ 𝑦𝑦1𝑜𝑜′′ + 2𝑦𝑦1′ + 𝑃𝑃𝑦𝑦1 𝑜𝑜′ = 0 Let 𝑤𝑤 = 𝑜𝑜𝑦 change of variables 𝑤𝑤′ = 𝑜𝑜𝑦𝑦 𝑦𝑦1𝑤𝑤′ + 2𝑦𝑦1′ + 𝑃𝑃𝑦𝑦1 𝑤𝑤 = 0 linear and separable
⇒ 𝑦𝑦1𝑤𝑤′ = − 2𝑦𝑦1′ + 𝑃𝑃𝑦𝑦1 𝑤𝑤 𝑓𝑓𝑓𝑓 𝑦𝑦1𝑑𝑑𝑤𝑤𝑑𝑑𝑥𝑥
= − 2𝑑𝑑𝑦𝑦1𝑑𝑑𝑥𝑥
+ 𝑃𝑃 𝑦𝑦1 𝑤𝑤
⇒𝑑𝑑𝑤𝑤𝑤𝑤
= −1𝑦𝑦1
2𝑑𝑑𝑦𝑦1𝑑𝑑𝑥𝑥
+ 𝑃𝑃 𝑦𝑦1 𝑑𝑑𝑥𝑥
=0 since 𝑦𝑦1 is a solution
𝑜𝑜𝑦𝑦1′′ + 2𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′′𝑦𝑦1 + 𝑃𝑃 𝑥𝑥 𝑜𝑜𝑦𝑦1′ + 𝑜𝑜′𝑦𝑦1 + 𝑄𝑄 𝑥𝑥 𝑜𝑜𝑦𝑦1 = 0
Reduction of Order
9/30/2014 Dr. Eli Saber 52
�𝑑𝑑𝑤𝑤𝑤𝑤
= �−1𝑦𝑦1
2𝑑𝑑𝑦𝑦1𝑑𝑑𝑥𝑥
+ 𝑃𝑃 𝑦𝑦1 𝑑𝑑𝑥𝑥
⇒ �𝑑𝑑𝑤𝑤𝑤𝑤
= �−2𝑑𝑑𝑦𝑦1𝑦𝑦1
−�𝑃𝑃𝑑𝑑𝑥𝑥
⇒ ln 𝑤𝑤 = −2 ln 𝑦𝑦1 − �𝑃𝑃𝑑𝑑𝑥𝑥 + 𝑐𝑐
⇒ ln 𝑤𝑤 + 2 ln 𝑦𝑦1 = −�𝑃𝑃𝑑𝑑𝑥𝑥 + 𝑐𝑐 ⇒ ln 𝑤𝑤 + ln 𝑦𝑦12 = −�𝑃𝑃𝑑𝑑𝑥𝑥 + 𝑐𝑐
⇒ ln 𝑤𝑤𝑦𝑦12 = −�𝑃𝑃𝑑𝑑𝑥𝑥 + 𝑐𝑐
⇒ 𝑤𝑤𝑦𝑦12 = 𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥 +𝑐𝑐 = 𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥 𝑒𝑒𝑐𝑐 = 𝑐𝑐1𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥
Reduction of Order
9/30/2014 Dr. Eli Saber 53
𝑤𝑤𝑦𝑦12 = 𝑐𝑐1𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥 ⇒ 𝑤𝑤 = 𝑐𝑐1𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥 /𝑦𝑦12 But,
𝑤𝑤 = 𝑜𝑜′ ⇒ 𝑤𝑤 = 𝑜𝑜′ =𝑑𝑑𝑜𝑜𝑑𝑑𝑥𝑥 =
𝑐𝑐1𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥
𝑦𝑦12
⇒ 𝑑𝑑𝑜𝑜 =𝑐𝑐1𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥
𝑦𝑦12 𝑑𝑑𝑥𝑥 ⇒ �𝑑𝑑𝑜𝑜 = �
𝑐𝑐1𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥
𝑦𝑦12 𝑑𝑑𝑥𝑥
⇒ 𝑜𝑜 = 𝑐𝑐1 �𝑒𝑒− ∫ 𝑃𝑃𝑑𝑑𝑥𝑥
𝑦𝑦12𝑑𝑑𝑥𝑥 + 𝑐𝑐2
Let 𝑐𝑐1 = 1 & 𝑐𝑐2 = 0 & note: 𝑦𝑦2 𝑥𝑥 = 𝑜𝑜 𝑥𝑥 𝑦𝑦1(𝑥𝑥)
⇒ 𝒚𝒚𝟐𝟐 𝟐𝟐 = 𝒚𝒚𝟏𝟏 𝟐𝟐 �𝒆𝒆− ∫ 𝑷𝑷𝒅𝒅𝟐𝟐
𝒚𝒚𝟏𝟏𝟐𝟐(𝟐𝟐)𝒅𝒅𝟐𝟐 ; 𝑷𝑷 𝟐𝟐 =
𝒂𝒂𝟏𝟏 𝟐𝟐𝒂𝒂𝟐𝟐 𝟐𝟐
Reduction of Order
9/30/2014 Dr. Eli Saber 54
E.g. 𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 4𝑦𝑦 = 0; 𝐼𝐼 ≡ 0,∞ Let 𝑦𝑦1 𝑥𝑥 = 𝑥𝑥2 be a solution. Find a 2nd solution 𝑦𝑦2(𝑥𝑥) and the general solution 𝑦𝑦(𝑥𝑥) Solution: 𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 4𝑦𝑦 = 0
⇒ 𝑦𝑦′′ −3𝑥𝑥𝑥𝑥2 𝑦𝑦′ +
4𝑥𝑥2 𝑦𝑦 = 0
⇒ 𝑦𝑦′′ + − 3𝑥𝑥 𝑦𝑦′ +
4𝑥𝑥2 𝑦𝑦 = 0
P(x) Q(x)
Reduction of Order
9/30/2014 Dr. Eli Saber 55
According to our derivation, 𝑦𝑦2 𝑥𝑥 = 𝑦𝑦1 𝑥𝑥 ∫ 𝑒𝑒− ∫ 𝑃𝑃𝑃𝑃𝑃𝑃
𝑦𝑦12(𝑥𝑥)𝑑𝑑𝑥𝑥
= 𝑥𝑥2 �𝑒𝑒− ∫ −3𝑥𝑥 𝑑𝑑𝑥𝑥
𝑥𝑥2 2 𝑑𝑑𝑥𝑥
= 𝑥𝑥2 �𝑒𝑒∫
3𝑥𝑥 𝑑𝑑𝑥𝑥
𝑥𝑥4𝑑𝑑𝑥𝑥
= 𝑥𝑥2 �𝑒𝑒3 ln 𝑥𝑥
𝑥𝑥4𝑑𝑑𝑥𝑥 = 𝑥𝑥2 �
𝑒𝑒ln 𝑥𝑥3
𝑥𝑥4𝑑𝑑𝑥𝑥 = 𝑥𝑥2 �
𝑥𝑥3
𝑥𝑥4𝑑𝑑𝑥𝑥
= 𝑥𝑥2 �1𝑥𝑥𝑑𝑑𝑥𝑥 = 𝑥𝑥2 ln 𝑥𝑥
⇒ 𝒚𝒚𝟐𝟐 𝟐𝟐 = 𝟐𝟐𝟐𝟐 𝐥𝐥𝐥𝐥 𝟐𝟐 General solution: 𝒚𝒚 𝟐𝟐 = 𝒄𝒄𝟏𝟏𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟐𝟐 𝟐𝟐𝟐𝟐𝐥𝐥𝐥𝐥 𝟐𝟐
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 4𝑦𝑦 = 0
Reduction of Order
9/30/2014 Dr. Eli Saber 56
Check: 𝑦𝑦 = 𝑐𝑐1𝑥𝑥2 + 𝑐𝑐2 𝑥𝑥2 ln 𝑥𝑥
⇒ 𝑦𝑦′ = 2 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2 2𝑥𝑥 ln 𝑥𝑥 +𝑥𝑥2
𝑥𝑥
⇒ 𝑦𝑦′ = 2𝑐𝑐1𝑥𝑥 + 2𝑐𝑐2𝑥𝑥 ln 𝑥𝑥 + 𝑐𝑐2𝑥𝑥
𝑦𝑦′′ = 2𝑐𝑐1 + 2𝑐𝑐2 ln 𝑥𝑥 +𝑥𝑥𝑥𝑥 + 𝑐𝑐2
⇒ 𝑦𝑦′′ = 2𝑐𝑐1 + 2𝑐𝑐2 ln 𝑥𝑥 + 2𝑐𝑐2 + 𝑐𝑐2 = 2𝑐𝑐1 + 3𝑐𝑐2 + 2𝑐𝑐2 ln 𝑥𝑥
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 4𝑦𝑦 = 0
Reduction of Order
9/30/2014 Dr. Eli Saber 57
We know, 𝑦𝑦 = 𝑐𝑐1𝑥𝑥2 + 𝑐𝑐2 𝑥𝑥2 ln 𝑥𝑥 𝑦𝑦′ = 2𝑐𝑐1𝑥𝑥 + 2𝑐𝑐2𝑥𝑥 ln 𝑥𝑥 + 𝑐𝑐2𝑥𝑥 & 𝑦𝑦′′ = 2𝑐𝑐1 + 2𝑐𝑐2 ln 𝑥𝑥 + 2𝑐𝑐2 + 𝑐𝑐2 = 2𝑐𝑐1 + 3𝑐𝑐2 + 2𝑐𝑐2 ln 𝑥𝑥 Replace in D.E.: 𝑥𝑥2 2𝑐𝑐1 + 3𝑐𝑐2 + 2𝑐𝑐2 ln 𝑥𝑥 − 3x 2𝑐𝑐1𝑥𝑥 + 2𝑐𝑐2𝑥𝑥 ln 𝑥𝑥 + 𝑐𝑐2𝑥𝑥 + 4 𝑐𝑐1𝑥𝑥2 + 𝑐𝑐2𝑥𝑥2 ln 𝑥𝑥 = 2𝑐𝑐1𝑥𝑥2 + 3𝑐𝑐2𝑥𝑥2 + 2𝑐𝑐2 ln 𝑥𝑥 𝑥𝑥2 − 6𝑐𝑐1𝑥𝑥2 − 6𝑐𝑐2𝑥𝑥2 ln 𝑥𝑥 − 3𝑐𝑐2𝑥𝑥2 + 4𝑐𝑐1𝑥𝑥2
+ 4𝑐𝑐2𝑥𝑥2 ln 𝑥𝑥 = 𝟎𝟎
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 4𝑦𝑦 = 0
Section 3.3 Homogeneous Linear Eq. with
Constant Coefficients
9/30/2014 Dr. Eli Saber 58
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 59
Introduction
𝒂𝒂𝒏𝒏𝒚𝒚 𝒏𝒏 + 𝒂𝒂𝒏𝒏−𝟏𝟏𝒚𝒚(𝒏𝒏−𝟏𝟏) + ⋯+ 𝒂𝒂𝟏𝟏𝒚𝒚′ + 𝒂𝒂𝟎𝟎𝒚𝒚 = 𝟎𝟎 • 𝑎𝑎𝑖𝑖; 𝑠𝑠 = 0,1, … ,𝑛𝑛 are real constant coefficients and 𝑎𝑎𝑛𝑛 ≠ 0 Objective: To find a solution to the above homogeneous solution
𝑛𝑛𝑡𝑡𝑡 order Linear Constant Coefficients Differential Equation
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 60
Auxiliary Equation
Consider the special Case ( 2nd order LCCDE) given as: 𝑎𝑎𝑦𝑦′′ + 𝑏𝑏𝑦𝑦′ + 𝑐𝑐𝑦𝑦 = 0 Try a solution of the form 𝑦𝑦 = 𝑒𝑒𝑚𝑚𝑥𝑥 ⇒ 𝑦𝑦′ = 𝑚𝑚𝑒𝑒𝑚𝑚𝑥𝑥 ⇒ 𝑦𝑦′′ = 𝑚𝑚2𝑒𝑒𝑚𝑚𝑥𝑥 Substituting back in the given D.E., 𝑎𝑎 𝑚𝑚2𝑒𝑒𝑚𝑚𝑥𝑥 + 𝑏𝑏 𝑚𝑚𝑒𝑒𝑚𝑚𝑥𝑥 + 𝑐𝑐 𝑒𝑒𝑚𝑚𝑥𝑥 = 0 ⇒ 𝑎𝑎𝑚𝑚2 + 𝑏𝑏𝑚𝑚 + 𝑐𝑐 𝑒𝑒𝑚𝑚𝑥𝑥 = 0 Now, 𝒆𝒆𝒎𝒎𝟐𝟐 ≠ 𝟎𝟎 ∀𝟐𝟐𝒓𝒓𝒆𝒆𝒂𝒂𝒓𝒓 ⇒ 𝒂𝒂𝒎𝒎𝟐𝟐 + 𝒃𝒃𝒎𝒎 + 𝒄𝒄 = 𝟎𝟎
Auxiliary Eqn. of the LCCDE
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 61
Introduction
𝒂𝒂𝒎𝒎𝟐𝟐 + 𝒃𝒃𝒎𝒎 + 𝒄𝒄 = 𝟎𝟎 Auxiliary Eqn. of the LCCDE The only way that 𝑦𝑦 = 𝑒𝑒𝑚𝑚𝑥𝑥 can satisfy the D.E. is if 𝑎𝑎𝑚𝑚2 + 𝑏𝑏𝑚𝑚 + 𝑐𝑐 = 0 Hence, choose 𝒎𝒎 as the root of the equation to solve the problem
⇒ 𝑚𝑚1,2 =−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐
2𝑎𝑎
The 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 leads to 3 cases:
1) 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 > 0
2) 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = 0
3) 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 < 0
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 62
Introduction
Case 1: 𝒃𝒃𝟐𝟐−𝟒𝟒𝒂𝒂𝒄𝒄 > 𝟎𝟎 Here, 𝑚𝑚1& 𝑚𝑚2 are real and distinct 2 solutions: 𝑦𝑦1 = 𝑒𝑒𝑚𝑚1𝑥𝑥 & 𝑦𝑦2 = 𝑒𝑒𝑚𝑚2𝑥𝑥
𝑦𝑦1&𝑦𝑦2 are linearly independent
𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝑚𝑚1𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝑚𝑚2𝑥𝑥 is the general solution
𝑚𝑚1,2 =−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐
2𝑎𝑎
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 63
Introduction
Case 2: 𝒃𝒃𝟐𝟐−𝟒𝟒𝒂𝒂𝒄𝒄 = 𝟎𝟎
𝑚𝑚1 = 𝑚𝑚2 = −𝑏𝑏2𝑎𝑎
⇒ 𝑦𝑦1 = 𝑒𝑒𝑚𝑚1𝑥𝑥 & 𝑦𝑦2 = 𝑥𝑥𝑒𝑒𝑚𝑚1𝑥𝑥
Digression:
𝑎𝑎𝑦𝑦′′ + 𝑏𝑏𝑦𝑦′ + 𝑐𝑐𝑦𝑦 = 0 ⇒ 𝑦𝑦′′ +𝑏𝑏𝑎𝑎 𝑦𝑦′ +
𝑐𝑐𝑎𝑎 𝑦𝑦 = 0
⇒ 𝑦𝑦2 𝑥𝑥 = 𝑦𝑦1 𝑥𝑥 �𝑒𝑒− ∫ 𝑃𝑃 𝑥𝑥 𝑑𝑑𝑥𝑥
𝑦𝑦1 𝑥𝑥 2 𝑑𝑑𝑥𝑥 = 𝑒𝑒𝑚𝑚1𝑥𝑥 �𝑒𝑒− ∫𝑏𝑏𝑎𝑎 𝑑𝑑𝑥𝑥
𝑒𝑒2𝑚𝑚1𝑥𝑥𝑑𝑑𝑥𝑥 = 𝑒𝑒𝑚𝑚1𝑥𝑥 �
𝑒𝑒∫ 2𝑚𝑚1 𝑑𝑑𝑥𝑥
𝑒𝑒2𝑚𝑚1𝑥𝑥𝑑𝑑𝑥𝑥
𝑚𝑚1,2 =−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐
2𝑎𝑎
P(x) Q(x)
(Note: 𝑚𝑚1 = − 𝑏𝑏2𝑎𝑎⇒ − 𝑏𝑏
𝑎𝑎= 2𝑚𝑚1)
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 64
Introduction
𝑦𝑦2 𝑥𝑥 = 𝑒𝑒𝑚𝑚1𝑥𝑥 �𝑒𝑒∫ 2𝑚𝑚1 𝑑𝑑𝑥𝑥
𝑒𝑒2𝑚𝑚1𝑥𝑥𝑑𝑑𝑥𝑥 = 𝑒𝑒𝑚𝑚1𝑥𝑥 �
𝑒𝑒2𝑚𝑚1𝑥𝑥
𝑒𝑒2𝑚𝑚1𝑥𝑥𝑑𝑑𝑥𝑥
= 𝑒𝑒𝑚𝑚1𝑥𝑥 �𝑑𝑑𝑥𝑥 = 𝑥𝑥𝑒𝑒𝑚𝑚1𝑥𝑥
𝑦𝑦2 𝑥𝑥 = 𝑥𝑥𝑒𝑒𝑚𝑚1𝑥𝑥 General solution: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝑚𝑚1𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒𝑚𝑚1𝑥𝑥
𝑚𝑚1,2 =−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐
2𝑎𝑎
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 65
Introduction
Case 3: 𝒃𝒃𝟐𝟐−𝟒𝟒𝒂𝒂𝒄𝒄 < 𝟎𝟎 𝑚𝑚1 & 𝑚𝑚2 are complex conjugate numbers
𝑚𝑚1 = 𝛼𝛼 + 𝑗𝑗𝛽𝛽 & 𝑚𝑚2 = 𝛼𝛼 − 𝑗𝑗𝛽𝛽
• 𝛼𝛼,𝛽𝛽 > 0 and are real • 𝑗𝑗2 = −1
General solution: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝑚𝑚1𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝑚𝑚2𝑥𝑥
𝑦𝑦 = 𝑐𝑐1𝑒𝑒 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥
𝑚𝑚1,2 =−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐
2𝑎𝑎
Since 𝑦𝑦 = 𝑐𝑐1𝑒𝑒 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥 is a solution ∀𝑐𝑐1 &∀𝑐𝑐2
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 66
Introduction
𝑦𝑦1 = 𝑐𝑐1𝑒𝑒 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥 = 𝑒𝑒 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥 + 𝑒𝑒 𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥 = 𝑒𝑒𝛼𝛼𝑥𝑥 𝑒𝑒𝑗𝑗𝑗𝑗𝑥𝑥 + 𝑒𝑒−𝑗𝑗𝑗𝑗𝑥𝑥 = 𝑒𝑒𝛼𝛼𝑥𝑥2 cos𝛽𝛽𝑥𝑥 𝑦𝑦1 = 2𝑒𝑒𝛼𝛼𝑥𝑥 cos𝛽𝛽𝑥𝑥
𝑦𝑦2 = 𝑐𝑐1𝑒𝑒 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥 = 𝑒𝑒 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥 − 𝑒𝑒 𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥 = 𝑒𝑒𝛼𝛼𝑥𝑥 𝑒𝑒𝑗𝑗𝑗𝑗𝑥𝑥 − 𝑒𝑒−𝑗𝑗𝑗𝑗𝑥𝑥 = 𝑒𝑒𝛼𝛼𝑥𝑥2𝑗𝑗 sin𝛽𝛽𝑥𝑥 𝑦𝑦2 = 2𝑗𝑗𝑒𝑒𝛼𝛼𝑥𝑥 sin𝛽𝛽𝑥𝑥
Choose 𝑐𝑐1 = 𝑐𝑐2 = 1 Choose 𝑐𝑐1 = 1 & 𝑐𝑐2 = −1
General solution: 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝜶𝜶𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜𝜷𝜷𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆𝜶𝜶𝟐𝟐 𝐜𝐜𝐬𝐬𝐥𝐥𝜷𝜷𝟐𝟐
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 67
Alternate Derivation: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥 = 𝑐𝑐1𝑒𝑒𝛼𝛼𝑥𝑥𝑒𝑒𝑗𝑗𝑗𝑗𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝛼𝛼𝑥𝑥𝑒𝑒−𝑗𝑗𝑗𝑗𝑥𝑥 = 𝑐𝑐1𝑒𝑒𝛼𝛼𝑥𝑥 cos𝛽𝛽𝑥𝑥 + 𝑗𝑗 sin𝛽𝛽𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝛼𝛼𝑥𝑥 cos𝛽𝛽𝑥𝑥 − 𝑗𝑗 sin𝛽𝛽𝑥𝑥 = 𝑐𝑐1𝑒𝑒𝛼𝛼𝑥𝑥 cos𝛽𝛽𝑥𝑥 + 𝑗𝑗𝑐𝑐1𝑒𝑒𝛼𝛼𝑥𝑥 sin𝛽𝛽𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝛼𝛼𝑥𝑥 cos𝛽𝛽𝑥𝑥 − 𝑗𝑗𝑐𝑐2𝑒𝑒𝛼𝛼𝑥𝑥 sin𝛽𝛽𝑥𝑥 = 𝑒𝑒𝛼𝛼𝑥𝑥 𝑐𝑐1 + 𝑐𝑐2 cos𝛽𝛽𝑥𝑥 + 𝑒𝑒𝛼𝛼𝑥𝑥 𝑗𝑗𝑐𝑐1 − 𝑗𝑗𝑐𝑐2 sin𝛽𝛽𝑥𝑥 Hence, 𝒚𝒚 = ∝𝟏𝟏 𝒆𝒆𝜶𝜶𝟐𝟐 𝒄𝒄𝒄𝒄𝒄𝒄𝜷𝜷𝟐𝟐+∝𝟐𝟐 𝒆𝒆𝜶𝜶𝟐𝟐 𝒄𝒄𝒔𝒔𝒏𝒏𝜷𝜷𝟐𝟐
∝1 ∝2
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 68
Example: a) 2𝑦𝑦′′ − 5𝑦𝑦′ − 3𝑦𝑦 = 0 Now, 2𝑚𝑚2 − 5𝑚𝑚 − 3 = 0 ⇒ 2𝑚𝑚 + 1 𝑚𝑚 − 3 = 0
⇒ 𝑚𝑚1 = − 12 ;𝑚𝑚2 = 3
General solution: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒− 12𝑥𝑥 + 𝑐𝑐2𝑒𝑒3𝑥𝑥
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 69
b) 𝑦𝑦′′ − 10𝑦𝑦′ + 25𝑦𝑦 = 0 𝑚𝑚2 − 10𝑚𝑚 + 25 = 0 ⇒ 𝑚𝑚 − 5 2 = 0 ⇒ 𝑚𝑚1 = 𝑚𝑚2 = 5 General solution: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒5𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒5𝑥𝑥
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 70
c) 𝑦𝑦′′ + 4𝑦𝑦′ + 7𝑦𝑦 = 0 ⇒ 𝑚𝑚2 + 4𝑚𝑚 + 7 = 0
⇒ 𝑚𝑚 =−4 ± 4 2 − 4(1)(7)
2(1)=−4 ± 16− 28
2
⇒ 𝑚𝑚 = −4 ± −12
2=−4 ± 12 −1
2=−4 ± 𝑗𝑗 12
2
⇒ 𝑚𝑚 =−4 ± 𝑗𝑗 2 3
2= −2 ± 𝑗𝑗 3
General solution: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒 −2+𝑗𝑗 3 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 −2−𝑗𝑗 3 𝑥𝑥 or 𝑦𝑦 = 𝑒𝑒−2𝑥𝑥 𝑐𝑐1 cos 3𝑥𝑥 + 𝑐𝑐2 sin 3𝑥𝑥
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 71
𝑦𝑦′′ + 𝐾𝐾2𝑦𝑦 = 0 & 𝑦𝑦′′ − 𝐾𝐾2𝑦𝑦 = 0 𝐾𝐾:real Where do we see these equations??
D.E. Free of Undamped Motion: 𝑑𝑑2𝑥𝑥𝑑𝑑𝑡𝑡2
+ 𝜔𝜔2𝑥𝑥 = 0 With the solution: 𝟐𝟐 = 𝒄𝒄𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜𝒘𝒘𝒘𝒘 + 𝒄𝒄𝟐𝟐 𝐜𝐜𝐬𝐬𝐥𝐥𝒘𝒘𝒘𝒘
Two important Equations
HOW?
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed. Ref. D. Zill & W. Wright, Advanced
Engineering Mathematics. 5th Ed.
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 72
𝑦𝑦′′ + 𝐾𝐾2𝑦𝑦 = 0 & 𝑦𝑦′′ − 𝐾𝐾2𝑦𝑦 = 0 𝐾𝐾:real 𝑚𝑚2 + 𝐾𝐾2 = 0 ⇒ 𝑚𝑚2 = −𝐾𝐾2 = 𝐾𝐾2𝑗𝑗2 ⇒ 𝑚𝑚 = ±𝐾𝐾𝑗𝑗 Which results in: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝐾𝐾𝑗𝑗𝑥𝑥 + 𝑐𝑐2𝑒𝑒−𝐾𝐾𝑗𝑗𝑥𝑥 or 𝑦𝑦 = 𝑐𝑐1 cos𝐾𝐾𝑥𝑥 + 𝑐𝑐2 sin𝐾𝐾𝑥𝑥
Two important Equations
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 73
𝑦𝑦′′ + 𝐾𝐾2𝑦𝑦 = 0 & 𝑦𝑦′′ − 𝐾𝐾2𝑦𝑦 = 0 𝐾𝐾:real 𝑚𝑚2 − 𝐾𝐾2 = 0 ⇒ 𝑚𝑚2 = 𝐾𝐾2 ⇒ 𝑚𝑚 = ±𝐾𝐾 Which results in: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝐾𝐾𝑥𝑥 + 𝑐𝑐2𝑒𝑒−𝐾𝐾𝑥𝑥
Two important Equations
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 74
Note: 𝑦𝑦′′ − 𝐾𝐾2𝑦𝑦 = 0 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝑲𝑲𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆−𝑲𝑲𝟐𝟐
• If 𝑐𝑐1 = 𝑐𝑐2 = 12⇒ 𝑦𝑦 = 1
2 𝑒𝑒𝐾𝐾𝑥𝑥 + 1
2 𝑒𝑒−𝐾𝐾𝑥𝑥 = cosh𝐾𝐾𝑥𝑥
• If 𝑐𝑐1 = 12
& 𝑐𝑐2 = −12⇒ 𝑦𝑦 = 1
2 𝑒𝑒𝐾𝐾𝑥𝑥 − 1
2 𝑒𝑒−𝐾𝐾𝑥𝑥 = sinh𝐾𝐾𝑥𝑥
• Since cosh𝐾𝐾𝑥𝑥 & sinh𝐾𝐾𝑥𝑥 are linearly independent
– Alternate solution of 𝑦𝑦′′ − 𝐾𝐾2𝑦𝑦 = 0 is 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝑲𝑲𝟐𝟐 + 𝒄𝒄𝟐𝟐 𝐜𝐜𝐬𝐬𝐥𝐥𝐜𝐜𝑲𝑲𝟐𝟐
Two important Equations
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 75
𝑎𝑎𝑛𝑛 𝑥𝑥 𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛
+ 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑑𝑑𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1
+ … + 𝑎𝑎1 𝑥𝑥 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 0
,where 𝑎𝑎𝑖𝑖 , 𝑠𝑠 = 0,1, … ,𝑛𝑛 are real constants Auxiliary Equation: 𝑎𝑎𝑛𝑛𝑚𝑚𝑛𝑛 + 𝑎𝑎𝑛𝑛−1𝑚𝑚𝑛𝑛−1 + ⋯+ 𝑎𝑎2𝑚𝑚2 + 𝑎𝑎1𝑚𝑚 + 𝑎𝑎0 𝑚𝑚0 = 0 Case 1: If all roots are distinct – general solution is given by:
𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝑚𝑚1𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝑚𝑚2𝑥𝑥 + ⋯+ 𝑐𝑐𝑛𝑛𝑒𝑒𝑚𝑚𝑛𝑛𝑥𝑥 (similar to a 2nd order D.E.)
Higher Order Equations
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 76
Case 2: For multiple roots, if 𝑚𝑚1 is a root with multiplicity 𝐾𝐾 i.e. 𝐾𝐾 roots equal to 𝑚𝑚1
Then the general solution will have terms: 𝑒𝑒𝑚𝑚1𝑥𝑥, 𝑥𝑥𝑒𝑒𝑚𝑚1𝑥𝑥, 𝑥𝑥2𝑒𝑒𝑚𝑚1𝑥𝑥,…, 𝑥𝑥𝑘𝑘−1𝑒𝑒𝑚𝑚1𝑥𝑥 Case 3: Complex roots appear in conjugate pairs when the coefficients of the D.E. are real
Higher Order Equations
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 77
E.g. 𝑦𝑦′′′ + 3𝑦𝑦′′ − 4𝑦𝑦 = 0 Auxiliary equation: 𝑚𝑚3 + 3𝑚𝑚2 − 4 = 0 By inspection, 𝑚𝑚1 = 1 is a root since 1 3 + 3 1 2 − 4 = 1 + 3 − 4 = 4 − 4 = 𝟎𝟎 Dividing the Auxiliary equation 𝑚𝑚3 + 3𝑚𝑚2 − 4 = 0 by 𝑚𝑚 − 1 , we get 𝑚𝑚2 + 4𝑚𝑚 + 4 ⇒ 𝑚𝑚− 1 𝑚𝑚2 + 4𝑚𝑚 + 4 = 𝑚𝑚3 + 3𝑚𝑚2 − 4 ⇒ 𝑚𝑚− 1 𝑚𝑚2 + 4𝑚𝑚 + 4 = 0 ⇒ 𝑚𝑚− 1 𝑚𝑚 + 2 2 = 0 Roots: 𝑚𝑚1 = 1,𝑚𝑚2 = 𝑚𝑚3 = −2 General solution: 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆−𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟑𝟑𝟐𝟐𝒆𝒆−𝟐𝟐𝟐𝟐
Higher Order Equations
Section 3.4 Undetermined Coefficients
9/30/2014 Dr. Eli Saber 78
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 79
By: 1. Finding a complementary solution 𝑦𝑦𝑐𝑐 for the homogeneous equation. 2. Finding a particular solution 𝑦𝑦𝑝𝑝.
⇒ 𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑓𝑓𝑎𝑎𝑔𝑔 𝑠𝑠𝑓𝑓𝑔𝑔𝑜𝑜𝑡𝑡𝑠𝑠𝑓𝑓𝑛𝑛: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
Solve a non-homogeneous Linear Differential Equation: 𝑎𝑎𝑛𝑛𝑦𝑦 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1𝑦𝑦 𝑛𝑛−1 + ⋯+ 𝑎𝑎1𝑦𝑦1 + 𝑎𝑎0𝑦𝑦0 = 𝑔𝑔 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 80
Method of undetermined coefficient 𝒖𝒖𝒄𝒄 “Educated guess about the form of 𝒚𝒚𝒑𝒑” Method is limited to non-homogeneous linear D.E. such that: 1. The coefficient 𝑎𝑎𝑖𝑖 , 𝑠𝑠 = 0,1,2, … ,𝑛𝑛 are constant. 2. 𝑔𝑔 𝑥𝑥 is a constant, polynomial function, exponential function, sin or cos or
finite sums and products of these functions. E.g.: 𝑔𝑔 𝑥𝑥 = 10;𝑔𝑔 𝑥𝑥 = 𝑥𝑥2 − 5𝑥𝑥, … …
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 81
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 82
E.g. 1: 𝑦𝑦′′ + 4𝑦𝑦′ − 2𝑦𝑦 = 2𝑥𝑥2 − 3𝑥𝑥 + 6 Step 1: Solve the associated Homogeneous equation. 𝑦𝑦′′ + 4𝑦𝑦′ − 2𝑦𝑦 = 0
𝑚𝑚2 + 4𝑚𝑚 − 2 = 0 ⇒ 𝑚𝑚 =−4 ± 16 − 4 1 −2
2
⇒ 𝑚𝑚 =−4 ± 24
2 =−4 ± 2 6
2
⇒ 𝑚𝑚 = −2 ± 6 ⇒ 𝑚𝑚1 = −2 − 6 𝑎𝑎𝑛𝑛𝑑𝑑 𝑚𝑚2 = −2 + 6
𝑔𝑔 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 83
𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒𝑚𝑚1𝑥𝑥 + 𝑐𝑐2𝑒𝑒𝑚𝑚2𝑥𝑥
⇒ 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒 −2− 6 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 −2+ 6 𝑥𝑥 Step 2: Note 𝑔𝑔(𝑥𝑥) is a quadratic ⇒ assume a particular solution of quadratic form. (See Table 3.4.1) ⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥2 + 𝐵𝐵𝑥𝑥 + 𝐶𝐶 ⇒ 𝑦𝑦𝑝𝑝′ = 2𝐴𝐴𝑥𝑥 + 𝐵𝐵 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦𝑝𝑝′′ = 2𝐴𝐴 Substitute into D.E. 𝑦𝑦𝑝𝑝′′ + 4𝑦𝑦𝑝𝑝′ − 2𝑦𝑦𝑝𝑝 = 3𝑥𝑥2 − 3𝑥𝑥 + 6 ⇒ 2𝐴𝐴 + 4 2𝐴𝐴𝑥𝑥 + 𝐵𝐵 − 2 𝐴𝐴𝑥𝑥2 + 𝐵𝐵𝑥𝑥 + 𝐶𝐶 = 2𝑥𝑥2 − 3𝑥𝑥 + 6
𝑚𝑚1 = −2 − 6 𝑎𝑎𝑛𝑛𝑑𝑑 𝑚𝑚2 = −2 + 6
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 84
⇒ 2𝐴𝐴 + 8𝐴𝐴𝑥𝑥 + 4𝐵𝐵 − 2𝐴𝐴𝑥𝑥2 − 2𝐵𝐵𝑥𝑥 − 2𝐶𝐶 = 2𝑥𝑥2 − 3𝑥𝑥 + 6 ⇒ −𝟐𝟐𝑨𝑨𝑥𝑥2 + 𝟖𝟖𝑨𝑨 − 𝟐𝟐𝑩𝑩 𝑥𝑥 + 𝟐𝟐𝑨𝑨 + 𝟒𝟒𝑩𝑩 − 𝟐𝟐𝑪𝑪 = 2𝑥𝑥2 − 3𝑥𝑥 + 6 • −2𝐴𝐴 = 2 ⇒ 𝐴𝐴 = −1
• 8𝐴𝐴 − 2𝐵𝐵 = −3 ⇒ 2𝐵𝐵 = 8𝐴𝐴 + 3 = 8 −1 + 3 = −5
⇒ 2𝐵𝐵 = −5 ⇒ 𝐵𝐵 =−52
• 2𝐴𝐴 + 4𝐵𝐵 − 2𝐶𝐶 = 6 ⇒ 2𝐶𝐶 = 2𝐴𝐴 + 4𝐵𝐵 − 6 = 2 −1 + 4 −52
− 6
2𝐶𝐶 = −2 − 10 − 6 ⇒ 2𝐶𝐶 = −18 ⇒ 𝐶𝐶 = −9
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 85
⇒ 𝑦𝑦𝑝𝑝 = −𝑥𝑥2 −52𝑥𝑥 − 9
⇒ 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒 −2− 6 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 −2+ 6 𝑥𝑥 ⇒ 𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑓𝑓𝑎𝑎𝑔𝑔 𝑠𝑠𝑓𝑓𝑔𝑔𝑜𝑜𝑡𝑡𝑠𝑠𝑓𝑓𝑛𝑛: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆− 𝟐𝟐+ 𝟔𝟔 𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆 −𝟐𝟐+ 𝟔𝟔 𝟐𝟐 − 𝟐𝟐𝟐𝟐 −𝟓𝟓𝟐𝟐𝟐𝟐 − 𝟗𝟗
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 86
E.g. 2: 𝑦𝑦′′ − 𝑦𝑦′ + 𝑦𝑦 = 2 sin 3𝑥𝑥 Step 1: Find 𝑦𝑦𝑐𝑐 𝑓𝑓𝑓𝑓𝑓𝑓 𝑦𝑦′′ − 𝑦𝑦′ + 𝑦𝑦 = 0
𝑚𝑚2 −𝑚𝑚 + 1 = 0 ⇒ 𝑚𝑚 =1 ± 1 − 4 1 1
2
⇒ 𝑚𝑚 =1 ± 3𝑗𝑗2
2 ⇒ 𝑚𝑚 =12 ± 𝑗𝑗
32
⇒ 𝑚𝑚1 =12 + 𝑗𝑗
32 𝑎𝑎𝑛𝑛𝑑𝑑 𝑚𝑚2 =
12 − 𝑗𝑗
32
⇒ 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒12+𝑗𝑗
32 𝑥𝑥 + 𝑐𝑐2𝑒𝑒
12 − 𝑗𝑗 3
2 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 87
Step 2: 𝐹𝐹𝑠𝑠𝑛𝑛𝑑𝑑 𝑦𝑦𝑝𝑝. 𝐴𝐴𝑠𝑠𝑠𝑠𝑜𝑜𝑚𝑚𝑒𝑒 𝑦𝑦𝑝𝑝 = 𝐴𝐴 cos3𝑥𝑥 + 𝐵𝐵 sin 3𝑥𝑥 (see Table 3.4.1) ⇒ 𝑦𝑦𝑝𝑝′ = −3𝐴𝐴 sin3𝑥𝑥 + 3𝐵𝐵 cos3𝑥𝑥 𝑦𝑦𝑝𝑝′′ = −9𝐴𝐴 cos3𝑥𝑥 − 9𝐵𝐵 sin 3𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝′′ − 𝑦𝑦𝑝𝑝′ + 𝑦𝑦 = 2 sin 3𝑥𝑥 ⇒ −9𝐴𝐴 cos3𝑥𝑥 − 9𝐵𝐵 sin 3𝑥𝑥 − −3𝐴𝐴 sin 3𝑥𝑥 + 3𝐵𝐵 cos3𝑥𝑥 + 𝐴𝐴 cos3𝑥𝑥 + 𝐵𝐵 sin 3𝑥𝑥
= 2 sin 3𝑥𝑥 ⇒ −9𝐴𝐴 cos3𝑥𝑥 − 9𝐵𝐵 sin3𝑥𝑥 + 3𝐴𝐴 sin 3𝑥𝑥 − 3𝐵𝐵 cos3𝑥𝑥 + 𝐴𝐴 cos3𝑥𝑥 + 𝐵𝐵 sin3𝑥𝑥 = 2 sin 𝑥𝑥 ⇒ −9𝐴𝐴 cos3𝑥𝑥 − 3𝐵𝐵 cos3𝑥𝑥 + 𝐴𝐴 cos3𝑥𝑥 − 9𝐵𝐵 sin 3𝑥𝑥 + 3𝐴𝐴 sin 3𝑥𝑥 + 𝐵𝐵 sin3𝑥𝑥 = 2 sin 𝑥𝑥 ⇒ 𝐜𝐜𝐜𝐜𝐜𝐜𝟑𝟑𝟐𝟐 −𝟖𝟖𝑨𝑨 − 𝟑𝟑𝑩𝑩 + 𝐜𝐜𝐬𝐬𝐥𝐥𝟑𝟑𝟐𝟐 −𝟖𝟖𝑩𝑩 + 𝟑𝟑𝑨𝑨 = 2 sin 𝑥𝑥 ⇒ 3𝐴𝐴 − 8𝐵𝐵 = 2 𝑎𝑎𝑛𝑛𝑑𝑑 − 8𝐴𝐴 − 3𝐵𝐵 = 0
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 88
⇒ 𝐴𝐴 =6
73 𝑎𝑎𝑛𝑛𝑑𝑑 𝐵𝐵 = −
1673
⇒ 𝑦𝑦𝑝𝑝 =6
73cos3𝑥𝑥 −
1673
sin 3𝑥𝑥
⇒ 𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑓𝑓𝑎𝑎𝑔𝑔 𝑠𝑠𝑓𝑓𝑔𝑔𝑜𝑜𝑡𝑡𝑠𝑠𝑓𝑓𝑛𝑛: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝟏𝟏𝟐𝟐+𝒋𝒋
𝟑𝟑𝟐𝟐 𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆
𝟏𝟏𝟐𝟐−𝒋𝒋
𝟑𝟑𝟐𝟐 𝟐𝟐 +
𝟔𝟔𝟐𝟐𝟑𝟑𝐜𝐜𝐜𝐜𝐜𝐜𝟑𝟑𝟐𝟐 −
𝟏𝟏𝟔𝟔𝟐𝟐𝟑𝟑 𝐜𝐜𝐬𝐬𝐥𝐥𝟑𝟑𝟐𝟐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 89
E.g. 3: Using superposition 𝑦𝑦′′ − 2𝑦𝑦′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥
Given:
𝑦𝑦′′ − 2𝑦𝑦′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥
Step 1: 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒−𝑥𝑥 + 𝑐𝑐2𝑒𝑒3𝑥𝑥 Step 2: Find 𝑦𝑦𝑝𝑝
polynomial exponential
𝒈𝒈𝟏𝟏(𝟐𝟐) 𝒈𝒈𝟐𝟐(𝟐𝟐)
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 90
𝑦𝑦′′ − 2𝑦𝑦′ − 3𝑦𝑦 = 0 𝑚𝑚2 − 2𝑚𝑚 − 3 = 0 ⇒ (𝑚𝑚 − 3)(𝑚𝑚 + 1) = 0 ⇒ 𝑚𝑚1 = 3,𝑚𝑚2 = −1 Complimentary Solution: 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒−𝑥𝑥 + 𝑐𝑐2𝑒𝑒3𝑥𝑥
𝑦𝑦′′ − 2𝑦𝑦′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 91
⇒ 𝑎𝑎𝑠𝑠𝑠𝑠𝑜𝑜𝑚𝑚𝑒𝑒 𝑦𝑦𝑝𝑝 = 𝑦𝑦𝑝𝑝1 + 𝑦𝑦𝑝𝑝2 𝑠𝑠𝑜𝑜𝑠𝑠𝑒𝑒𝑓𝑓𝑠𝑠𝑓𝑓𝑠𝑠𝑠𝑠𝑡𝑡𝑠𝑠𝑓𝑓𝑛𝑛 ⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥 + 𝐵𝐵 + 𝐶𝐶𝑥𝑥𝑒𝑒2𝑥𝑥 + 𝐷𝐷𝑒𝑒2𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝′ = 𝐴𝐴 + 𝐶𝐶 𝑒𝑒2𝑥𝑥 + 2𝑥𝑥𝑒𝑒2𝑥𝑥 + 2𝐷𝐷𝑒𝑒2𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝′′ = 2𝐶𝐶𝑒𝑒2𝑥𝑥 + 2𝐶𝐶 𝑒𝑒2𝑥𝑥 + 2𝑥𝑥𝑒𝑒2𝑥𝑥 + 4𝐷𝐷𝑒𝑒2𝑥𝑥 ⇒ 𝑦𝑦′′ − 2𝑦𝑦′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥 Substitute 𝑦𝑦′′ 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦′𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦: ⇒ 2𝐶𝐶𝑒𝑒2𝑥𝑥 + 2𝐶𝐶𝑒𝑒2𝑥𝑥 + 4𝐶𝐶𝑥𝑥𝑒𝑒2𝑥𝑥 + 4𝐷𝐷𝑒𝑒2𝑥𝑥 − 2𝐴𝐴 − 2𝐶𝐶𝑒𝑒2𝑥𝑥 − 4𝐶𝐶𝑥𝑥𝑒𝑒2𝑥𝑥 − 4𝐷𝐷𝑒𝑒2𝑥𝑥 − 3𝐴𝐴𝑥𝑥
− 3𝐵𝐵 − 3𝐶𝐶𝑥𝑥𝑒𝑒2𝑥𝑥 − 3𝐷𝐷𝑒𝑒2𝑥𝑥 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥
𝑦𝑦′′ − 2𝑦𝑦′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥
𝒚𝒚𝒑𝒑𝟏𝟏 :for 𝒈𝒈𝟏𝟏(𝟐𝟐)
𝒚𝒚𝒑𝒑𝟐𝟐 :for 𝒈𝒈𝟐𝟐(𝟐𝟐)
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 92
⇒ −3𝐴𝐴𝑥𝑥 − 2𝐴𝐴 − 3𝐵𝐵 − 3𝐶𝐶𝑥𝑥𝑒𝑒2𝑥𝑥 + 𝑒𝑒2𝑥𝑥 2𝐶𝐶 − 3𝐷𝐷 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥
• −3𝐴𝐴 = 4 ⇒ 𝐴𝐴 = −43
• −2𝐴𝐴 − 3𝐵𝐵 = −5 ⇒ −3𝐵𝐵 = −5 + 2𝐴𝐴 = −5 + 2 −43
⇒ −3𝐵𝐵 = −5 −83 = −
153 −
83 = −
233 ⇒ 𝐵𝐵 =
239
• −3𝐶𝐶 = 6 ⇒ 𝐶𝐶 = −2
• 2𝐶𝐶 − 3𝐷𝐷 = 0 ⇒ 3𝐷𝐷 = 2𝐶𝐶 = 2 −2 = −4 ⇒ D = −43
𝑦𝑦′′ − 2𝑦𝑦′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 93
𝑦𝑦𝑝𝑝 = −43𝑥𝑥 +
239− 2𝑥𝑥𝑒𝑒2𝑥𝑥 −
43𝑒𝑒2𝑥𝑥
𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑓𝑓𝑎𝑎𝑔𝑔 𝑠𝑠𝑓𝑓𝑔𝑔𝑜𝑜𝑡𝑡𝑠𝑠𝑓𝑓𝑛𝑛: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆−𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆𝟑𝟑𝟐𝟐 −𝟒𝟒𝟑𝟑𝟐𝟐 +
𝟐𝟐𝟑𝟑𝟗𝟗 − 𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐𝟐𝟐 −
𝟒𝟒𝟑𝟑𝒆𝒆
𝟐𝟐𝟐𝟐
𝑦𝑦′′ − 2𝑦𝑦′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒2𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 94
E.g. 4: 𝑦𝑦′′ − 5𝑦𝑦′ + 4𝑦𝑦 = 8𝑒𝑒𝑥𝑥 Step 1: Find 𝑦𝑦𝑐𝑐 → 𝑦𝑦𝑐𝑐 = 𝐶𝐶1𝑒𝑒𝑥𝑥 + 𝐶𝐶2𝑒𝑒4𝑥𝑥 Step 2: Find 𝑦𝑦𝑝𝑝 → 𝑎𝑎𝑠𝑠𝑠𝑠𝑜𝑜𝑚𝑚𝑒𝑒 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑒𝑒𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝′ = 𝐴𝐴𝑒𝑒𝑥𝑥 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦𝑝𝑝′′ = 𝐴𝐴𝑒𝑒𝑥𝑥 Re-substituting back ⇒ 𝐴𝐴𝑒𝑒𝑥𝑥 − 5𝐴𝐴𝑒𝑒𝑥𝑥 + 4𝐴𝐴𝑒𝑒𝑥𝑥 = 8𝑒𝑒𝑥𝑥 ⇒ 0𝐴𝐴𝑒𝑒𝑥𝑥 = 8𝑒𝑒𝑥𝑥 ⇒ 𝟎𝟎 = 𝟖𝟖𝒆𝒆𝟐𝟐 − 𝒏𝒏𝒄𝒄𝒘𝒘 𝒑𝒑𝒄𝒄𝒄𝒄𝒄𝒄𝒔𝒔𝒃𝒃𝒓𝒓𝒆𝒆 Note: 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒𝑥𝑥 + 𝑐𝑐2𝑒𝑒4𝑥𝑥 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑒𝑒𝑥𝑥 • 𝑒𝑒𝑥𝑥 is already present in 𝑦𝑦𝑐𝑐 ⇒ 𝑒𝑒𝑥𝑥 is a solution of the homogeneous equation. ⇒ 𝐴𝐴𝑒𝑒𝑥𝑥 when substituted into the D.E. produces zero ⇒(see case II in section 3.3)
Not Independent
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 95
𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 𝑦𝑦𝑝𝑝′ = 𝐴𝐴𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦𝑝𝑝′′ = 𝐴𝐴𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 = 2𝐴𝐴𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 ⇒ 2𝐴𝐴𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 − 5 𝐴𝐴𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 + 4𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 = 8𝑒𝑒𝑥𝑥 ⇒ 2𝐴𝐴𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 − 5𝐴𝐴𝑒𝑒𝑥𝑥 − 5𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 + 4𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 = 8𝑒𝑒𝑥𝑥
⇒ −3𝐴𝐴𝑒𝑒𝑥𝑥 = 8𝑒𝑒𝑥𝑥 ⇒ −3𝐴𝐴 = 8 ⇒ 𝐴𝐴 = −83
⇒ 𝑦𝑦𝑝𝑝 = −83𝑥𝑥𝑒𝑒𝑥𝑥
Now, 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆𝟒𝟒𝟐𝟐 −𝟖𝟖𝟑𝟑𝟐𝟐𝒆𝒆𝟐𝟐
𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒𝑥𝑥 + 𝑐𝑐2𝑒𝑒4𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 96
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 97
E.g. 5.1 𝑦𝑦′′ − 8𝑦𝑦′ + 25𝑦𝑦 = 5𝑥𝑥3𝑒𝑒−𝑥𝑥 − 7𝑒𝑒−𝑥𝑥 ⇒ 𝑦𝑦′′ − 8𝑦𝑦′ + 25𝑦𝑦 = 5𝑥𝑥3 − 7 𝑒𝑒−𝑥𝑥 Homogeneous solution: 𝑦𝑦𝑐𝑐 = 𝑒𝑒4𝑥𝑥 𝑐𝑐1 cos3𝑥𝑥 + 𝑐𝑐2 sin 3𝑥𝑥 Assume 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥3 + 𝐵𝐵𝑥𝑥2 + 𝐶𝐶𝑥𝑥 + 𝐸𝐸 𝑒𝑒−𝑥𝑥 Note no duplication of terms between 𝑦𝑦𝑝𝑝𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦𝑐𝑐
Case I: No function in the assumed particular solution 𝑦𝑦𝑝𝑝 is a solution of the associated Homogeneous Differential Equation.
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 98
E.g. 5.2 𝑦𝑦′′ + 4𝑦𝑦 = 𝑥𝑥 cos𝑥𝑥
𝑦𝑦𝑐𝑐 = 𝑐𝑐1 cos2𝑥𝑥 + 𝑐𝑐2 sin2𝑥𝑥 𝐴𝐴𝑠𝑠𝑠𝑠𝑜𝑜𝑚𝑚𝑒𝑒 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥 + 𝐵𝐵 cos𝑥𝑥 + 𝐶𝐶𝑥𝑥 + 𝐸𝐸 sin 𝑥𝑥 No duplication of terms between 𝑦𝑦𝑝𝑝 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦𝑐𝑐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 99
E.g. 6: 𝑦𝑦′′ − 9𝑦𝑦′ + 14𝑦𝑦 = 3𝑥𝑥2 − 5 sin 2𝑥𝑥 + 7𝑥𝑥𝑒𝑒6𝑥𝑥 Given 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒2𝑥𝑥 + 𝑐𝑐2𝑒𝑒7𝑥𝑥 (computer earlier) Since 𝑔𝑔 𝑥𝑥 has various terms, form 𝑦𝑦𝑝𝑝 by superposition 3𝑥𝑥2 → 𝑦𝑦𝑝𝑝1 = 𝐴𝐴𝑥𝑥2 + 𝐵𝐵𝑥𝑥 + 𝐶𝐶 −5 sin 2𝑥𝑥 → 𝑦𝑦𝑝𝑝2 = 𝐸𝐸 cos2𝑥𝑥 + 𝐹𝐹 sin 2𝑥𝑥 7𝑥𝑥𝑒𝑒6𝑥𝑥 → 𝑦𝑦𝑝𝑝3 = 𝐺𝐺𝑥𝑥 + 𝐻𝐻 𝑒𝑒6𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝 = 𝑦𝑦𝑝𝑝1 + 𝑦𝑦𝑝𝑝2 + 𝑦𝑦𝑝𝑝3 ⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥2 + 𝐵𝐵𝑥𝑥 + 𝐶𝐶 + 𝐸𝐸 cos2𝑥𝑥 + 𝐹𝐹 sin 2𝑥𝑥 + 𝐺𝐺𝑥𝑥 + 𝐻𝐻 𝑒𝑒6𝑥𝑥 Note: No duplication of terms between 𝑦𝑦𝑝𝑝 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦𝑐𝑐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 100
E.g. 7: 𝑦𝑦′′ − 2𝑦𝑦′ + 𝑦𝑦 = 𝑒𝑒𝑥𝑥 With 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒𝑥𝑥 (computed earlier) What do we assume for 𝑦𝑦𝑝𝑝? • 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑒𝑒𝑥𝑥 → will fail since 𝑒𝑒𝑥𝑥 is part of 𝑦𝑦𝑐𝑐
• 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 → will fail since 𝑥𝑥𝑒𝑒𝑥𝑥 is part of 𝑦𝑦𝑐𝑐
⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥2𝑒𝑒𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝′ = 2𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥2𝑒𝑒𝑥𝑥 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦𝑝𝑝′′ = 2𝐴𝐴𝑒𝑒𝑥𝑥 + 2𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 + 2𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥2𝑒𝑒𝑥𝑥
Case II: A function in the potential particular solution is also a solution of the associated Homogeneous Differential Equation.
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 101
⇒ 2𝐴𝐴𝑒𝑒𝑥𝑥 + 4𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥2𝑒𝑒𝑥𝑥 − 4𝐴𝐴𝑥𝑥𝑒𝑒𝑥𝑥 − 2𝐴𝐴𝑥𝑥2𝑒𝑒𝑥𝑥 + 𝐴𝐴𝑥𝑥2𝑒𝑒𝑥𝑥 = 𝑒𝑒𝑥𝑥
⇒ 2𝐴𝐴𝑒𝑒𝑥𝑥 = 𝑒𝑒𝑥𝑥 ⇒ 2𝐴𝐴 = 1 ⇒ 𝐴𝐴 =12
⇒ 𝒚𝒚𝒑𝒑 =𝟏𝟏𝟐𝟐𝟐𝟐
𝟐𝟐𝒆𝒆𝟐𝟐
⇒ 𝒚𝒚𝒄𝒄 = 𝒄𝒄𝟏𝟏𝒆𝒆𝟐𝟐 + 𝒄𝒄𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐 +𝟏𝟏𝟐𝟐𝟐𝟐
𝟐𝟐𝒆𝒆𝟐𝟐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 102
Hence if 𝑔𝑔 𝑥𝑥 consists of no terms similar to Table 3.4.1 and that:
𝑦𝑦𝑝𝑝 = 𝑦𝑦𝑝𝑝1 + 𝑦𝑦𝑝𝑝2 + ⋯+ 𝑦𝑦𝑝𝑝𝑚𝑚 (assumption) Where 𝑦𝑦𝑝𝑝𝑖𝑖 , 𝑠𝑠 = 1, 2, 3, … … ,𝑚𝑚 are potential particular solution Multiplication rule: If any 𝒚𝒚𝒑𝒑𝒔𝒔 contains terms that duplicate terms in 𝒚𝒚𝒄𝒄, then that 𝒚𝒚𝒑𝒑𝒔𝒔 must be multiplied by 𝟐𝟐𝒏𝒏, where n is the smallest positive integer that eliminates that duplication
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 103
E.g. 8: 𝑦𝑦′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥 Initial conditions: 𝑦𝑦 𝜋𝜋 = 0;𝑦𝑦′ 𝜋𝜋 = 2 Step 1: 𝑆𝑆𝑓𝑓𝑔𝑔𝑠𝑠𝑒𝑒 𝑦𝑦′′ + 𝑦𝑦 = 0 𝑚𝑚2 + 1 = 0 ⇒ 𝑚𝑚2 = −1 = 𝑗𝑗2 ⇒ 𝑚𝑚 = ±𝑗𝑗 ⇒ 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒𝑗𝑗𝑥𝑥 + 𝑐𝑐2𝑒𝑒−𝑗𝑗𝑥𝑥 = 𝑐𝑐1 cos𝑥𝑥 + 𝑗𝑗𝑐𝑐1 sin 𝑥𝑥 + 𝑐𝑐2 cos 𝑥𝑥 − 𝑗𝑗𝑐𝑐2 sin 𝑥𝑥 = 𝑐𝑐1 + 𝑐𝑐2 cos𝑥𝑥 + 𝑗𝑗 𝑐𝑐1 − 𝑐𝑐2 sin 𝑥𝑥 ⇒ 𝑦𝑦𝑐𝑐 = 𝛼𝛼1 cos 𝑥𝑥 + 𝛼𝛼2 sin 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 104
𝑔𝑔 𝑥𝑥 = 4𝑥𝑥 + 10 sin 𝑥𝑥
• 4𝑥𝑥 → 𝑦𝑦𝑐𝑐𝐴𝐴𝑥𝑥+𝐵𝐵
• 10 sin 𝑥𝑥 → 𝐶𝐶 cos 𝑥𝑥 + 𝐸𝐸 sin 𝑥𝑥
(but these are part of 𝑦𝑦𝑐𝑐)
= 𝐶𝐶𝑥𝑥 cos𝑥𝑥 + 𝐸𝐸𝑥𝑥 sin 𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥 + 𝐵𝐵 + 𝐶𝐶𝑥𝑥 cos 𝑥𝑥 + 𝐸𝐸𝑥𝑥 sin 𝑥𝑥 • 𝑦𝑦𝑝𝑝′ = 𝐴𝐴 + 𝐶𝐶 cos 𝑥𝑥 + 𝑥𝑥 − sin 𝑥𝑥 + 𝐸𝐸 sin 𝑥𝑥 + 𝑥𝑥 cos𝑥𝑥 = 𝐴𝐴 + 𝐶𝐶 cos𝑥𝑥 − 𝐶𝐶𝑥𝑥 sin 𝑥𝑥 + 𝐸𝐸 sin 𝑥𝑥 + 𝐸𝐸𝑥𝑥 cos 𝑥𝑥
𝑦𝑦′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 105
• 𝑦𝑦𝑝𝑝′′ = −𝐶𝐶 sin 𝑥𝑥 − 𝐶𝐶 sin 𝑥𝑥 + 𝑥𝑥 cos𝑥𝑥 + 𝐸𝐸 cos 𝑥𝑥 + 𝐸𝐸 cos𝑥𝑥 − 𝑥𝑥 sin 𝑥𝑥 = −𝐶𝐶 sin 𝑥𝑥 − 𝐶𝐶 sin 𝑥𝑥 − 𝐶𝐶𝑥𝑥 cos 𝑥𝑥 + 𝐸𝐸 cos 𝑥𝑥 + 𝐸𝐸 cos 𝑥𝑥 − 𝐸𝐸𝑥𝑥 sin 𝑥𝑥 = −2𝐶𝐶 sin 𝑥𝑥 + 2𝐸𝐸 cos 𝑥𝑥 − 𝐶𝐶𝑥𝑥 cos 𝑥𝑥 − 𝐸𝐸𝑥𝑥 sin 𝑥𝑥 We have: 𝑦𝑦′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥 ⇒ −2𝐶𝐶 sin 𝑥𝑥 + 2𝐸𝐸 cos 𝑥𝑥 − 𝐶𝐶𝑥𝑥 cos𝑥𝑥 − 𝐸𝐸𝑥𝑥 sin 𝑥𝑥 + 𝐴𝐴𝑥𝑥 + 𝐵𝐵 + 𝐶𝐶𝑥𝑥 cos 𝑥𝑥 + 𝐸𝐸𝑥𝑥 sin 𝑥𝑥
= 4𝑥𝑥 + 10 sin 𝑥𝑥
𝑦𝑦′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 106
⇒ −2𝐶𝐶 sin 𝑥𝑥 + 2𝐸𝐸 cos 𝑥𝑥 − 𝐶𝐶𝑥𝑥 cos 𝑥𝑥 − 𝐸𝐸𝑥𝑥 sin 𝑥𝑥 + 𝐴𝐴𝑥𝑥 + 𝐵𝐵 + 𝐶𝐶𝑥𝑥 cos 𝑥𝑥 + 𝐸𝐸𝑥𝑥 sin 𝑥𝑥
= 4𝑥𝑥 + 10 sin 𝑥𝑥 ⇒ 𝐴𝐴𝑥𝑥 + 𝐵𝐵 − 2𝐶𝐶 sin 𝑥𝑥 − 𝐶𝐶𝑥𝑥 cos 𝑥𝑥 + 𝐶𝐶𝑥𝑥 cos 𝑥𝑥 + 2𝐸𝐸 cos 𝑥𝑥 − 𝐸𝐸𝑥𝑥 sin 𝑥𝑥 + 𝐸𝐸𝑥𝑥 sin 𝑥𝑥
= 4𝑥𝑥 + 10 sin 𝑥𝑥 ⇒ 𝑨𝑨𝟐𝟐 + 𝑩𝑩 + −𝟐𝟐𝑪𝑪𝐜𝐜𝐬𝐬𝐥𝐥𝟐𝟐 + 𝟐𝟐𝟐𝟐𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐 = 𝟒𝟒𝟐𝟐 + 𝟎𝟎 + 𝟏𝟏𝟎𝟎𝐜𝐜𝐬𝐬𝐥𝐥𝟐𝟐 + 𝟎𝟎 ⇒ 𝐴𝐴𝑥𝑥 = 4𝑥𝑥 ⇒ 𝐴𝐴 = 4 ⇒ 𝐵𝐵 = 0 ⇒ −2𝐶𝐶 sin 𝑥𝑥 = 10 sin 𝑥𝑥 ⇒ −2𝐶𝐶 = 10 ⇒ 𝐶𝐶 = −5 ⇒ 2𝐸𝐸 cos 𝑥𝑥 = 0 ⇒ 𝐸𝐸 = 0
𝑦𝑦′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 107
WE know, 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥 + 𝐵𝐵 + 𝐶𝐶𝑥𝑥 cos𝑥𝑥 + 𝐸𝐸𝑥𝑥 sin 𝑥𝑥 & 𝐴𝐴 = 4,𝐵𝐵 = 0,𝐶𝐶 = −5,𝐷𝐷 = 0 ⇒ 𝒚𝒚𝒑𝒑 = 𝟒𝟒𝟐𝟐 − 𝟓𝟓𝟐𝟐𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐 We know: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝 ⇒ 𝑦𝑦 = 𝛼𝛼1 cos 𝑥𝑥 + 𝛼𝛼2 sin 𝑥𝑥 + 4𝑥𝑥 − 5𝑥𝑥 cos𝑥𝑥 Initial conditions: 𝑦𝑦 𝑥𝑥 = 0 𝑎𝑎𝑛𝑛𝑑𝑑 𝑦𝑦′ 𝑥𝑥 = 2 • 𝑦𝑦 𝜋𝜋 = 0 ⇒ 0 = 𝛼𝛼1 cos𝜋𝜋 + 𝛼𝛼2 sin𝜋𝜋 + 4𝜋𝜋 − 5𝜋𝜋 cos𝜋𝜋
⇒ 0 = −𝛼𝛼1 + 0 + 4𝜋𝜋 + 5𝜋𝜋 ⇒ 𝛼𝛼1 = 9𝜋𝜋
𝑦𝑦′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 108
• 𝑦𝑦′ 𝜋𝜋 = 2
⇒ 𝑦𝑦′ = −𝛼𝛼1 sin 𝑥𝑥 + 𝛼𝛼2 cos 𝑥𝑥 + 4 − 5 cos𝑥𝑥 − 𝑥𝑥 sin 𝑥𝑥 ⇒ 2 = −9𝜋𝜋 sin𝜋𝜋 + 𝛼𝛼2 cos𝜋𝜋 + 4 − 5 cos𝜋𝜋 + 5𝜋𝜋 sin𝜋𝜋 ⇒ 2 = −𝛼𝛼2 + 4 + 5 ⇒ 𝛼𝛼2 = 9 − 2 ⇒ 𝛼𝛼2 = 7 Therefore:
𝒚𝒚 = 𝟗𝟗𝟗𝟗𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟐𝟐 𝐜𝐜𝐬𝐬𝐥𝐥 𝟐𝟐 + 𝟒𝟒𝟐𝟐 − 𝟓𝟓𝟐𝟐𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐
𝑦𝑦′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥 𝑦𝑦𝑝𝑝 = 4𝑥𝑥 − 5𝑥𝑥 𝑐𝑐𝑓𝑓𝑠𝑠 𝑥𝑥
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 109
• To solve a non-Homogeneous D.E.
𝑎𝑎𝑛𝑛𝑦𝑦 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1𝑦𝑦 𝑛𝑛−1 + ⋯+ 𝑎𝑎1𝑦𝑦1 + 𝑎𝑎0𝑦𝑦0= 𝑔𝑔 𝑥𝑥
Step1: Finding a complementary solution 𝑦𝑦𝑐𝑐 by equating it to 0. Step2: Finding a particular solution 𝑦𝑦𝑝𝑝. Step3: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
• In case of multiple additive terms in the right
hand side that constitute 𝑔𝑔(𝑥𝑥), take into account all factors contributing to 𝑦𝑦𝑝𝑝
• Multiplication rule: If any 𝒚𝒚𝒑𝒑𝒔𝒔 contains terms
that duplicate terms in 𝒚𝒚𝒄𝒄, then that 𝒚𝒚𝒑𝒑𝒔𝒔 must be multiplied by 𝟐𝟐𝒏𝒏, where n is the smallest positive integer that eliminates that duplication
Summary
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.
Section 3.5 Variation of Parameters
9/30/2014 Dr. Eli Saber 110
Variation of Parameters
9/30/2014 Dr. Eli Saber 111
• See also section 2.3 for first order differential equations
Advantages: • Always yields a particular solution 𝑦𝑦𝑝𝑝 assuming 𝑦𝑦𝑐𝑐 can be found. • Not limited to cases such as the described in Table 3.4.1 (slide 109) • Not limited to differential equation with constant coefficients.
Variation of Parameters
9/30/2014 Dr. Eli Saber 112
Given 𝑎𝑎2 𝑥𝑥 𝑦𝑦′′ + 𝑎𝑎1 𝑥𝑥 𝑦𝑦′ + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔 𝑥𝑥
Divide by 𝑎𝑎2 𝑥𝑥
⟹ 𝑦𝑦′′ +𝑎𝑎1𝑎𝑎2
𝑥𝑥 𝑦𝑦′ +𝑎𝑎0𝑎𝑎2
𝑥𝑥 𝑦𝑦 =𝑔𝑔 𝑥𝑥𝑎𝑎2
⟹ 𝑦𝑦′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦 = 𝑓𝑓 𝑥𝑥 (similar to 𝑦𝑦𝑦 + 𝑃𝑃(𝑥𝑥)𝑦𝑦 = 𝑓𝑓(𝑥𝑥)) Assumptions: • 𝑃𝑃(𝑥𝑥),𝑄𝑄(𝑥𝑥), 𝑓𝑓(𝑥𝑥) are continuous on some interval 𝐼𝐼 • 𝑦𝑦𝑐𝑐 can be found
𝑷𝑷(𝟐𝟐) 𝑸𝑸(𝟐𝟐) 𝒇𝒇(𝟐𝟐)
Variation of Parameters
9/30/2014 Dr. Eli Saber 113
Method: • For first order differential equation 𝑦𝑦′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦 = 𝑓𝑓 𝑥𝑥 , seek a solution 𝒚𝒚𝒑𝒑 = 𝝁𝝁𝟏𝟏(𝟐𝟐)𝒚𝒚𝟏𝟏(𝟐𝟐) 𝒚𝒚𝟏𝟏(𝟐𝟐): fundamental solution for homogeneous D.E • For second order D.E 𝑦𝑦′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦 = 𝑓𝑓 𝑥𝑥 , seek a solution 𝒚𝒚𝒑𝒑 = 𝝁𝝁𝟏𝟏 𝟐𝟐 𝒚𝒚𝟏𝟏 𝟐𝟐 + 𝝁𝝁𝟐𝟐(𝟐𝟐)𝒚𝒚𝟐𝟐(𝟐𝟐)
𝒚𝒚𝟏𝟏 𝟐𝟐 , 𝒚𝒚𝟐𝟐(𝟐𝟐): fundamental solution for homogeneous D.E
Variation of Parameters
9/30/2014 Dr. Eli Saber 114
𝑦𝑦𝑝𝑝 = 𝜇𝜇1𝑦𝑦1 + 𝜇𝜇2𝑦𝑦2 ⟹ 𝑦𝑦𝑝𝑝′ = 𝜇𝜇1𝑦𝑦1′ + 𝜇𝜇1′ 𝑦𝑦1 + 𝜇𝜇2𝑦𝑦2′ + 𝜇𝜇2′ 𝑦𝑦2 ⇒ 𝑦𝑦𝑝𝑝𝑦𝑦 = 𝜇𝜇1𝑦𝑦1′′ + 𝜇𝜇1′ 𝑦𝑦1′ + 𝜇𝜇1′ 𝑦𝑦1′ + 𝜇𝜇1′′𝑦𝑦1 + 𝜇𝜇2𝑦𝑦2′′ + 𝜇𝜇2′ 𝑦𝑦2′ + 𝜇𝜇2′ 𝑦𝑦2′ + 𝜇𝜇2′′𝑦𝑦2 Substitute into D.E: 𝑦𝑦𝑝𝑝′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦𝑝𝑝′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦𝑝𝑝 = 𝑓𝑓 𝑥𝑥 ⟹ 𝜇𝜇1𝑦𝑦1′′ + 𝜇𝜇1′ 𝑦𝑦1′ + 𝜇𝜇1′ 𝑦𝑦1′ + 𝜇𝜇1′′𝑦𝑦1 + 𝜇𝜇2𝑦𝑦2′′ + 𝜇𝜇2′ 𝑦𝑦2′ + 𝜇𝜇2′ 𝑦𝑦2′ + 𝜇𝜇2′′𝑦𝑦2 𝑃𝑃 𝑥𝑥 𝜇𝜇1𝑦𝑦1′ + 𝜇𝜇1′ 𝑦𝑦1 + 𝜇𝜇2𝑦𝑦2′ + 𝜇𝜇2′ 𝑦𝑦2 +𝑄𝑄 𝑥𝑥 𝜇𝜇1𝑦𝑦1 + 𝜇𝜇2𝑦𝑦2 = 𝑓𝑓(𝑥𝑥)
Variation of Parameters
9/30/2014 Dr. Eli Saber 115
Rearranging the equations, 𝑜𝑜1 𝑦𝑦1′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦1′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦1 + 𝑜𝑜2 𝑦𝑦2′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦2′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦2 +𝑜𝑜1′′𝑦𝑦1 + 𝑜𝑜1′ 𝑦𝑦1′ + 𝑜𝑜2′′𝑦𝑦2 + 𝑜𝑜2′ 𝑦𝑦2′ + 𝑃𝑃 𝑜𝑜1′ 𝑦𝑦1 + 𝑜𝑜2′ 𝑦𝑦2 + 𝑜𝑜1′ 𝑦𝑦1′ + 𝑜𝑜2′ 𝑦𝑦2′ = 𝑓𝑓(𝑥𝑥)
= 𝟎𝟎 = 𝟎𝟎
Since 𝑦𝑦1 & 𝑦𝑦2 are the solutions to the homogeneous equation
Variation of Parameters
9/30/2014 Dr. Eli Saber 116
𝑜𝑜1′′𝑦𝑦1 + 𝑜𝑜1
′𝑦𝑦1′ + 𝑜𝑜2
′′𝑦𝑦2 + 𝑜𝑜2′𝑦𝑦2
′ + 𝑃𝑃 𝑜𝑜1′𝑦𝑦1 + 𝑜𝑜2
′𝑦𝑦2 + 𝑜𝑜1′𝑦𝑦1
′ + 𝑜𝑜2′𝑦𝑦2
′ = 𝑓𝑓(𝑥𝑥)
⟹𝒅𝒅𝒅𝒅𝟐𝟐
𝒖𝒖𝟏𝟏′𝒚𝒚𝟏𝟏 +𝒅𝒅𝒅𝒅𝟐𝟐
𝒖𝒖𝟐𝟐′𝒚𝒚𝟐𝟐 + 𝑃𝑃 𝑜𝑜1′𝑦𝑦1 + 𝑜𝑜2
′𝑦𝑦2 + 𝑜𝑜1′𝑦𝑦1
′ + 𝑜𝑜2′𝑦𝑦2
′ = 𝑓𝑓(𝑥𝑥)
⟹𝑑𝑑𝑑𝑑𝑥𝑥 𝑜𝑜1
′𝑦𝑦1 + 𝑜𝑜2′𝑦𝑦2
+ 𝑃𝑃[𝑜𝑜1′𝑦𝑦1 + 𝑜𝑜2
′𝑦𝑦2] + 𝑜𝑜1′𝑦𝑦1
′ + 𝑜𝑜2′𝑦𝑦2
′ = 𝑓𝑓(𝑥𝑥)
• Have two unknown functions 𝑜𝑜1 & 𝑜𝑜2 ⟹ Need two equations ⟹ make further assumption that 𝒖𝒖𝟏𝟏’𝒚𝒚𝟏𝟏 + 𝒖𝒖𝟐𝟐’𝒚𝒚𝟐𝟐 = 𝟎𝟎 ⟹ 𝒖𝒖𝟏𝟏’𝒚𝒚𝟏𝟏’ + 𝒖𝒖𝟐𝟐’𝒚𝒚𝟐𝟐’ = 𝒇𝒇(𝟐𝟐)
Variation of Parameters
9/30/2014 Dr. Eli Saber 117
• Hence, we have two equations with two unknowns:
𝑦𝑦1𝑜𝑜1’ + 𝑦𝑦2𝑜𝑜2’ = 0
𝑦𝑦1𝑦𝑜𝑜1’ + 𝑦𝑦2′𝑜𝑜2’ = 𝑓𝑓(𝑥𝑥)
• Solve for 𝑜𝑜1′ & 𝑜𝑜2′ & then integrate to get 𝑜𝑜1 & 𝑜𝑜2
• Using Cramer’s rule: 𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2𝑦
𝑜𝑜1𝑦𝑜𝑜2𝑦
= 0𝑓𝑓(𝑥𝑥)
(1)
(2)
𝒖𝒖 𝑨𝑨 𝒃𝒃
Variation of Parameters
9/30/2014 Dr. Eli Saber 118
We have, 𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2𝑦
𝑜𝑜1𝑦𝑜𝑜2𝑦
= 0𝑓𝑓(𝑥𝑥)
𝒖𝒖𝟏𝟏′ =
𝟎𝟎 𝒚𝒚𝟐𝟐𝒇𝒇(𝟐𝟐) 𝒚𝒚𝟐𝟐𝑦𝒚𝒚𝟏𝟏 𝒚𝒚𝟐𝟐𝒚𝒚𝟏𝟏𝑦 𝒚𝒚𝟐𝟐𝑦
& 𝒖𝒖𝟐𝟐′ =
𝒚𝒚𝟏𝟏 𝟎𝟎𝒚𝒚𝟏𝟏𝑦 𝒇𝒇(𝟐𝟐)𝒚𝒚𝟏𝟏 𝒚𝒚𝟐𝟐𝒚𝒚𝟏𝟏𝑦 𝒚𝒚𝟐𝟐𝑦
Note:
W≡𝑦𝑦1 𝑦𝑦2𝑦𝑦1′ 𝑦𝑦2′
⇒ 𝑡𝑡ℎ𝑒𝑒 𝑾𝑾𝒓𝒓𝒄𝒄𝒏𝒏𝒄𝒄𝒓𝒓𝒔𝒔𝒂𝒂𝒏𝒏 𝑓𝑓𝑓𝑓 𝑦𝑦1 & 𝑦𝑦2
Hence, Since y1 & y2 are independent ⟹ W≠ 𝟎𝟎 𝒇𝒇𝒄𝒄𝒓𝒓 ∀𝟐𝟐 ∈ 𝑰𝑰
Variation of Parameters
9/30/2014 Dr. Eli Saber 119
Summary: Given 𝑎𝑎2𝑦𝑦𝑦𝑦 + 𝑎𝑎1𝑦𝑦𝑦 + 𝑎𝑎0𝑦𝑦 = 𝑔𝑔(𝑥𝑥)
1. Put Eq. into standard form by dividing throughout by a2(x)
𝑦𝑦′′ +𝑎𝑎1𝑎𝑎2
𝑦𝑦′ +𝑎𝑎0𝑎𝑎2𝑦𝑦 =
𝑔𝑔 𝑥𝑥𝑎𝑎2(𝑥𝑥)
2. Find the complementary solution 𝑦𝑦𝑐𝑐
= 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2
3. Compute Wronskian of 𝑦𝑦1 & 𝑦𝑦2 𝑊𝑊 =𝑦𝑦1 𝑦𝑦2𝑦𝑦1′ 𝑦𝑦2′
𝑷𝑷(𝟐𝟐) 𝑸𝑸(𝟐𝟐) 𝒇𝒇(𝟐𝟐)
Variation of Parameters
9/30/2014 Dr. Eli Saber 120
4. Compute 𝑜𝑜1’ & 𝑜𝑜2’ using:
𝑜𝑜1′ =
0 𝑦𝑦2𝑓𝑓(𝑥𝑥) 𝑦𝑦2𝑦𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2𝑦
;𝑜𝑜2′ =
𝑦𝑦1 0𝑦𝑦1𝑦 𝑓𝑓(𝑥𝑥)𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2𝑦
5. Find 𝑜𝑜1 & 𝑜𝑜2 by integrating 𝑜𝑜1𝑦 & 𝑜𝑜2𝑦 respectively. 6. Form 𝑦𝑦𝑝𝑝 = 𝑜𝑜1(𝑥𝑥)𝑦𝑦1(𝑥𝑥) + 𝑜𝑜2(𝑥𝑥)𝑦𝑦2(𝑥𝑥) 7. General Solution: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
Variation of Parameters
9/30/2014 Dr. Eli Saber 121
Note: When integrating 𝑜𝑜1’ & 𝑜𝑜2’, you don’t need to introduce any constants because:
• 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2
�𝑜𝑜1𝑦 𝑑𝑑𝑥𝑥 = 𝑜𝑜1 + 𝑎𝑎1,�𝑜𝑜2𝑦
𝑑𝑑𝑥𝑥 = 𝑜𝑜2 + 𝑎𝑎2
⟹ 𝑦𝑦𝑝𝑝 = (𝑜𝑜1 + 𝑎𝑎1)𝑦𝑦1 + (𝑜𝑜2 + 𝑎𝑎2)𝑦𝑦2 ⟹ 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝 = 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2 + 𝑜𝑜1𝑦𝑦1 + 𝑎𝑎1𝑦𝑦1 + 𝑜𝑜2𝑦𝑦2 + 𝑎𝑎2𝑦𝑦2
Rearranging, 𝑦𝑦 = (𝑐𝑐1 + 𝑎𝑎1)𝑦𝑦1 + (𝑐𝑐2 + 𝑎𝑎2)𝑦𝑦2 + 𝑜𝑜1𝑦𝑦1 + 𝑜𝑜2𝑦𝑦2
⟹ 𝑦𝑦 = 𝜶𝜶𝟏𝟏𝑦𝑦1 + 𝜶𝜶𝟐𝟐𝑦𝑦2 + 𝑜𝑜1𝑦𝑦1 + 𝑜𝑜2𝑦𝑦2
Where, 𝛼𝛼1&𝛼𝛼2: constants computed using initial conditions or boundary conditions
𝑎𝑎1, 𝑎𝑎2 are constants
Variation of Parameters
9/30/2014 Dr. Eli Saber 122
E.g.1: 𝑦𝑦′′ − 4𝑦𝑦′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒2𝑥𝑥
1. Equation is already in standard form: 𝑦𝑦′′ − 4𝑦𝑦′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒2𝑥𝑥 2. Find 𝑦𝑦𝑐𝑐: ⇒ 𝑚𝑚2 − 4𝑚𝑚 + 4 = 0 → 𝑚𝑚− 2 2 = 0 ⇒ 𝑚𝑚1 = 𝑚𝑚2 = 2 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒2𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒2𝑥𝑥 3. Compute 𝑊𝑊
𝑊𝑊 =𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2𝑦
= 𝑒𝑒2𝑥𝑥 𝑥𝑥𝑒𝑒2𝑥𝑥2𝑒𝑒2𝑥𝑥 𝑒𝑒2𝑥𝑥 + 2𝑥𝑥 𝑒𝑒2𝑥𝑥
= 𝑒𝑒2𝑥𝑥 𝑒𝑒2𝑥𝑥 + 2𝑥𝑥 𝑒𝑒2𝑥𝑥 − 𝑥𝑥𝑒𝑒2𝑥𝑥 2𝑒𝑒2𝑥𝑥 = 𝑒𝑒4𝑥𝑥 + 2𝑥𝑥 𝑒𝑒4𝑥𝑥 − 2𝑥𝑥 𝑒𝑒4𝑥𝑥 ⇒ 𝑊𝑊 = 𝑒𝑒4𝑥𝑥
𝑷𝑷(𝟐𝟐) 𝑸𝑸(𝟐𝟐) 𝒇𝒇(𝟐𝟐)
𝒚𝒚𝟏𝟏 𝒚𝒚𝟐𝟐
Variation of Parameters
9/30/2014 Dr. Eli Saber 123
4. Compute 𝑜𝑜1′& 𝑜𝑜2𝑦
𝑊𝑊1 = 0 𝑥𝑥𝑒𝑒2𝑥𝑥𝑥𝑥 + 1 𝑒𝑒2𝑥𝑥 𝑒𝑒2𝑥𝑥 + 2𝑥𝑥𝑒𝑒2𝑥𝑥 = −𝟐𝟐 𝟐𝟐 + 𝟏𝟏 𝒆𝒆𝟒𝟒𝟐𝟐
𝑊𝑊2 = 𝑒𝑒2𝑥𝑥 02𝑒𝑒2𝑥𝑥 𝑥𝑥 + 1 𝑒𝑒2𝑥𝑥 = 𝟐𝟐 + 𝟏𝟏 𝒆𝒆𝟒𝟒𝟐𝟐
Now,
𝑜𝑜1′ =𝑊𝑊1𝑊𝑊 = −
𝑥𝑥 + 1 𝑥𝑥𝑒𝑒4𝑥𝑥
𝑒𝑒4𝑥𝑥 = −𝑥𝑥 𝑥𝑥 + 1 ⇒ 𝑜𝑜1′ = −𝑥𝑥2 − 𝑥𝑥
𝑜𝑜1 = � −𝑥𝑥2 − 𝑥𝑥 𝑑𝑑𝑥𝑥
𝑜𝑜1 = −13 𝑥𝑥
3 −𝑥𝑥2
2
𝑦𝑦′′ − 4𝑦𝑦′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒2𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒2𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒2𝑥𝑥
𝑊𝑊 = 𝑒𝑒4𝑥𝑥
Variation of Parameters
9/30/2014 Dr. Eli Saber 124
Now,
𝑜𝑜2′ =𝑊𝑊2𝑊𝑊 =
𝑥𝑥 + 1 𝑒𝑒4𝑥𝑥
𝑒𝑒4𝑥𝑥 = 𝑥𝑥 + 1 → 𝑜𝑜2𝑦
𝑜𝑜2 = � 𝑥𝑥 + 1 𝑑𝑑𝑥𝑥
𝑜𝑜2 =𝑥𝑥2
2 + 𝑥𝑥
Hence, 𝑦𝑦𝑝𝑝 = 𝑜𝑜1 𝑥𝑥 𝑦𝑦1 𝑥𝑥 + 𝑜𝑜2 𝑥𝑥 𝑦𝑦2 𝑥𝑥
𝑦𝑦𝑝𝑝 𝑥𝑥 = −13𝑥𝑥3 −
𝑥𝑥2
2𝑒𝑒2𝑥𝑥 +
𝑥𝑥2
𝑥𝑥+ 𝑥𝑥 𝑥𝑥𝑒𝑒2𝑥𝑥
𝑦𝑦′′ − 4𝑦𝑦′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒2𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒2𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒2𝑥𝑥
𝑊𝑊 = 𝑒𝑒4𝑥𝑥 𝑊𝑊2 = 𝑥𝑥 + 1 𝑒𝑒4𝑥𝑥
𝑜𝑜1 = −13𝑥𝑥3 −
𝑥𝑥2
2
Variation of Parameters
9/30/2014 Dr. Eli Saber 125
𝑦𝑦𝑝𝑝 𝑥𝑥 = −13𝑥𝑥3 −
𝑥𝑥2
2𝑒𝑒2𝑥𝑥 +
𝑥𝑥2
𝑥𝑥+ 𝑥𝑥 𝑥𝑥𝑒𝑒2𝑥𝑥
= −13
𝑥𝑥3𝑒𝑒2𝑥𝑥 −12𝑥𝑥2𝑒𝑒2𝑥𝑥 +
12𝑥𝑥3𝑒𝑒2𝑥𝑥 + 𝑥𝑥2𝑒𝑒2𝑥𝑥
𝑦𝑦𝑝𝑝 =16 𝑥𝑥3𝑒𝑒2𝑥𝑥 +
12 𝑥𝑥2𝑒𝑒2𝑥𝑥
And, 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐𝟐𝟐 +𝟏𝟏𝟔𝟔𝟐𝟐
𝟑𝟑𝒆𝒆𝟐𝟐𝟐𝟐 +𝟏𝟏𝟐𝟐𝟐𝟐
𝟐𝟐𝒆𝒆𝟐𝟐𝟐𝟐
𝑦𝑦′′ − 4𝑦𝑦′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒2𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒2𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒2𝑥𝑥
𝑊𝑊 = 𝑒𝑒4𝑥𝑥 𝑊𝑊2 = 𝑥𝑥 + 1 𝑒𝑒4𝑥𝑥
𝑜𝑜1 = −13𝑥𝑥3 −
𝑥𝑥2
2
𝑜𝑜2 =𝑥𝑥2
2+ 𝑥𝑥
Variation of Parameters
9/30/2014 Dr. Eli Saber 126
E.g.2: 𝑦𝑦′′ − 5𝑦𝑦′ + 4𝑦𝑦 = 8𝑒𝑒𝑥𝑥
1. Equation is already in standard form: 𝑦𝑦′′ − 5𝑦𝑦′ + 4𝑦𝑦 = 8𝑒𝑒𝑥𝑥 2. Find 𝑦𝑦𝑐𝑐: ⇒ 𝑚𝑚2 − 5𝑚𝑚 + 4 = 0 → (𝑚𝑚− 1)(𝑚𝑚− 4) = 0 ⇒ 𝑚𝑚1 = 1&𝑚𝑚2 = 4 𝑦𝑦𝑐𝑐 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒4𝑥𝑥 3. Compute 𝑊𝑊
𝑊𝑊 =𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2𝑦
= 𝑒𝑒𝑥𝑥 𝑒𝑒4𝑥𝑥𝑒𝑒𝑥𝑥 4𝑒𝑒4𝑥𝑥
= 𝑒𝑒𝑥𝑥 4𝑒𝑒4𝑥𝑥 − 𝑒𝑒𝑥𝑥𝑒𝑒4𝑥𝑥 = 4𝑒𝑒5𝑥𝑥 − 𝑒𝑒5𝑥𝑥 ⇒ 𝑊𝑊 = 3𝑒𝑒5𝑥𝑥
𝑷𝑷(𝟐𝟐) 𝑸𝑸(𝟐𝟐) 𝒇𝒇(𝟐𝟐)
𝒚𝒚𝟏𝟏 𝒚𝒚𝟐𝟐
Sec. 3.4. E.g. 4 (slide 94-95)
Variation of Parameters
9/30/2014 Dr. Eli Saber 127
4. Compute 𝑜𝑜1′& 𝑜𝑜2𝑦
𝑜𝑜1′ =
0 𝑦𝑦2𝑓𝑓(𝑥𝑥) 𝑦𝑦2𝑦
𝑊𝑊 =0 𝑒𝑒4𝑥𝑥
8𝑒𝑒𝑥𝑥 4𝑒𝑒4𝑥𝑥3𝑒𝑒5𝑥𝑥 = −
8𝑒𝑒5𝑥𝑥
3𝑒𝑒5𝑥𝑥 = −83
𝑜𝑜2′ =
𝑦𝑦1 0𝑦𝑦1𝑦 𝑓𝑓(𝑥𝑥)
𝑊𝑊 =𝑒𝑒𝑥𝑥 0𝑒𝑒𝑥𝑥 8𝑒𝑒𝑥𝑥
3𝑒𝑒5𝑥𝑥 =8𝑒𝑒2𝑥𝑥
3𝑒𝑒5𝑥𝑥 =83 𝑒𝑒−3𝑥𝑥
5. Find 𝑜𝑜1 𝑎𝑎𝑛𝑛𝑑𝑑 𝑜𝑜2
𝑜𝑜1 = �−83𝑑𝑑𝑥𝑥 = −
83 𝑥𝑥
𝑜𝑜2 = �83 𝑒𝑒
−3𝑥𝑥 𝑑𝑑𝑥𝑥 = −89 𝑒𝑒−3𝑥𝑥
𝑦𝑦′′ − 5𝑦𝑦′ + 4𝑦𝑦 = 8𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒4𝑥𝑥
𝑊𝑊 = 3𝑒𝑒5𝑥𝑥
Variation of Parameters
9/30/2014 Dr. Eli Saber 128
Hence, 𝑦𝑦𝑝𝑝 = 𝑜𝑜1 𝑥𝑥 𝑦𝑦1 𝑥𝑥 + 𝑜𝑜2 𝑥𝑥 𝑦𝑦2 𝑥𝑥
𝑦𝑦𝑝𝑝 = −83𝑥𝑥 𝑒𝑒𝑥𝑥 + −
89
𝑒𝑒−3𝑥𝑥 𝑒𝑒4𝑥𝑥
𝑦𝑦𝑝𝑝 = −83 𝑥𝑥 𝑒𝑒𝑥𝑥 −
89 𝑒𝑒
𝑥𝑥
And, 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
⇒ 𝒚𝒚 = 𝜶𝜶𝟏𝟏𝒆𝒆𝟐𝟐 + 𝜶𝜶𝟐𝟐𝒆𝒆𝟒𝟒𝟐𝟐 −𝟖𝟖𝟑𝟑𝟐𝟐𝒆𝒆
𝟐𝟐 −𝟖𝟖𝟗𝟗𝒆𝒆
𝟐𝟐
𝑦𝑦′′ − 5𝑦𝑦′ + 4𝑦𝑦 = 8𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒4𝑥𝑥
𝑊𝑊 = 3𝑒𝑒5𝑥𝑥
𝑜𝑜1 = −83𝑥𝑥 & 𝑜𝑜2 = −
89𝑒𝑒−3𝑥𝑥
Variation of Parameters
9/30/2014 Dr. Eli Saber 129
We got,
⇒ 𝒚𝒚 = 𝜶𝜶𝟏𝟏𝒆𝒆𝟐𝟐 + 𝜶𝜶𝟐𝟐𝒆𝒆𝟒𝟒𝟐𝟐 −𝟖𝟖𝟑𝟑𝟐𝟐𝒆𝒆𝟐𝟐 −
𝟖𝟖𝟗𝟗𝒆𝒆𝟐𝟐
From Sec. 3.4. E.g. 4 (slide 94-95), we have:
𝒚𝒚 = 𝒄𝒄𝟏𝟏𝒆𝒆𝟐𝟐 + 𝒄𝒄𝟐𝟐𝒆𝒆𝟒𝟒𝟐𝟐 −𝟖𝟖𝟑𝟑𝟐𝟐𝒆𝒆𝟐𝟐
as the solution.
Notice, 𝑦𝑦 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒4𝑥𝑥 −83𝑥𝑥𝑒𝑒𝑥𝑥 − 8
9𝑒𝑒𝑥𝑥 = 𝛼𝛼1𝑒𝑒𝑥𝑥 −
89𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒4𝑥𝑥 −
83𝑥𝑥𝑒𝑒𝑥𝑥
= 𝛼𝛼1 −89
𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒4𝑥𝑥 −83𝑥𝑥𝑒𝑒𝑥𝑥
= 𝒄𝒄𝟏𝟏𝑒𝑒𝑥𝑥 + 𝒄𝒄𝟐𝟐𝑒𝑒4𝑥𝑥 −83𝑥𝑥𝑒𝑒𝑥𝑥
Variation of Parameters
9/30/2014 Dr. Eli Saber 130
• Higher Order Equations Generalize method to linear nth order D.E.
𝑦𝑦(𝑛𝑛) + 𝑃𝑃𝑛𝑛−1 𝑥𝑥 𝑦𝑦(𝑛𝑛−1) + ⋯+ 𝑃𝑃1 𝑥𝑥 𝑦𝑦′ + 𝑃𝑃0 𝑥𝑥 𝑦𝑦 = 𝑓𝑓 𝑥𝑥
If 𝑦𝑦c = 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2+. . . +𝑐𝑐𝑛𝑛𝑦𝑦𝑛𝑛 is the complementary function , then a particular solution is:
𝑦𝑦𝑝𝑝 = 𝑜𝑜1(𝑥𝑥)𝑦𝑦1(𝑥𝑥) + 𝑜𝑜2(𝑥𝑥)𝑦𝑦2(𝑥𝑥)+. . . +𝑜𝑜𝑛𝑛(𝑥𝑥)𝑦𝑦𝑛𝑛 (𝑥𝑥)
,where the 𝑜𝑜𝑘𝑘′ , 𝑘𝑘 = 1,2, … ,𝑛𝑛 are determined by the 𝑛𝑛 eqn.
𝑦𝑦1𝑜𝑜1′ + 𝑦𝑦2𝑜𝑜2′ + ⋯+ 𝑦𝑦𝑛𝑛𝑜𝑜𝑛𝑛′ = 0 𝑦𝑦1′𝑜𝑜1′ + 𝑦𝑦2′𝑜𝑜2′ + ⋯+ 𝑦𝑦𝑛𝑛′𝑜𝑜𝑛𝑛′ = 0
⋮ ⋮
𝑦𝑦1(𝑛𝑛−1)𝑜𝑜1′ + 𝑦𝑦2
(𝑛𝑛−1)𝑜𝑜2′ + ⋯+ 𝑦𝑦𝑛𝑛𝑛𝑛−1 𝑜𝑜𝑛𝑛′ = 𝑓𝑓(𝑥𝑥)
Variation of Parameters
9/30/2014 Dr. Eli Saber 131
⇒
𝑦𝑦1 𝑦𝑦2 … 𝑦𝑦𝑛𝑛𝑦𝑦1𝑦 𝑦𝑦2′ … 𝑦𝑦𝑛𝑛𝑦⋮ ⋮ ⋮ ⋮
𝑦𝑦1𝑛𝑛−1 𝑦𝑦2
𝑛𝑛−1 … 𝑦𝑦𝑛𝑛𝑛𝑛−1
𝑜𝑜1′
𝑜𝑜2′⋮𝑜𝑜𝑛𝑛′
=
00⋮
𝑓𝑓(𝑥𝑥)
And, 𝑜𝑜𝑘𝑘′ =𝑊𝑊𝑘𝑘𝑊𝑊 ; 𝑘𝑘 = 1,2, … ,𝑛𝑛
Where, 𝑊𝑊1 =
0 𝑦𝑦2 ⋯ 𝑦𝑦𝑛𝑛0 𝑦𝑦2′ ⋯ 𝑦𝑦𝑛𝑛′⋮ ⋮ ⋮ ⋮
𝑓𝑓(𝑥𝑥) 𝑦𝑦2(𝑛𝑛−1) ⋯ 𝑦𝑦𝑛𝑛 (𝑛𝑛−1)
𝑜𝑜𝑘𝑘 can be computed by integrating 𝑜𝑜𝑘𝑘′ ;𝑘𝑘 = 1,2, … ,𝑛𝑛 𝑦𝑦𝑝𝑝 = 𝑜𝑜1(𝑥𝑥)𝑦𝑦1(𝑥𝑥) + 𝑜𝑜2(𝑥𝑥)𝑦𝑦2(𝑥𝑥)+. . . +𝑜𝑜𝑛𝑛(𝑥𝑥)𝑦𝑦𝑛𝑛
(𝑥𝑥)
Variation of Parameters
9/30/2014 Dr. Eli Saber 132
Summary
Given 𝑎𝑎2𝑦𝑦𝑦𝑦 + 𝑎𝑎1𝑦𝑦𝑦 + 𝑎𝑎0𝑦𝑦 = 𝑔𝑔(𝑥𝑥) 1. Put Eq. into standard form by dividing throughout by a2(x)
𝑦𝑦′′ +𝑎𝑎1𝑎𝑎2
𝑦𝑦′ +𝑎𝑎0𝑎𝑎2𝑦𝑦 =
𝑔𝑔 𝑥𝑥𝑎𝑎2(𝑥𝑥)
2. Find 𝑦𝑦𝑐𝑐
= 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2
Verify 𝑦𝑦𝑐𝑐
for the D.E.
3. Compute 𝑊𝑊 =𝑦𝑦1 𝑦𝑦2𝑦𝑦1′ 𝑦𝑦2′
4. Compute 𝑜𝑜1’ & 𝑜𝑜2’ using:
𝑜𝑜1′ =
0 𝑦𝑦2𝑓𝑓(𝑥𝑥) 𝑦𝑦2𝑦𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2𝑦
;𝑜𝑜2′ =
𝑦𝑦1 0𝑦𝑦1𝑦 𝑓𝑓(𝑥𝑥)𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2𝑦
5. Find 𝑜𝑜1 & 𝑜𝑜2 by integrating 𝑜𝑜1′ & 𝑜𝑜2′ respectively. 6. Form 𝑦𝑦𝑝𝑝 = 𝑜𝑜1(𝑥𝑥)𝑦𝑦1(𝑥𝑥) + 𝑜𝑜2(𝑥𝑥)𝑦𝑦2(𝑥𝑥) 7. General Solution: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝 verify the solution for the D.E.
Section 3.6 Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 133
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 134
Any linear Differential Equation of the form:
𝑎𝑎𝑛𝑛𝑥𝑥𝑛𝑛𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛
+ 𝑎𝑎𝑛𝑛−1𝑥𝑥𝑛𝑛−1 𝑑𝑑𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1
+ … + 𝑎𝑎1𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑎𝑎0 𝑦𝑦 = 𝑔𝑔(𝑥𝑥)
, where 𝑎𝑎𝑛𝑛,𝑎𝑎𝑛𝑛−1, … ,𝑎𝑎1,𝑎𝑎0 are constants
And the degree 𝑛𝑛 at 𝑥𝑥𝑛𝑛 matches the order 𝑛𝑛 of the differentiation 𝑑𝑑𝑛𝑛𝑦𝑦
𝑑𝑑𝑥𝑥𝑛𝑛
is called a Cauchy-Euler Equation E.g.
1) 𝑥𝑥2𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
− 2𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
− 4𝑦𝑦 = 0
2) 𝑥𝑥2𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
− 3𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥
same same
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 135
General 2nd order: 𝑎𝑎𝑥𝑥2 𝑑𝑑2𝑦𝑦
𝑑𝑑𝑥𝑥2+ 𝑏𝑏𝑥𝑥 𝑑𝑑𝑦𝑦
𝑑𝑑𝑥𝑥+ 𝑐𝑐𝑦𝑦 = 0
Proceed to develop solution for 2nd order and then generalize.
𝑎𝑎𝑥𝑥2𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 + 𝑏𝑏𝑥𝑥
𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 + 𝑐𝑐𝑦𝑦 = 0
Note: 𝑎𝑎𝑥𝑥2 = 0 @ 𝑥𝑥 = 0 confine attention to interval 𝐼𝐼 ≡ 0,∞ For (−∞, 0), let 𝑡𝑡 = −𝑥𝑥
Homogeneous
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 136
Try a solution of the form 𝑦𝑦 = 𝑥𝑥𝑚𝑚
⇒𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
= 𝑚𝑚𝑥𝑥𝑚𝑚−1 &𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
= 𝑚𝑚 𝑚𝑚− 1 𝑥𝑥𝑚𝑚−2
⇒ 𝑎𝑎𝑥𝑥2 𝑚𝑚 𝑚𝑚− 1 𝑥𝑥𝑚𝑚−2 + 𝑏𝑏𝑥𝑥 𝑚𝑚𝑥𝑥𝑚𝑚−1 + 𝑐𝑐𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑎𝑎𝑚𝑚 𝑚𝑚− 1 𝑥𝑥𝑚𝑚 + 𝑏𝑏𝑚𝑚𝑥𝑥𝑚𝑚 + 𝑐𝑐𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑎𝑎𝑚𝑚 𝑚𝑚 − 1 + 𝑏𝑏𝑚𝑚 + 𝑐𝑐 𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑎𝑎𝑚𝑚2 − 𝑎𝑎𝑚𝑚 + 𝑏𝑏𝑚𝑚 + 𝑐𝑐 𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑎𝑎𝑚𝑚2 + 𝑏𝑏 − 𝑎𝑎 𝑚𝑚 + 𝑐𝑐 𝑥𝑥𝑚𝑚 = 0 Thus, 𝒚𝒚 = 𝟐𝟐𝒎𝒎 is a solution of the D.E. whenever 𝑚𝑚 is a solution to the auxiliary equation
𝒂𝒂𝒎𝒎𝟐𝟐 + 𝒃𝒃 − 𝒂𝒂 𝒎𝒎 + 𝒄𝒄 = 𝟎𝟎
𝑎𝑎𝑥𝑥2𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
+ 𝑏𝑏𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑐𝑐𝑦𝑦 = 0
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 137
Case 1: Distinct Real Roots If 𝑚𝑚1 & 𝑚𝑚2 are the real roots of 𝑎𝑎𝑚𝑚2 + 𝑏𝑏 − 𝑎𝑎 𝑚𝑚 + 𝑐𝑐 = 0 with 𝑚𝑚1 ≠ 𝑚𝑚2 ⇒ 𝑦𝑦1 = 𝑥𝑥𝑚𝑚1 & 𝑦𝑦2 = 𝑥𝑥𝑚𝑚2 form a fundamental set of solutions and the general solution is 𝑦𝑦 = 𝑐𝑐1𝑥𝑥𝑚𝑚1 + 𝑐𝑐2𝑥𝑥𝑚𝑚2 General case: 𝑦𝑦 = 𝑐𝑐1𝑥𝑥𝑚𝑚1 + 𝑐𝑐2𝑥𝑥𝑚𝑚2 + ⋯+ 𝑐𝑐𝑛𝑛𝑥𝑥𝑚𝑚𝑛𝑛 𝑛𝑛𝑡𝑡𝑡 order
Case 1: Distinct Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 138
E.g. 𝑥𝑥2 𝑑𝑑2𝑦𝑦
𝑑𝑑𝑥𝑥2− 2𝑥𝑥 𝑑𝑑𝑦𝑦
𝑑𝑑𝑥𝑥 − 4𝑦𝑦 = 0
Assume 𝑦𝑦 = 𝑥𝑥𝑚𝑚 as the solution. ⇒ 𝑥𝑥2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚−2 − 2𝑥𝑥 𝑚𝑚𝑥𝑥𝑚𝑚−1 − 4𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚 − 2𝑚𝑚 𝑥𝑥𝑚𝑚 − 4𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑚𝑚 𝑚𝑚 − 1 − 2𝑚𝑚 − 4 𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑚𝑚2 −𝑚𝑚 − 2𝑚𝑚 − 4 𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑚𝑚2 − 3𝑚𝑚 − 4 𝑥𝑥𝑚𝑚 = 0
𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 = 𝑚𝑚𝑥𝑥𝑚𝑚−1
𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 = 𝑚𝑚 𝑚𝑚− 1 𝑥𝑥𝑚𝑚−2
Auxiliary Equation
Case 1: Distinct Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 139
𝑚𝑚2 − 3𝑚𝑚 − 4 = 0 ⇒ 𝑚𝑚 + 1 𝑚𝑚 − 4 = 0 ⇒ 𝑚𝑚 = −1 𝑓𝑓𝑓𝑓 𝑚𝑚 = 4 Hence,
𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐−𝟏𝟏 + 𝒄𝒄𝟐𝟐𝟐𝟐𝟒𝟒
Case 1: Distinct Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 140
Case 2: Repeated Real Roots If the roots are repeated i.e. 𝑚𝑚1 = 𝑚𝑚2, only one solution: 𝒚𝒚 = 𝟐𝟐𝒎𝒎𝟏𝟏 ⇒ 𝑎𝑎𝑚𝑚2 + 𝑏𝑏 − 𝑎𝑎 𝑚𝑚 + 𝑐𝑐 = 0
⇒ 𝑚𝑚 =− 𝑏𝑏 − 𝑎𝑎 ± 𝑏𝑏 − 𝑎𝑎 2 − 4𝑎𝑎𝑐𝑐
2𝑎𝑎
For 𝑚𝑚1 = 𝑚𝑚2,⇒ 𝑏𝑏 − 𝑎𝑎 2 − 4𝑎𝑎𝑐𝑐 = 0 ⇒ 𝑏𝑏 − 𝑎𝑎 2 = 4𝑎𝑎𝑐𝑐
Hence, 𝑚𝑚1 = 𝑚𝑚2 = − (𝑏𝑏−𝑎𝑎)2𝑎𝑎
Case 2: Repeated Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 141
Construct a second solution like Section 3.2.
𝑎𝑎𝑥𝑥2𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
+ 𝑏𝑏𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑐𝑐𝑦𝑦 = 0
⇒𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 +
𝑏𝑏𝑥𝑥𝑎𝑎𝑥𝑥2
𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 +
𝑐𝑐𝑎𝑎𝑥𝑥2 𝑦𝑦 = 0
⇒𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 +
𝑏𝑏𝑎𝑎𝑥𝑥
𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 +
𝑐𝑐𝑎𝑎𝑥𝑥2 𝑦𝑦 = 0
Let 𝑦𝑦 = 𝑜𝑜 𝑥𝑥 𝑦𝑦1 𝑥𝑥 ⇒ 𝑦𝑦′ = 𝑜𝑜𝑦𝑦1′ + 𝑜𝑜′𝑦𝑦1 & 𝑦𝑦′′ = 𝑜𝑜𝑦𝑦′′ + 2𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′′𝑦𝑦1
Case 2: Repeated Real Roots
𝑷𝑷(𝟐𝟐) Q(𝟐𝟐)
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 142
𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
+𝑏𝑏𝑎𝑎𝑥𝑥
𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+𝑐𝑐𝑎𝑎𝑥𝑥2
𝑦𝑦 = 0
Replace 𝑦𝑦1,𝑦𝑦1′ ,𝑦𝑦1′ 𝑦
⇒ 𝑜𝑜𝑦𝑦1′′ + 2𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′′𝑦𝑦1 +𝑏𝑏𝑎𝑎𝑥𝑥 𝑜𝑜𝑦𝑦1′ + 𝑜𝑜′𝑦𝑦1 +
𝑐𝑐𝑎𝑎𝑥𝑥2 𝑜𝑜𝑦𝑦1 = 0
⇒ 𝑜𝑜 𝑦𝑦1′′ +𝑏𝑏𝑎𝑎𝑥𝑥 𝑦𝑦1
′ +𝑐𝑐𝑎𝑎𝑥𝑥2 𝑦𝑦1 + 𝑦𝑦1𝑜𝑜′′ + 2𝑦𝑦1′ +
𝑏𝑏𝑎𝑎𝑥𝑥 𝑦𝑦1 𝑜𝑜′ = 0
= 𝑦𝑦1𝑜𝑜′′ + 2𝑦𝑦′ +𝑏𝑏𝑎𝑎𝑥𝑥 𝑦𝑦1 𝑜𝑜′ = 0
Case 2: Repeated Real Roots
𝑦𝑦1 = 𝑥𝑥𝑚𝑚1 𝑦𝑦 = 𝑜𝑜𝑦𝑦1
𝑦𝑦′ = 𝑜𝑜𝑦𝑦1′ + 𝑜𝑜′𝑦𝑦1 𝑦𝑦′′ = 𝑜𝑜𝑦𝑦′′ + 2𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′′𝑦𝑦1
=0 since 𝒚𝒚𝟏𝟏 = 𝟐𝟐𝒎𝒎 is a solution
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 143
𝑦𝑦1𝑜𝑜′′ + 2𝑦𝑦′ +𝑏𝑏𝑎𝑎𝑥𝑥
𝑦𝑦1 𝑜𝑜′ = 0
Let 𝑤𝑤 = 𝑜𝑜′ ⇒ 𝑦𝑦1𝑤𝑤′ + 2𝑦𝑦1′ + 𝑏𝑏𝑎𝑎𝑥𝑥
𝑦𝑦1 𝑤𝑤 = 0
⇒ 𝑦𝑦1𝑑𝑑𝑤𝑤𝑑𝑑𝑥𝑥 = − 2 𝑦𝑦1′ +
𝑏𝑏𝑎𝑎𝑥𝑥 𝑦𝑦1 𝑤𝑤
⇒𝑑𝑑𝑤𝑤𝑤𝑤 = −
1𝑦𝑦1
2𝑦𝑦1′ +𝑏𝑏𝑎𝑎𝑥𝑥 𝑦𝑦1 𝑑𝑑𝑥𝑥
⇒𝑑𝑑𝑤𝑤𝑤𝑤 = −2
𝑦𝑦1′
𝑦𝑦1 𝑑𝑑𝑥𝑥 −
𝑏𝑏𝑎𝑎𝑥𝑥 𝑑𝑑𝑥𝑥
𝑦𝑦1 = 𝑥𝑥𝑚𝑚1 𝑦𝑦 = 𝑜𝑜𝑦𝑦1
𝑦𝑦′ = 𝑜𝑜𝑦𝑦1′ + 𝑜𝑜′𝑦𝑦1 𝑦𝑦′′ = 𝑜𝑜𝑦𝑦′′ + 2𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′′𝑦𝑦1
Case 2: Repeated Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 144
⇒𝑑𝑑𝑤𝑤𝑤𝑤
= −2𝑚𝑚1𝑥𝑥𝑚𝑚1−1
𝑥𝑥𝑚𝑚1 𝑑𝑑𝑥𝑥 −
𝑏𝑏𝑎𝑎𝑥𝑥
𝑑𝑑𝑥𝑥
⇒ �𝑑𝑑𝑤𝑤𝑤𝑤
= �−2𝑚𝑚1𝑥𝑥 𝑑𝑑𝑥𝑥 − �
𝑏𝑏𝑎𝑎
1𝑥𝑥
𝑑𝑑𝑥𝑥
⇒ ln |𝑤𝑤| = −2 𝑚𝑚1 ln 𝑥𝑥 −𝑏𝑏𝑎𝑎 ln 𝑥𝑥 + 𝑐𝑐
⇒ ln |𝑤𝑤| + 2 𝑚𝑚1 ln 𝑥𝑥 +𝑏𝑏𝑎𝑎 ln 𝑥𝑥 = 𝑐𝑐
⇒ ln 𝑤𝑤 + ln 𝑥𝑥 2𝑚𝑚1 + ln 𝑥𝑥𝑏𝑏𝑎𝑎 = 𝑐𝑐
⇒ ln 𝑤𝑤𝑥𝑥2𝑚𝑚1𝑥𝑥𝑏𝑏𝑎𝑎 = 𝑐𝑐
Case 2: Repeated Real Roots
𝑑𝑑𝑤𝑤𝑤𝑤
= −2𝑦𝑦1′
𝑦𝑦1 𝑑𝑑𝑥𝑥 −
𝑏𝑏𝑎𝑎𝑥𝑥
𝑑𝑑𝑥𝑥
𝑦𝑦1 = 𝑥𝑥𝑚𝑚1
⇒ 𝑦𝑦1′ = 𝑚𝑚1𝑥𝑥𝑚𝑚1−1
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 145
⇒ ln 𝑤𝑤𝑥𝑥2𝑚𝑚1𝑥𝑥𝑏𝑏𝑎𝑎 = 𝑐𝑐
⇒ 𝑤𝑤𝑥𝑥2𝑚𝑚1𝑥𝑥𝑏𝑏𝑎𝑎 = 𝑒𝑒𝑐𝑐
But 𝑤𝑤 = 𝑜𝑜′ ⇒ 𝑜𝑜′𝑥𝑥2𝑚𝑚1𝑥𝑥𝑏𝑏𝑎𝑎 = 𝑒𝑒𝑐𝑐
⇒ 𝑜𝑜′ = 𝑒𝑒𝑐𝑐𝑥𝑥−2𝑚𝑚1𝑥𝑥− 𝑏𝑏𝑎𝑎
⇒ 𝑜𝑜 = �𝑒𝑒𝑐𝑐𝑥𝑥−2𝑚𝑚1𝑥𝑥− 𝑏𝑏𝑎𝑎 𝑑𝑑𝑥𝑥
Now, 𝑦𝑦2 = 𝑜𝑜𝑦𝑦1 = 𝑥𝑥𝑚𝑚1 ∫ 𝑒𝑒𝑐𝑐𝑥𝑥𝒃𝒃−𝒂𝒂𝒂𝒂 𝑥𝑥− 𝑏𝑏𝑎𝑎 𝑑𝑑𝑥𝑥
= 𝑥𝑥𝑚𝑚1 �𝑒𝑒𝑐𝑐 𝑥𝑥𝑏𝑏𝑎𝑎−1−
𝑏𝑏𝑎𝑎 𝑑𝑑𝑥𝑥 = 𝑥𝑥𝑚𝑚1 �𝑒𝑒𝑐𝑐 𝑥𝑥−1𝑑𝑑𝑥𝑥
Case 2: Repeated Real Roots
𝑚𝑚1 = −𝑏𝑏 − 𝑎𝑎2𝑎𝑎
⇒ 2𝑚𝑚1 = −𝑏𝑏 − 𝑎𝑎𝑎𝑎
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 146
𝑦𝑦2 = 𝑥𝑥𝑚𝑚1 �𝑒𝑒𝑐𝑐 𝑥𝑥−1𝑑𝑑𝑥𝑥
𝑦𝑦2 = 𝑒𝑒𝑐𝑐𝑥𝑥𝑚𝑚1 ln 𝑥𝑥 = 𝑐𝑐2𝑥𝑥𝑚𝑚1 ln 𝑥𝑥 General solution: 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐𝒎𝒎𝟏𝟏 + 𝒄𝒄𝟐𝟐𝟐𝟐𝒎𝒎𝟏𝟏 𝒓𝒓𝒏𝒏𝟐𝟐
Case 2: Repeated Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 147
E.g. 4𝑥𝑥2𝑦𝑦′′ + 8𝑥𝑥𝑦𝑦′ + 𝑦𝑦 = 0
Let 𝑦𝑦 = 𝑥𝑥𝑚𝑚 ⇒ 𝑦𝑦′ = 𝑚𝑚𝑥𝑥𝑚𝑚−1 ⇒ 𝑦𝑦′′ = 𝑚𝑚 𝑚𝑚− 1 𝑥𝑥𝑚𝑚−2 ⇒ 4𝑥𝑥2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚−2 + 8𝑥𝑥 𝑚𝑚𝑥𝑥𝑚𝑚−1 + 𝑥𝑥𝑚𝑚 = 0 ⇒ 4𝑚𝑚 𝑚𝑚− 1 𝑥𝑥𝑚𝑚 + 8𝑚𝑚 𝑥𝑥𝑚𝑚 + 𝑥𝑥𝑚𝑚 = 0 ⇒ 4𝑚𝑚2 − 4𝑚𝑚 + 8𝑚𝑚 + 1 𝑥𝑥𝑚𝑚 = 0 ⇒ 4𝑚𝑚2 + 4𝑚𝑚 + 1 = 0
⇒ 2𝑚𝑚 + 1 2 = 0 ⇒ 𝑚𝑚 = −12
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐− 𝟏𝟏𝟐𝟐 + 𝒄𝒄𝟐𝟐𝟐𝟐
− 𝟏𝟏𝟐𝟐 𝐥𝐥𝐥𝐥 𝟐𝟐
Case 2: Repeated Real Roots
𝑦𝑦 = 𝑐𝑐1𝑥𝑥𝑚𝑚1 + 𝑐𝑐2𝑥𝑥𝑚𝑚1 𝑔𝑔𝑛𝑛 𝑥𝑥
Repeated roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 148
Note: For higher order equations, if 𝑚𝑚1 is a root of multiplicity 𝐾𝐾
⇒ 𝑥𝑥𝑚𝑚1 , 𝑥𝑥𝑚𝑚1 ln𝑥𝑥 , 𝑥𝑥𝑚𝑚1 ln𝑥𝑥 2, … , 𝑥𝑥𝑚𝑚1 ln 𝑥𝑥 𝑘𝑘−1 are 𝐾𝐾 linearly independent solutions
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐𝒎𝒎𝟏𝟏 + 𝒄𝒄𝟐𝟐𝟐𝟐𝒎𝒎𝟏𝟏 𝒓𝒓𝒏𝒏𝟐𝟐 + 𝒄𝒄𝟑𝟑𝟐𝟐𝒎𝒎𝟏𝟏 𝒓𝒓𝒏𝒏𝟐𝟐 𝟐𝟐 + ⋯+ 𝒄𝒄𝒓𝒓𝟐𝟐𝒎𝒎𝟏𝟏 𝒓𝒓𝒏𝒏𝟐𝟐 𝒓𝒓−𝟏𝟏
Case 2: Repeated Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 149
Case 3: Conjugate Complex Roots If the roots are conjugate pairs i.e. 𝑚𝑚1 = 𝛼𝛼 + 𝑗𝑗𝛽𝛽 & 𝑚𝑚2 = 𝛼𝛼 − 𝑗𝑗𝛽𝛽 (𝛼𝛼,𝛽𝛽 > 0)
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐𝜶𝜶+𝒋𝒋𝜷𝜷 + 𝒄𝒄𝟐𝟐𝟐𝟐𝜶𝜶−𝒋𝒋𝜷𝜷 We can rewrite that in terms of 𝑐𝑐𝑓𝑓𝑠𝑠 & 𝑠𝑠𝑠𝑠𝑛𝑛 as:
𝑥𝑥𝑗𝑗𝑗𝑗 = 𝑒𝑒ln 𝑥𝑥 𝑗𝑗𝑗𝑗 = 𝑒𝑒ln 𝑥𝑥𝑗𝑗𝑗𝑗 = cos 𝛽𝛽 ln 𝑥𝑥 + 𝑗𝑗 sin(𝛽𝛽 ln 𝑥𝑥)
𝑥𝑥−𝑗𝑗𝑗𝑗 = 𝑒𝑒ln 𝑥𝑥 −𝑗𝑗𝑗𝑗 = 𝑒𝑒− ln 𝑥𝑥𝑗𝑗𝑗𝑗 = cos 𝛽𝛽 ln 𝑥𝑥 − 𝑗𝑗 sin(𝛽𝛽 ln 𝑥𝑥)
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 150
We have, 𝑦𝑦 = 𝑐𝑐1𝑥𝑥𝛼𝛼+𝑗𝑗𝑗𝑗 + 𝑐𝑐2𝑥𝑥𝛼𝛼−𝑗𝑗𝑗𝑗 ⇒ 𝑦𝑦 = 𝑐𝑐1𝑥𝑥𝛼𝛼 𝑐𝑐𝑓𝑓𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 + 𝑗𝑗 𝑠𝑠𝑠𝑠𝑛𝑛(𝛽𝛽 ln 𝑥𝑥) + 𝑐𝑐2𝑥𝑥𝛼𝛼 𝑐𝑐𝑓𝑓𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 − 𝑗𝑗 𝑠𝑠𝑠𝑠𝑛𝑛(𝛽𝛽 ln 𝑥𝑥)
= 𝑐𝑐1𝑥𝑥𝛼𝛼 𝑐𝑐𝑓𝑓𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 + 𝑗𝑗𝑐𝑐1𝑥𝑥𝛼𝛼 𝑠𝑠𝑠𝑠𝑛𝑛 𝛽𝛽 ln 𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝛼𝛼 𝑐𝑐𝑓𝑓𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 − 𝑗𝑗𝑐𝑐2𝑥𝑥𝛼𝛼 𝑠𝑠𝑠𝑠𝑛𝑛 𝛽𝛽 ln 𝑥𝑥 = 𝑥𝑥𝛼𝛼 𝑐𝑐1 𝑐𝑐𝑓𝑓𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 + 𝑐𝑐2 𝑐𝑐𝑓𝑓𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 + 𝑗𝑗𝑐𝑐1 𝑠𝑠𝑠𝑠𝑛𝑛 𝛽𝛽 ln 𝑥𝑥 − 𝑗𝑗𝑐𝑐2 𝑠𝑠𝑠𝑠𝑛𝑛 𝛽𝛽 ln 𝑥𝑥 = 𝑥𝑥𝛼𝛼 𝑐𝑐𝑓𝑓𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 {𝑐𝑐1 + 𝑐𝑐2} + 𝑗𝑗{𝑐𝑐1 − 𝑐𝑐2} 𝑠𝑠𝑠𝑠𝑛𝑛 𝛽𝛽 ln 𝑥𝑥 𝒚𝒚 = 𝟐𝟐𝜶𝜶 ∝𝟏𝟏 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷 𝒓𝒓𝒏𝒏𝟐𝟐 +∝𝟐𝟐 𝒄𝒄𝒔𝒔𝒏𝒏 𝜷𝜷 𝒓𝒓𝒏𝒏𝟐𝟐
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 151
E.g.
4𝑥𝑥2𝑦𝑦′′ + 17𝑦𝑦 = 0 with I.C. 𝑦𝑦 1 = −1; 𝑦𝑦′ 1 = −12
Let 𝑦𝑦 = 𝑥𝑥𝑚𝑚 ⇒ 𝑦𝑦′ = 𝑚𝑚𝑥𝑥𝑚𝑚−1 ⇒ 𝑦𝑦′′ = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚−2 ⇒ 4𝑥𝑥2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚−2 + 17𝑥𝑥𝑚𝑚 = 0 ⇒ 4𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚 + 17𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑥𝑥𝑚𝑚 4𝑚𝑚2 − 4𝑚𝑚 + 17 = 0 Auxiliary Eqn. : 𝟒𝟒𝒎𝒎𝟐𝟐 − 𝟒𝟒𝒎𝒎 + 𝟏𝟏𝟐𝟐 = 𝟎𝟎
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 152
Auxiliary Eqn. : 4𝑚𝑚2 − 4𝑚𝑚 + 17 = 0
𝑚𝑚 =− −4 ± 16 − 4(4)(17)
8=
4 ± 16 − 2728
=4 ± 256 𝑗𝑗2
8
𝑚𝑚 =4 ± 4 ∗ 64 𝑗𝑗2
8 =4 ± 𝑗𝑗2(8)
8 =12 ± 2𝑗𝑗
⇒ 𝒎𝒎𝟏𝟏 =𝟏𝟏𝟐𝟐 + 𝟐𝟐𝒋𝒋 & 𝒎𝒎𝟐𝟐 =
𝟏𝟏𝟐𝟐 − 𝟐𝟐𝒋𝒋
𝛼𝛼 =12 & 𝛽𝛽 = 2
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐+𝟐𝟐𝒋𝒋 + 𝒄𝒄𝟐𝟐𝟐𝟐
𝟏𝟏𝟐𝟐−𝟐𝟐𝒋𝒋 𝒄𝒄𝒓𝒓 𝒚𝒚 = 𝟐𝟐
𝟏𝟏𝟐𝟐 ∝𝟏𝟏 𝒄𝒄𝒄𝒄𝒄𝒄 𝟐𝟐 𝒓𝒓𝒏𝒏 𝟐𝟐 +∝𝟐𝟐 𝒄𝒄𝒔𝒔𝒏𝒏 𝟐𝟐 𝒓𝒓𝒏𝒏 𝟐𝟐
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 153
Now,
I.C. 𝑦𝑦 1 = −1;𝑦𝑦′ 1 = −12
𝑦𝑦 1 = −1 ⇒ −1 = 112 ∝1 𝑐𝑐𝑓𝑓𝑠𝑠 2 𝑔𝑔𝑛𝑛 1 +∝2 𝑠𝑠𝑠𝑠𝑛𝑛 2 𝑔𝑔𝑛𝑛 1
⇒ −1 = 1 ∝1 1 +∝2 0 ⇒ ∝1= −1
𝑦𝑦′ 1 = −12
𝑦𝑦′ =∝112 𝑥𝑥
−12 cos 2 ln 𝑥𝑥 + 𝑥𝑥12 − sin 2 ln 𝑥𝑥
2𝑥𝑥 +
∝212𝑥𝑥−
12 sin 2 ln 𝑥𝑥 + 𝑥𝑥
12 cos 2 ln 𝑥𝑥 2
𝑥𝑥
𝑦𝑦 = 𝑥𝑥12 ∝1 𝑐𝑐𝑓𝑓𝑠𝑠 2 𝑔𝑔𝑛𝑛 𝑥𝑥 +∝2 𝑠𝑠𝑠𝑠𝑛𝑛 2 𝑔𝑔𝑛𝑛 𝑥𝑥
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 154
𝑦𝑦′ =∝112𝑥𝑥−
12 cos 2 ln 𝑥𝑥 + 𝑥𝑥
12 − sin 2 ln 𝑥𝑥
2𝑥𝑥
+
∝212𝑥𝑥−
12 sin 2 ln 𝑥𝑥 + 𝑥𝑥
12 cos 2 ln 𝑥𝑥 2
𝑥𝑥
⇒ −12
= −1 1
2 112
cos 0 + 112 − sin 0
21
+
+∝2 1
2 112
sin 0 + 112 cos 0 2
1
⇒ −12
= −112
+∝2⇒∝2= −12
+12
= 0 ⇒∝2= 0
⇒ 𝒚𝒚 = −𝟐𝟐𝟏𝟏𝟐𝟐𝐜𝐜𝐜𝐜𝐜𝐜 (𝟐𝟐 𝐥𝐥𝐥𝐥 𝟐𝟐)
𝑦𝑦 = 𝑥𝑥12 ∝1 𝑐𝑐𝑓𝑓𝑠𝑠 2 𝑔𝑔𝑛𝑛 𝑥𝑥 +∝2 𝑠𝑠𝑠𝑠𝑛𝑛 2 𝑔𝑔𝑛𝑛 𝑥𝑥
∝1= −1
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 155
E.g.
𝑥𝑥3𝑑𝑑3𝑦𝑦𝑑𝑑𝑥𝑥3
+ 5𝑥𝑥2𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2
+ 7𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 8𝑦𝑦 = 0
Assume 𝑦𝑦 = 𝑥𝑥𝑚𝑚 ⇒ 𝑦𝑦′ = 𝑚𝑚𝑥𝑥𝑚𝑚−1 ⇒ 𝑦𝑦′′ = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚−2 ⇒ 𝑦𝑦′′′ = 𝑚𝑚 𝑚𝑚− 1 𝑚𝑚− 2 𝑥𝑥𝑚𝑚−3 ⇒ 𝑥𝑥3 𝑚𝑚 𝑚𝑚 − 1 𝑚𝑚− 2 𝑥𝑥𝑚𝑚−3 + 5𝑥𝑥2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚−2 +7𝑥𝑥 𝑚𝑚𝑥𝑥𝑚𝑚−1 + 8𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚 𝑚𝑚 − 1 𝑚𝑚− 2 + 5𝑚𝑚 𝑚𝑚 − 1 + 7𝑚𝑚 + 8 = 0 ⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚 𝑚𝑚2 − 3𝑚𝑚 + 2 + 5𝑚𝑚2 − 5𝑚𝑚 + 7𝑚𝑚 + 8 = 0 ⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚3 − 3𝑚𝑚2 + 2𝑚𝑚 + 5𝑚𝑚2 + 2𝑚𝑚 + 8 = 0 ⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚3 + 2𝑚𝑚2 + 4𝑚𝑚 + 8 = 0
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 156
⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚3 + 2𝑚𝑚2 + 4𝑚𝑚 + 8 = 0 ⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚 + 2 𝑚𝑚2 + 4 = 0 𝑚𝑚 + 2 𝑚𝑚2 + 4 = 0
⇒ 𝑚𝑚 + 2 𝑚𝑚 + 2𝑗𝑗 𝑚𝑚 − 2𝑗𝑗 = 0 ⇒ 𝑚𝑚1 = −2,𝑚𝑚2 = −2𝑗𝑗,𝑚𝑚3 = 2𝑗𝑗 Solution: 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐−𝟐𝟐 + 𝒄𝒄𝟐𝟐𝟐𝟐𝟐𝟐𝒋𝒋 + 𝒄𝒄𝟑𝟑𝟐𝟐−𝟐𝟐𝒋𝒋 or 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐−𝟐𝟐 + 𝒄𝒄𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜(𝟐𝟐 𝐥𝐥𝐥𝐥 𝟐𝟐) + 𝒄𝒄𝟑𝟑 𝐜𝐜𝐬𝐬𝐥𝐥(𝟐𝟐 𝐥𝐥𝐥𝐥 𝟐𝟐)
Case 3: Conjugate Complex Roots
Auxiliary Equation
𝑚𝑚2 + 4 = 0 ⇒ 𝑚𝑚2 = −4 ⇒ 𝑚𝑚2 = 4𝑗𝑗2 ⇒ 𝑚𝑚 = ±2𝑗𝑗
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 157
E.g. 𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥
• Non Homogeneous Eqn. solve associated Homogeneous Eqn.
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 0 Assume 𝑦𝑦 = 𝑥𝑥𝑚𝑚 ⇒ 𝑦𝑦′ = 𝑚𝑚𝑥𝑥𝑚𝑚−1 ⇒ 𝑦𝑦′′ = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚−2 ⇒ 𝑥𝑥2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥𝑚𝑚−2 − 3𝑥𝑥 𝑚𝑚𝑥𝑥𝑚𝑚−1 + 3𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑚𝑚2 −𝑚𝑚 𝑥𝑥𝑚𝑚 − 3𝑚𝑚𝑥𝑥𝑚𝑚 + 3𝑥𝑥𝑚𝑚 = 0 ⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚2 −𝑚𝑚 − 3𝑚𝑚 + 3 = 0 ⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚2 − 4𝑚𝑚 + 3 = 0
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 158
⇒ 𝑥𝑥𝑚𝑚 𝑚𝑚2 − 4𝑚𝑚 + 3 = 0 Auxiliary Eqn. 𝑚𝑚2 − 4𝑚𝑚 + 3 = 0 ⇒ 𝑚𝑚 − 1 𝑚𝑚 − 3 = 0 ⇒ 𝑚𝑚1 = 1 & 𝑚𝑚2 = 3
⇒ 𝒚𝒚𝒄𝒄 = 𝒄𝒄𝟏𝟏𝟐𝟐 + 𝒄𝒄𝟐𝟐𝟐𝟐𝟑𝟑 • Utilize Variation of Parameters to solve for particular solution 𝑦𝑦𝑝𝑝
𝑦𝑦𝑝𝑝 = 𝑜𝑜1𝑦𝑦1 + 𝑜𝑜2𝑦𝑦2
,where 𝑦𝑦1 = 𝑥𝑥 & 𝑦𝑦2 = 𝑥𝑥3
Case 3: Conjugate Complex Roots
Auxiliary Equation
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 159
Note: To use Variation of Parameters must transform the equation
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥
Divide by 𝑥𝑥2,
𝑦𝑦′′ −3𝑥𝑥𝑥𝑥2 𝑦𝑦′ +
3𝑥𝑥2 𝑦𝑦 =
2𝑥𝑥4𝑒𝑒𝑥𝑥
𝑥𝑥2
⇒ 𝑦𝑦′′ −3𝑥𝑥 𝑦𝑦′ +
3𝑥𝑥2 𝑦𝑦 = 2𝑥𝑥2𝑒𝑒𝑥𝑥
Case 3: Conjugate Complex Roots
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2𝑥𝑥3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥3
𝑷𝑷(𝟐𝟐) 𝑸𝑸(𝟐𝟐) 𝒇𝒇(𝟐𝟐)
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 160
Form all Wronskians: 𝑊𝑊 =
𝑦𝑦1 𝑦𝑦2𝑦𝑦1𝑦 𝑦𝑦2′
= 𝑥𝑥 𝑥𝑥31 3𝑥𝑥2
= 3𝑥𝑥3 − 𝑥𝑥3 = 2𝑥𝑥3
𝑊𝑊1 =0 𝑦𝑦2
𝑓𝑓(𝑥𝑥) 𝑦𝑦2′= 0 𝑥𝑥3
2𝑥𝑥2𝑒𝑒𝑥𝑥 3𝑥𝑥2= −2𝑥𝑥5𝑒𝑒𝑥𝑥
𝑊𝑊2 = 𝑥𝑥 01 2𝑥𝑥2𝑒𝑒𝑥𝑥 = 2𝑥𝑥3𝑒𝑒𝑥𝑥
⇒ 𝑜𝑜1′ =𝑊𝑊1𝑊𝑊 = −
2𝑥𝑥5𝑒𝑒𝑥𝑥
2𝑥𝑥3 = −𝑥𝑥2𝑒𝑒𝑥𝑥
⇒ 𝑜𝑜2′ =𝑊𝑊2𝑊𝑊 =
2𝑥𝑥3𝑒𝑒𝑥𝑥
2𝑥𝑥3 = 𝑒𝑒𝑥𝑥
Case 3: Conjugate Complex Roots
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2𝑥𝑥3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥2𝑒𝑒𝑥𝑥
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 161
Integrate 𝑜𝑜1′ & 𝑜𝑜2′ to get 𝑜𝑜1&𝑜𝑜2 :
𝑜𝑜2 = �𝑜𝑜2′ 𝑑𝑑𝑥𝑥 = �𝑒𝑒𝑥𝑥𝑑𝑑𝑥𝑥 = 𝒆𝒆𝟐𝟐
𝑜𝑜1 = �𝑜𝑜1′ 𝑑𝑑𝑥𝑥 = −�𝑥𝑥2𝑒𝑒𝑥𝑥𝑑𝑑𝑥𝑥
Let 𝛼𝛼 = 𝑥𝑥2 ⇒ 𝑑𝑑𝛼𝛼 = 2𝑥𝑥 𝑑𝑑𝑥𝑥 ;𝑑𝑑𝛽𝛽 = 𝑒𝑒𝑥𝑥𝑑𝑑𝑥𝑥 ⇒ 𝛽𝛽 = 𝑒𝑒𝑥𝑥
⇒ 𝑜𝑜1 = − 𝑥𝑥2𝑒𝑒𝑥𝑥 − �𝑒𝑒𝑥𝑥2𝑥𝑥𝑑𝑑𝑥𝑥
⇒ 𝑜𝑜1 = −𝑥𝑥2𝑒𝑒𝑥𝑥 + 2�𝑒𝑒𝑥𝑥𝑥𝑥𝑑𝑑𝑥𝑥
Case 3: Conjugate Complex Roots
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2𝑥𝑥3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥2𝑒𝑒𝑥𝑥 𝑜𝑜1′ = −𝑥𝑥2𝑒𝑒𝑥𝑥
𝑜𝑜2′ = 𝑒𝑒𝑥𝑥
Integration by parts
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 162
⇒ 𝑜𝑜1 = −𝑥𝑥2𝑒𝑒𝑥𝑥 + 2�𝑒𝑒𝑥𝑥𝑥𝑥𝑑𝑑𝑥𝑥
⇒ 𝑜𝑜1 = −𝑥𝑥2𝑒𝑒𝑥𝑥 + 2 𝑥𝑥𝑒𝑒𝑥𝑥 − 𝑒𝑒𝑥𝑥 ⇒ 𝒖𝒖𝟏𝟏 = −𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐 + 𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐 − 𝟐𝟐𝒆𝒆𝟐𝟐 Now, 𝑦𝑦𝑝𝑝 = 𝑜𝑜1𝑦𝑦1 + 𝑜𝑜2𝑦𝑦2 = −𝑥𝑥2𝑒𝑒𝑥𝑥 + 2𝑥𝑥𝑒𝑒𝑥𝑥 − 2𝑒𝑒𝑥𝑥 𝑥𝑥 + 𝑒𝑒𝑥𝑥𝑥𝑥3 = −𝑥𝑥3𝑒𝑒𝑥𝑥 + 2𝑥𝑥2𝑒𝑒𝑥𝑥 − 2𝑒𝑒𝑥𝑥𝑥𝑥 + 𝑒𝑒𝑥𝑥𝑥𝑥3 = 𝟐𝟐𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐 − 𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐 ⇒ 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏𝟐𝟐 + 𝒄𝒄𝟐𝟐𝟐𝟐𝟑𝟑 + 𝟐𝟐𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐 − 𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐
Case 3: Conjugate Complex Roots
𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2𝑥𝑥3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥2𝑒𝑒𝑥𝑥 𝑜𝑜1′ = −𝑥𝑥2𝑒𝑒𝑥𝑥
𝑜𝑜2 = 𝑒𝑒𝑥𝑥
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 163
Summary
• Identified when 𝟐𝟐𝒏𝒏 matches the order of the differentiation 𝒅𝒅
𝒏𝒏𝒚𝒚𝒅𝒅𝟐𝟐𝒏𝒏
𝑎𝑎𝑛𝑛𝑥𝑥𝑛𝑛𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛
+ … + 𝑎𝑎1𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥
+ 𝑎𝑎0 𝑦𝑦 = 𝑔𝑔(𝑥𝑥)
Step1: Obtain Complementary Solution(𝑦𝑦𝑐𝑐) • Consider 𝒈𝒈 𝟐𝟐 = 𝟎𝟎
• Try the form 𝒚𝒚 = 𝟐𝟐𝒎𝒎
• Auxiliary Equation:
𝒂𝒂𝒎𝒎𝟐𝟐 + 𝒃𝒃 − 𝒂𝒂 𝒎𝒎 + 𝒄𝒄 = 𝟎𝟎
• Obtain roots for the equation – Case 1: Distinct Real Roots – 𝑦𝑦 = 𝑐𝑐1𝑥𝑥𝑚𝑚1 + 𝑐𝑐2𝑥𝑥𝑚𝑚2 + ⋯+ 𝑐𝑐𝑛𝑛𝑥𝑥𝑚𝑚𝑛𝑛
– Case 2: Repeated Real Roots – 𝑦𝑦 = 𝑐𝑐1𝑥𝑥𝑚𝑚1 + 𝑐𝑐2𝑥𝑥𝑚𝑚1 𝑔𝑔𝑛𝑛 𝑥𝑥 +
𝑐𝑐3𝑥𝑥𝑚𝑚1 𝑔𝑔𝑛𝑛 𝑥𝑥 2 + ⋯+𝑐𝑐𝑘𝑘𝑥𝑥𝑚𝑚1 𝑔𝑔𝑛𝑛 𝑥𝑥 𝑘𝑘−1
– Case 3: Conjugate Complex Roots – 𝑦𝑦 =
𝑥𝑥𝛼𝛼 ∝1 𝑐𝑐𝑓𝑓𝑠𝑠 𝛽𝛽 𝑔𝑔𝑛𝑛 𝑥𝑥 +∝2 𝑠𝑠𝑠𝑠𝑛𝑛 𝛽𝛽 𝑔𝑔𝑛𝑛 𝑥𝑥
Step2: Obtain Particular Solution (𝑦𝑦𝑝𝑝) • Use either Undetermined Coefficients
(3.4) or Variation of Parameters (3.5)
Step3: Combine to obtain general solution • 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
Step4: Verify the solution
Section 3.8 Linear Models
9/30/2014 Dr. Eli Saber 164
Linear Models
9/30/2014 Dr. Eli Saber 165
Note: 𝑉𝑉𝑅𝑅 = 𝑅𝑅𝑠𝑠
𝑉𝑉𝐿𝐿 = 𝐿𝐿𝑑𝑑𝑠𝑠𝑑𝑑𝑡𝑡
𝑠𝑠𝑐𝑐 = 𝑐𝑐𝑑𝑑𝑉𝑉𝑐𝑐𝑑𝑑𝑡𝑡
; 𝑠𝑠 =𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
⇒𝑑𝑑𝑠𝑠𝑑𝑑𝑡𝑡
=𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
𝑑𝑑𝑑𝑑 = 𝑠𝑠 𝑑𝑑𝑡𝑡 ⇒ 𝑑𝑑 = ∫ 𝑠𝑠 𝑑𝑑𝑡𝑡 𝑑𝑑 charge
Kirchoff’s Voltage Law:
𝐸𝐸 = 𝑅𝑅𝑠𝑠 + 𝐿𝐿𝑑𝑑𝑠𝑠𝑑𝑑𝑡𝑡
+1𝐶𝐶
�𝑠𝑠 𝑑𝑑𝑡𝑡
C
L
R
E
3.8.4. : Series Circuit (LRC)
𝑠𝑠(𝑡𝑡)
Linear Models
9/30/2014 Dr. Eli Saber 166
𝐸𝐸 = 𝑅𝑅𝑠𝑠 + 𝐿𝐿𝑑𝑑𝑠𝑠𝑑𝑑𝑡𝑡
+1𝐶𝐶
�𝑠𝑠 𝑑𝑑𝑡𝑡
⇒ 𝐸𝐸 = 𝑅𝑅𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
+ 𝐿𝐿𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
+𝑑𝑑𝐶𝐶
⇒ 𝑹𝑹𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡 + 𝑳𝑳
𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2 +
𝟏𝟏𝑪𝑪𝑑𝑑 = 𝐸𝐸
• 𝑠𝑠 = 𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
;𝑉𝑉𝑅𝑅 = 𝑅𝑅𝑠𝑠
• 𝑉𝑉𝐿𝐿 = 𝐿𝐿 𝑑𝑑𝑖𝑖𝑑𝑑𝑡𝑡
• 𝑉𝑉𝐶𝐶 = 1𝐶𝐶 ∫ 𝑠𝑠 𝑑𝑑𝑡𝑡
• 𝐸𝐸(𝑡𝑡): forcing function
C
L
R
E
3.8.4. : Series Circuit (LRC)
𝑠𝑠(𝑡𝑡)
Linear Models
9/30/2014 Dr. Eli Saber 167
𝑅𝑅𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
+ 𝐿𝐿𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
+𝑑𝑑𝐶𝐶
= 𝐸𝐸
Rearranging the equation, we get:
𝐿𝐿𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
+ 𝑅𝑅𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
+1𝐶𝐶𝑑𝑑 = 𝐸𝐸
Auxiliary Eqn: 𝐿𝐿𝑚𝑚2 + 𝑅𝑅𝑚𝑚 + 1𝐶𝐶
= 0
⇒ 𝑚𝑚 =−𝑅𝑅 ± 𝑅𝑅2 − 4 𝐿𝐿 1
𝐶𝐶2𝐿𝐿
⇒ 𝑚𝑚 =−𝑅𝑅 ± 𝑅𝑅2 − 4𝐿𝐿
𝐶𝐶2𝐿𝐿
C
L
R
E
3.8.4. : Series Circuit (LRC)
𝑠𝑠(𝑡𝑡)
(Assume 𝐸𝐸 𝑡𝑡 = 0)
Linear Models
9/30/2014 Dr. Eli Saber 168
𝑚𝑚 =−𝑅𝑅 ± 𝑅𝑅2 − 4𝐿𝐿
𝐶𝐶2𝐿𝐿
• If 𝑹𝑹𝟐𝟐 − 𝟒𝟒𝑳𝑳𝑪𝑪
> 𝟎𝟎 over damped
• If 𝑹𝑹𝟐𝟐 − 𝟒𝟒𝑳𝑳𝑪𝑪
= 𝟎𝟎 critically damped
• If 𝑹𝑹𝟐𝟐 − 𝟒𝟒𝑳𝑳𝑪𝑪
< 𝟎𝟎 under damped
Now,
𝑚𝑚 =−𝑅𝑅 ± 𝑅𝑅2 − 4𝐿𝐿
𝐶𝐶2𝐿𝐿 = −
𝑅𝑅2𝐿𝐿 ±
𝑅𝑅2 − 4𝐿𝐿𝐶𝐶
2𝐿𝐿
3.8.4. : Series Circuit (LRC)
𝜶𝜶 𝜷𝜷
C
L
R
E 𝑠𝑠(𝑡𝑡)
Linear Models
9/30/2014 Dr. Eli Saber 169
E.g. 𝐿𝐿 = 0.25𝐻𝐻;𝑅𝑅 = 10Ω;𝐶𝐶 = 0.001𝐹𝐹 𝐸𝐸 𝑡𝑡 = 0𝑉𝑉; 𝑑𝑑 0 = 𝑑𝑑0; 𝑠𝑠 0 = 0𝐴𝐴 Solution:
𝐿𝐿𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2 + 𝑅𝑅
𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡 +
1𝐶𝐶 𝑑𝑑 = 0 ⇒ 0.25𝑑𝑑′′ + 10𝑑𝑑′ + 1000𝑑𝑑 = 0
⇒ 𝑑𝑑′′ + 40𝑑𝑑′ + 4000𝑑𝑑 = 0 Aux. Eq.: 𝑚𝑚2 + 40𝑚𝑚 + 4000 = 0
𝑚𝑚 =−40 ± 1600 − 4(1)(4000)
2 =−40 ± 1600 − 16000
2
3.8.4. : Series Circuit (LRC)
C= 0.001𝐹𝐹
L= 0.25𝐻𝐻
R=10 Ω
E=0V 𝑠𝑠(𝑡𝑡)
Linear Models
9/30/2014 Dr. Eli Saber 170
𝑚𝑚 =−40 ± −14400
2
𝑚𝑚 =−40 ± 14400𝑗𝑗2
2 =−40 ± 16(900)𝑗𝑗2
2
=−40 ± 4 30 𝑗𝑗
2 = −𝟐𝟐𝟎𝟎 ± 𝟔𝟔𝟎𝟎𝒋𝒋
𝑚𝑚1 = −20 + 60𝑗𝑗 & 𝑚𝑚2 = −20 − 60𝑗𝑗 Hence: 𝛼𝛼 = −20 & 𝛽𝛽 = 60 ⇒ 𝑑𝑑 𝑡𝑡 = 𝑒𝑒−20𝑡𝑡 𝑐𝑐1 cos60𝑡𝑡 + 𝑐𝑐2 sin60𝑡𝑡
3.8.4. : Series Circuit (LRC) 𝐿𝐿 = 0.25𝐻𝐻; 𝑅𝑅 = 10Ω; 𝐶𝐶 = 0.001𝐹𝐹 𝐸𝐸 𝑡𝑡 = 0𝑉𝑉; 𝑑𝑑 0 = 𝑑𝑑0; 𝑠𝑠 0 = 0𝐴𝐴 Aux. Eqn. 𝑚𝑚2 + 40𝑚𝑚 + 4000 = 0
Linear Models
9/30/2014 Dr. Eli Saber 171
• 𝑑𝑑 0 = 𝑑𝑑0
⇒ 𝑑𝑑0 = 𝑒𝑒0 𝑐𝑐1 cos 0 + 𝑐𝑐2 sin 0 ⇒ 𝑑𝑑0 = 1 𝑐𝑐1 + 0 ⇒ 𝑐𝑐1 = 𝑑𝑑0 Hence, we now have:
𝒒𝒒 𝒘𝒘 = 𝒆𝒆−𝟐𝟐𝟎𝟎𝒘𝒘 𝒒𝒒𝟎𝟎 𝒄𝒄𝒄𝒄𝒄𝒄𝟔𝟔𝟎𝟎𝒘𝒘 + 𝒄𝒄𝟐𝟐 𝒄𝒄𝒔𝒔𝒏𝒏𝟔𝟔𝟎𝟎𝒘𝒘
⇒ 𝑠𝑠 𝑡𝑡 =𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡 = −20𝑒𝑒−20𝑡𝑡[𝑑𝑑0 cos60𝑡𝑡 + 𝑐𝑐2 sin 60𝑡𝑡]
+𝑒𝑒−20𝑡𝑡[−60𝑑𝑑0 sin 60𝑡𝑡 + 60𝑐𝑐2 cos60𝑡𝑡]
3.8.4. : Series Circuit (LRC) 𝐿𝐿 = 0.25𝐻𝐻; 𝑅𝑅 = 10Ω; 𝐶𝐶 = 0.001𝐹𝐹 𝐸𝐸 𝑡𝑡 = 0𝑉𝑉; 𝑑𝑑 0 = 𝑑𝑑0; 𝑠𝑠 0 = 0𝐴𝐴 Aux. Eqn. 𝑚𝑚2 + 40𝑚𝑚 + 4000 = 0
𝑑𝑑 𝑡𝑡 = 𝑒𝑒−20𝑡𝑡 𝑐𝑐1 cos60𝑡𝑡 + 𝑐𝑐2 sin 60𝑡𝑡
Linear Models
9/30/2014 Dr. Eli Saber 172
𝑠𝑠 𝑡𝑡 =−20𝑒𝑒−20𝑡𝑡[𝑑𝑑0 cos60𝑡𝑡 + 𝑐𝑐2 sin 60𝑡𝑡]
+𝑒𝑒−20𝑡𝑡[−60𝑑𝑑0 sin 60𝑡𝑡 + 60𝑐𝑐2 cos60𝑡𝑡]
But 𝑠𝑠 0 = 0 ⇒ 0 = −20 𝑑𝑑0 + 0 + 1 0 + 60 𝑐𝑐2 ⇒ 0 = −20 𝑑𝑑0 + 60𝑐𝑐2 ⇒ 60𝑐𝑐2 = 20𝑑𝑑0
⇒ 𝑐𝑐2 =2060 𝑑𝑑0 ⇒ 𝑐𝑐2 =
13 𝑑𝑑0
⇒ 𝑑𝑑 𝑡𝑡 = 𝑒𝑒−20𝑡𝑡 𝑑𝑑0 𝑐𝑐𝑓𝑓𝑠𝑠 60𝑡𝑡 +𝑑𝑑03 𝑠𝑠𝑠𝑠𝑛𝑛 60𝑡𝑡
⇒ 𝑑𝑑 𝑡𝑡 = 𝑑𝑑0𝑒𝑒−20𝑡𝑡 𝑐𝑐𝑓𝑓𝑠𝑠 60𝑡𝑡 +13 𝑠𝑠𝑠𝑠𝑛𝑛 60𝑡𝑡
3.8.4. : Series Circuit (LRC) 𝐿𝐿 = 0.25𝐻𝐻; 𝑅𝑅 = 10Ω; 𝐶𝐶 = 0.001𝐹𝐹 𝐸𝐸 𝑡𝑡 = 0𝑉𝑉; 𝑑𝑑 0 = 𝑑𝑑0; 𝑠𝑠 0 = 0𝐴𝐴 Aux. Eqn. 𝑚𝑚2 + 40𝑚𝑚 + 4000 = 0
𝑑𝑑 𝑡𝑡 = 𝑒𝑒−20𝑡𝑡 𝑑𝑑0 𝑐𝑐𝑓𝑓𝑠𝑠 60𝑡𝑡 + 𝑐𝑐2 𝑠𝑠𝑠𝑠𝑛𝑛 60𝑡𝑡
𝑑𝑑 𝑡𝑡 = 𝑑𝑑0𝑒𝑒−20𝑡𝑡 𝑐𝑐𝑓𝑓𝑠𝑠 60𝑡𝑡 +13𝑠𝑠𝑠𝑠𝑛𝑛 60𝑡𝑡
We know, sin 𝐴𝐴 + 𝐵𝐵 = sin𝐴𝐴 cos𝐵𝐵 + cos𝐴𝐴 sin𝐵𝐵 We can transform 𝑑𝑑(𝑡𝑡) into an alternate form:
𝑑𝑑 𝑡𝑡 = 𝑑𝑑0𝑒𝑒−20𝑡𝑡 (1) 𝑐𝑐𝑓𝑓𝑠𝑠 60𝑡𝑡 +13 𝑠𝑠𝑠𝑠𝑛𝑛 60𝑡𝑡
⇒ 𝑑𝑑 𝑡𝑡 = 𝑑𝑑0𝑒𝑒−20𝑡𝑡103
1103
𝑐𝑐𝑓𝑓𝑠𝑠 60𝑡𝑡 +13103
𝑠𝑠𝑠𝑠𝑛𝑛 60𝑡𝑡
Linear Models
9/30/2014 Dr. Eli Saber 173
3.8.4. : Series Circuit (LRC)
𝝓𝝓 13
1 1 2 +
13
2
=𝟏𝟏𝟎𝟎𝟑𝟑
sin𝜙𝜙 =1103
; cos𝜙𝜙 =13103
𝐜𝐜𝐬𝐬𝐥𝐥𝝓𝝓 𝐜𝐜𝐜𝐜𝐜𝐜𝝓𝝓
⇒ 𝑑𝑑 𝑡𝑡 = 𝑑𝑑0103
𝑒𝑒−20𝑡𝑡 sin[60𝑡𝑡 + 𝜙𝜙]
Note: sin𝜙𝜙 = 310⇒ 𝜙𝜙 = sin−1 3
10= 1.249 rad
⇒ 𝑑𝑑 𝑡𝑡 = 𝑑𝑑0103 𝑒𝑒−20𝑡𝑡 sin[60𝑡𝑡 + 1.249]
Linear Models
9/30/2014 Dr. Eli Saber 174
3.8.4. : Series Circuit (LRC)
𝝓𝝓 13
1 1 2 +
13
2
=𝟏𝟏𝟎𝟎𝟑𝟑
sin𝜙𝜙 =1103
; cos𝜙𝜙 =13103
Note: • 𝑑𝑑𝑐𝑐(𝑡𝑡): solution to the homogeneous equation is called the transient solution
• 𝑑𝑑𝑝𝑝 𝑡𝑡 : solution to the non-homogeneous equation (i.e. 𝐸𝐸(𝑡𝑡) ≠ 0) is called the
steady-state solution
Linear Models
9/30/2014 Dr. Eli Saber 175
3.8.4. : Series Circuit (LRC)
E.g. 𝐿𝐿 = 1𝐻𝐻;𝑅𝑅 = 2Ω;𝐶𝐶 = 0.25𝐹𝐹; 𝐸𝐸 𝑡𝑡 = 50 cos 𝑡𝑡 𝑠𝑠𝑓𝑓𝑔𝑔𝑡𝑡𝑠𝑠 Find the steady-state charge and the steady-state current in the LRC Circuit (Advanced Eng. Mathematics – 5th Edition – Ex. 3.8 Prob. 49) Solution:
𝐸𝐸 𝑡𝑡 = 𝐿𝐿𝑑𝑑𝑠𝑠𝑑𝑑𝑡𝑡
+ 𝑅𝑅𝑠𝑠 +𝑑𝑑𝐶𝐶
⇒ 𝐿𝐿𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
+ 𝑅𝑅𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
+𝑑𝑑𝐶𝐶
= 𝐸𝐸 𝑡𝑡
⇒ 1𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
+ 2𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
+𝑑𝑑
0.25= 50 cos 𝑡𝑡
Linear Models
9/30/2014 Dr. Eli Saber 176
3.8.4. : Series Circuit (LRC)
C= 0.25𝐹𝐹
L= 1𝐻𝐻
R=2 Ω
E=50 cos (t) V 𝑠𝑠(𝑡𝑡)
⇒ 1𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
+ 2𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
+ 4 𝑑𝑑 = 50 cos 𝑡𝑡
Homogeneous Eqn. 𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
+ 2 𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
+ 4 𝑑𝑑 = 0 ⇒ 𝑚𝑚2 + 2𝑚𝑚 + 4 = 0
⇒ 𝑚𝑚 =−2 ± 4 − 4(1)(4)
2 =−2 ± 12
2 =−2 ± 4 3 𝑗𝑗2
2
⇒ 𝑚𝑚 = −1 ± 𝑗𝑗 3 ⇒ 𝛼𝛼 = −1 & 𝛽𝛽 = 3
𝑑𝑑𝑐𝑐 𝑡𝑡 = 𝑐𝑐1𝑒𝑒 −1+𝑗𝑗 3 𝑡𝑡 + 𝑐𝑐2𝑒𝑒 −1−𝑗𝑗 3 𝑡𝑡 or
𝑑𝑑𝑐𝑐 𝑡𝑡 = 𝑒𝑒−𝑡𝑡 ∝1 cos 3𝑡𝑡+∝2 sin 3𝑡𝑡
Linear Models
9/30/2014 Dr. Eli Saber 177
3.8.4. : Series Circuit (LRC)
From table 3.4.1., we can write: 𝑑𝑑𝑝𝑝 𝑡𝑡 = 𝐴𝐴 cos 𝑡𝑡 + 𝐵𝐵 sin 𝑡𝑡 𝑑𝑑′ 𝑡𝑡 = −𝐴𝐴 sin 𝑡𝑡 + 𝐵𝐵 cos 𝑡𝑡 𝑑𝑑′′ 𝑡𝑡 = −𝐴𝐴 cos 𝑡𝑡 − 𝐵𝐵 sin 𝑡𝑡 Re−Substituting back in eqn. ⇒ −𝐴𝐴 cos 𝑡𝑡 − 𝐵𝐵 sin 𝑡𝑡 + 2 −𝐴𝐴 sin 𝑡𝑡 + 𝐵𝐵 cos 𝑡𝑡 + 4 𝐴𝐴 cos 𝑡𝑡 + 𝐵𝐵 sin 𝑡𝑡 = 50 cos 𝑡𝑡 ⇒ −𝐴𝐴 cos 𝑡𝑡 − 𝐵𝐵 sin 𝑡𝑡 − 2𝐴𝐴 sin 𝑡𝑡 + 2𝐵𝐵 cos 𝑡𝑡 + 4𝐴𝐴 cos 𝑡𝑡 + 4𝐵𝐵 sin 𝑡𝑡 = 50 cos 𝑡𝑡 ⇒ cos 𝑡𝑡 3𝐴𝐴 + 2𝐵𝐵 + sin 𝑡𝑡 −2𝐴𝐴 + 3𝐵𝐵 = 50 cos 𝑡𝑡 ⇒ 3𝐴𝐴 + 2𝐵𝐵 = 50
⇒ −2𝐴𝐴 + 3𝐵𝐵 = 0 ⇒ 2𝐴𝐴 = 3𝐵𝐵 ⇒ 𝐴𝐴 =32𝐵𝐵
Linear Models
9/30/2014 Dr. Eli Saber 178
3.8.4. : Series Circuit (LRC)
1𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2 + 2
𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡 + 4 𝑑𝑑 = 50 cos 𝑡𝑡
3𝐴𝐴 + 2𝐵𝐵 = 50
332𝐵𝐵 + 2𝐵𝐵 = 50 ⇒
92𝐵𝐵 + 2𝐵𝐵 = 50 ⇒
132𝐵𝐵 = 50 ⇒ 𝐵𝐵 =
10013
𝐴𝐴 =32𝐵𝐵 =
32∗
10013
=30026
⇒ 𝐴𝐴 =15013
𝑑𝑑𝑝𝑝 𝑡𝑡 =15013
cos 𝑡𝑡 +10013
sin 𝑡𝑡
We already have: 𝑑𝑑𝑐𝑐 𝑡𝑡 = 𝑒𝑒−𝑡𝑡 ∝1 cos 3𝑡𝑡+∝2 sin 3𝑡𝑡 Hence,
𝑑𝑑 𝑡𝑡 = 𝑒𝑒−𝑡𝑡 𝑐𝑐1 cos 3𝑡𝑡 + 𝑐𝑐2 sin 3𝑡𝑡 +15013
cos 𝑡𝑡 +10013
sin 𝑡𝑡
Linear Models
9/30/2014 Dr. Eli Saber 179
3.8.4. : Series Circuit (LRC)
1𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2 + 2
𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡 + 4 𝑑𝑑 = 50 cos 𝑡𝑡
3𝐴𝐴 + 2𝐵𝐵 = 50
𝐴𝐴 =32𝐵𝐵
Transient Solution Steady-State Solution
𝑑𝑑𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝑑𝑑𝑝𝑝 𝑡𝑡 ⇒ 𝑑𝑑𝑠𝑠𝑠𝑠 𝑡𝑡 =15013
cos 𝑡𝑡 +10013
sin 𝑡𝑡
𝑠𝑠 𝑡𝑡 =𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
= −𝑒𝑒−𝑡𝑡 𝑐𝑐1 cos 3𝑡𝑡 + 𝑐𝑐2 sin 3𝑡𝑡 + 𝑒𝑒−𝑡𝑡 −𝑐𝑐1 3 sin 3𝑡𝑡 + 𝑐𝑐2 3 cos 3𝑡𝑡 −15013
sin 𝑡𝑡 +10013
cos 𝑡𝑡
⇒ 𝑠𝑠 𝑡𝑡 = −𝑐𝑐1𝑒𝑒−𝑡𝑡 cos 3𝑡𝑡 − 𝑐𝑐2𝑒𝑒−𝑡𝑡 sin 3𝑡𝑡 − 𝑐𝑐1𝑒𝑒−𝑡𝑡 3 sin 3𝑡𝑡 + 𝑐𝑐2𝑒𝑒−𝑡𝑡 3 cos 3𝑡𝑡 −15013
sin 𝑡𝑡
+10013
cos 𝑡𝑡
⇒ 𝑠𝑠 𝑡𝑡 = 𝑒𝑒−𝑡𝑡 −𝑐𝑐1 + 3𝑐𝑐2 cos 3𝑡𝑡 + 𝑒𝑒−𝑡𝑡 −𝑐𝑐2 − 3𝑐𝑐1 sin 3𝑡𝑡 −15013
sin 𝑡𝑡 +10013
cos 𝑡𝑡
⇒ 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 = −15013
sin 𝑡𝑡 + 10013
cos 𝑡𝑡
Linear Models
9/30/2014 Dr. Eli Saber 180
3.8.4. : Series Circuit (LRC)
𝑑𝑑 𝑡𝑡 = 𝑒𝑒−𝑡𝑡 𝑐𝑐1 cos 3𝑡𝑡 + 𝑐𝑐2 sin 3𝑡𝑡 +15013
cos 𝑡𝑡 +10013
sin 𝑡𝑡
Linear Models
9/30/2014 Dr. Eli Saber 181
Summary
𝐿𝐿𝑑𝑑𝑠𝑠𝑑𝑑𝑡𝑡
+ 𝑅𝑅𝑠𝑠 +1𝐶𝐶� 𝑠𝑠 𝑑𝑑𝑡𝑡 = 𝐸𝐸(𝑡𝑡)
⇒ 𝐿𝐿𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2
+ 𝑅𝑅𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡
+𝑑𝑑𝐶𝐶
= 𝐸𝐸 𝑡𝑡
Auxiliary Eqn: 𝐿𝐿𝑚𝑚2 + 𝑅𝑅𝑚𝑚 + 1𝐶𝐶
= 0
𝑚𝑚 =−𝑅𝑅 ± 𝑅𝑅2 − 4𝐿𝐿
𝐶𝐶2𝐿𝐿
= −𝑅𝑅2𝐿𝐿
±𝑅𝑅2 − 4𝐿𝐿
𝐶𝐶2𝐿𝐿
Use already known methods to obtain 𝑦𝑦𝑝𝑝
𝒚𝒚 = 𝒚𝒚𝒄𝒄 + 𝒚𝒚𝒑𝒑
C
L
R
E 𝑠𝑠(𝑡𝑡)
𝜶𝜶 𝜷𝜷
⇒ obtain 𝑦𝑦𝑐𝑐
Section 3.12 Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 182
Newton’s 2nd Law:
𝑚𝑚1𝑑𝑑2𝑥𝑥1𝑑𝑑𝑡𝑡2 = −𝑘𝑘1𝑥𝑥1 + 𝑘𝑘2 𝑥𝑥2 − 𝑥𝑥1
𝑚𝑚2𝑑𝑑2𝑥𝑥2𝑑𝑑𝑡𝑡2 = −𝑘𝑘2 𝑥𝑥2 − 𝑥𝑥1
Also can be written as: 𝑚𝑚1𝑥𝑥1′′ = −𝑘𝑘1𝑥𝑥1 + 𝑘𝑘2 𝑥𝑥2 − 𝑥𝑥1 𝑚𝑚2𝑥𝑥2′′ = −𝑘𝑘2 𝑥𝑥2 − 𝑥𝑥1
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 183
Coupled Spring/Mass Systems
A coupled system of Differential Equations
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.
Given 𝑎𝑎𝑛𝑛𝑦𝑦 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1𝑦𝑦 𝑛𝑛−1 + ⋯+ 𝑎𝑎1𝑦𝑦′ + 𝑎𝑎0𝑦𝑦 = 𝑔𝑔 𝑥𝑥 ,where 𝑎𝑎𝑖𝑖 , 𝑠𝑠 = 0,1,2,3, … ,𝑛𝑛 are constants Rewrite as: 𝑎𝑎𝑛𝑛𝐷𝐷𝑛𝑛 + 𝑎𝑎𝑛𝑛−1𝐷𝐷𝑛𝑛−1 + ⋯+ 𝑎𝑎1𝐷𝐷 + 𝑎𝑎0 𝑦𝑦 = 𝑔𝑔 𝑥𝑥 Then group like terms for solving.
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 184
Systematic Elimination
Given: 𝑥𝑥′′ + 2𝑥𝑥′ + 𝑦𝑦′′ = 𝑥𝑥 + 3𝑦𝑦 + sin 𝑡𝑡 𝑥𝑥′ + 𝑦𝑦′ = −4𝑥𝑥 + 2𝑦𝑦 + 𝑒𝑒−𝑡𝑡 ⇒ 𝑥𝑥′′ + 2𝑥𝑥′ + 𝑦𝑦′′ − 𝑥𝑥 − 3𝑦𝑦 = sin 𝑡𝑡 ⇒ 𝑥𝑥′ + 𝑦𝑦′ + 4𝑥𝑥 − 2𝑦𝑦 = 𝑒𝑒−𝑡𝑡 ⇒ 𝐷𝐷2𝑥𝑥 + 2𝐷𝐷𝑥𝑥 + 𝐷𝐷2𝑦𝑦 − 𝑥𝑥 − 3𝑦𝑦 = sin 𝑡𝑡 ⇒ 𝐷𝐷𝑥𝑥 + 𝐷𝐷𝑦𝑦 + 4𝑥𝑥 − 2𝑦𝑦 = 𝑒𝑒−𝑡𝑡 ⇒ 𝑫𝑫𝟐𝟐 + 𝟐𝟐𝑫𝑫 − 𝟏𝟏 𝑥𝑥 + 𝑫𝑫𝟐𝟐 − 𝟑𝟑 𝑦𝑦 = sin 𝑡𝑡 ⇒ 𝑫𝑫 + 𝟒𝟒 𝑥𝑥 + 𝑫𝑫− 𝟐𝟐 𝑦𝑦 = 𝑒𝑒−𝑡𝑡
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 185
A solution of a system of D.E. is a set of sufficiently differentiable functions
𝑥𝑥 = ∅1 𝑡𝑡 𝑦𝑦 = ∅2 𝑡𝑡 𝑧𝑧 = ∅3 𝑡𝑡
⋮ 𝑎𝑎𝑛𝑛𝑑𝑑 𝑠𝑠𝑓𝑓 𝑓𝑓𝑛𝑛
that satisfies each equation in the system on some common interval 𝐼𝐼
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 186
Solution of System
E.g. Linear 1st order equations: 𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡
= 3𝑦𝑦
𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡
= 2𝑥𝑥
Solution: 𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡 = 3𝑦𝑦 ⇒ 𝐷𝐷𝑥𝑥 − 3𝑦𝑦 = 0
𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡 = 2𝑥𝑥 ⇒ 𝐷𝐷𝑦𝑦 − 2𝑥𝑥 = 0
𝐷𝐷𝑥𝑥 − 3𝑦𝑦 = 0 → 𝑓𝑓𝑠𝑠𝑒𝑒𝑓𝑓𝑎𝑎𝑡𝑡𝑒𝑒 𝑤𝑤𝑠𝑠𝑡𝑡ℎ 𝐷𝐷 → 𝐷𝐷2𝑥𝑥 − 3𝐷𝐷𝑦𝑦 = 0 𝐷𝐷𝑦𝑦 − 2𝑥𝑥 = 0 → 𝑚𝑚𝑜𝑜𝑔𝑔𝑡𝑡𝑠𝑠𝑠𝑠𝑔𝑔𝑦𝑦 𝑏𝑏𝑦𝑦 3 → +3𝐷𝐷𝑦𝑦 − 6𝑥𝑥 = 0
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 187
𝑫𝑫𝟐𝟐𝟐𝟐 − 𝟔𝟔𝟐𝟐 = 𝟎𝟎
⇒ 𝐷𝐷2𝑥𝑥 − 6𝑥𝑥 = 0 Auxiliary Equation: 𝑚𝑚2 − 6 = 0 ⇒ 𝑚𝑚2 = 6 ⇒ 𝑚𝑚 = ± 6
𝟐𝟐 𝒘𝒘 = 𝒄𝒄𝟏𝟏𝒆𝒆− 𝟔𝟔𝒘𝒘 + 𝒄𝒄𝟐𝟐𝒆𝒆 𝟔𝟔𝒘𝒘 Now, to obtain 𝑦𝑦(𝑡𝑡): 𝐷𝐷𝑥𝑥 − 3𝑦𝑦 = 0 → 𝑚𝑚𝑜𝑜𝑔𝑔𝑡𝑡𝑠𝑠𝑠𝑠𝑔𝑔𝑦𝑦 𝑏𝑏𝑦𝑦 2 → 2𝐷𝐷𝑥𝑥 − 6𝑦𝑦 = 0 𝐷𝐷𝑦𝑦 − 2𝑥𝑥 = 0 → 𝑓𝑓𝑠𝑠𝑒𝑒𝑓𝑓𝑎𝑎𝑡𝑡𝑒𝑒 𝑤𝑤𝑠𝑠𝑡𝑡ℎ 𝐷𝐷 → 𝐷𝐷2𝑦𝑦 − 2𝐷𝐷𝑥𝑥 = 0 Auxiliary Equation: 𝑚𝑚2 − 6 = 0 ⇒ 𝑚𝑚2 = 6 ⇒ 𝑚𝑚 = ± 6
𝒚𝒚 𝒘𝒘 = 𝒄𝒄𝟑𝟑𝒆𝒆− 𝟔𝟔𝒘𝒘 + 𝒄𝒄𝟒𝟒𝒆𝒆 𝟔𝟔𝒘𝒘
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 188
𝑫𝑫𝟐𝟐𝒚𝒚 − 𝟔𝟔𝒚𝒚 = 𝟎𝟎
𝑥𝑥𝑦 𝑡𝑡 = − 6𝑐𝑐1𝑒𝑒− 6𝑡𝑡 + 6𝑐𝑐2𝑒𝑒 6𝑡𝑡
We know: 𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡
= 3𝑦𝑦
⇒ − 6𝑐𝑐1𝑒𝑒− 6𝑡𝑡 + 6𝑐𝑐2𝑒𝑒 6𝑡𝑡 = 3𝑐𝑐3𝑒𝑒− 6𝑡𝑡 + 3𝑐𝑐4𝑒𝑒 6𝑡𝑡 ⇒ − 6𝑐𝑐1 − 3𝑐𝑐3 𝑒𝑒− 6𝑡𝑡 + 6𝑐𝑐2 − 3𝑐𝑐4 𝑒𝑒 6𝑡𝑡 = 0 ∀𝑡𝑡
⇒ − 6𝑐𝑐1 − 3𝑐𝑐3 = 0 ⇒ 𝑐𝑐3 = −6
3𝑐𝑐1
⇒ 6𝑐𝑐2 − 3𝑐𝑐4 = 0 ⇒ 𝑐𝑐4 =6
3c2
𝟐𝟐 𝒘𝒘 = 𝒄𝒄𝟏𝟏𝒆𝒆− 𝟔𝟔𝒘𝒘 + 𝒄𝒄𝟐𝟐𝒆𝒆 𝟔𝟔𝒘𝒘 & 𝒚𝒚 𝒘𝒘 = −𝟔𝟔𝟑𝟑𝒄𝒄𝟏𝟏𝒆𝒆− 𝟔𝟔𝒘𝒘 +
𝟔𝟔𝟑𝟑𝒄𝒄𝟐𝟐𝒆𝒆 𝟔𝟔𝒘𝒘
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 189
𝑥𝑥 𝑡𝑡 = 𝑐𝑐1𝑒𝑒− 6𝑡𝑡 + 𝑐𝑐2𝑒𝑒 6𝑡𝑡 𝑦𝑦 𝑡𝑡 = 𝑐𝑐3𝑒𝑒− 6𝑡𝑡 + 𝑐𝑐4𝑒𝑒 6𝑡𝑡
E.g. 𝑥𝑥′ − 4𝑥𝑥 + 𝑦𝑦′′ = 𝑡𝑡2 𝑥𝑥′ + 𝑥𝑥 + 𝑦𝑦′ = 0
Solution: 𝐷𝐷𝑥𝑥 − 4𝑥𝑥 + 𝐷𝐷2𝑦𝑦 = 𝑡𝑡2 ⇒ 𝐷𝐷 − 4 𝑥𝑥 + 𝐷𝐷2𝑦𝑦 = 𝑡𝑡2 𝐷𝐷𝑥𝑥 + 𝑥𝑥 + 𝐷𝐷𝑦𝑦 = 0 ⇒ 𝐷𝐷 + 1 𝑥𝑥 + 𝐷𝐷𝑦𝑦 = 0 Solving for 𝑦𝑦 first: 1 ∗ 𝐷𝐷 + 1 ⇒ 𝐷𝐷 − 4 𝐷𝐷 + 1 𝑥𝑥 + 𝐷𝐷2 𝐷𝐷 + 1 𝑦𝑦 = 𝐷𝐷 + 1 𝑡𝑡2
2 ∗ 𝐷𝐷 − 4 ⇒ 𝐷𝐷 + 1 𝐷𝐷 − 4 𝑥𝑥 + 𝐷𝐷 𝐷𝐷 − 4 𝑦𝑦 = 0
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 190
(𝟏𝟏)
(𝟐𝟐)
𝑫𝑫𝟐𝟐 𝑫𝑫 + 𝟏𝟏 𝒚𝒚 − 𝑫𝑫 𝑫𝑫− 𝟒𝟒 𝒚𝒚 = 𝑫𝑫 + 𝟏𝟏 𝒘𝒘𝟐𝟐 (−) (−) (−)
⇒ 𝐷𝐷2 𝐷𝐷 + 1 𝑦𝑦 − 𝐷𝐷 𝐷𝐷 − 4 𝑦𝑦 = 𝐷𝐷 + 1 𝑡𝑡2 ⇒ 𝐷𝐷3 + 𝐷𝐷2 − 𝐷𝐷2 + 4𝐷𝐷 𝑦𝑦 = 𝐷𝐷𝑡𝑡2 + 𝑡𝑡2 ⇒ 𝐷𝐷3 + 4𝐷𝐷 𝑦𝑦 = 2𝑡𝑡 + 𝑡𝑡2
→𝑑𝑑3𝑦𝑦𝑑𝑑𝑡𝑡3 + 4
𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡 = 2𝑡𝑡 + 𝑡𝑡2
⇒ 𝐷𝐷3 + 4𝐷𝐷 𝑦𝑦 = 2𝑡𝑡 + 𝑡𝑡2 Aux. Equation: 𝑚𝑚3 + 4𝑚𝑚 = 0 ⇒ 𝑚𝑚 𝑚𝑚2 + 4 = 0 ⇒ 𝑚𝑚1 = 0;𝑚𝑚2 = 2𝑗𝑗;𝑚𝑚3 = −2𝑗𝑗
𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑒𝑒0𝑡𝑡 + 𝑐𝑐2 cos2𝑡𝑡 + 𝑐𝑐3 sin 2𝑡𝑡
𝑦𝑦𝑐𝑐 = 𝑐𝑐1 + 𝑐𝑐2 𝑐𝑐𝑓𝑓𝑠𝑠 2𝑡𝑡 + 𝑐𝑐3 𝑠𝑠𝑠𝑠𝑛𝑛 2𝑡𝑡
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 191
Here, 𝐷𝐷𝑡𝑡2 = 𝑑𝑑𝑑𝑑𝑡𝑡
𝑡𝑡2 = 2𝑡𝑡
Particular Solution 𝒚𝒚: use undetermined coefficient ⇒ 𝐴𝐴𝑠𝑠𝑠𝑠𝑜𝑜𝑚𝑚𝑒𝑒 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑡𝑡3 + 𝐵𝐵𝑡𝑡2 + 𝐶𝐶𝑡𝑡 ⇒ 𝑦𝑦𝑝𝑝′ = 3𝐴𝐴𝑡𝑡2 + 2𝐵𝐵𝑡𝑡 + 𝐶𝐶; 𝑦𝑦𝑝𝑝′′ = 6𝐴𝐴𝑡𝑡 + 2𝐵𝐵; 𝑦𝑦𝑝𝑝′′′ = 6𝐴𝐴 ⇒ 6𝐴𝐴 + 4 3𝐴𝐴𝑡𝑡2 + 2𝐵𝐵𝑡𝑡 + 𝐶𝐶 = 𝑡𝑡2 + 2𝑡𝑡 ⇒ 6𝐴𝐴 + 12𝐴𝐴𝑡𝑡2 + 8𝐵𝐵𝑡𝑡 + 4𝐶𝐶 = 𝑡𝑡2 + 2𝑡𝑡 ⇒ 12𝐴𝐴𝑡𝑡2 + 8𝐵𝐵𝑡𝑡 + 6𝐴𝐴 + 4𝐶𝐶 = 𝑡𝑡2 + 2𝑡𝑡
⇒ 12𝐴𝐴 = 1 → 𝐴𝐴 =1
12
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 192
𝑑𝑑3𝑦𝑦𝑑𝑑𝑡𝑡3
+ 4𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡
= 𝑡𝑡2 + 2𝑡𝑡 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 + 𝑐𝑐2 𝑐𝑐𝑓𝑓𝑠𝑠 2𝑡𝑡 + 𝑐𝑐3 𝑠𝑠𝑠𝑠𝑛𝑛 2𝑡𝑡
Note: 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑡𝑡3 + 𝐵𝐵𝑡𝑡2 + 𝐶𝐶𝑡𝑡 Here, +𝐷𝐷 is not considered since 𝑦𝑦𝑐𝑐 already has a constant term
⇒ 8𝐵𝐵 = 2 → 𝐵𝐵 =14
⇒ 6𝐴𝐴 + 4𝐶𝐶 = 0 ⇒ 4𝐶𝐶 = −6𝐴𝐴 = −61
12
⇒ 𝐶𝐶 = −18
Hence,
𝑦𝑦𝑝𝑝 =1
12 𝑡𝑡3 +
14 𝑡𝑡
2 −18 𝑡𝑡
𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 + 𝒄𝒄𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐𝒘𝒘 + 𝒄𝒄𝟑𝟑 𝐜𝐜𝐬𝐬𝐥𝐥𝟐𝟐𝒘𝒘 +𝟏𝟏𝟏𝟏𝟐𝟐 𝒘𝒘
𝟑𝟑 +𝟏𝟏𝟒𝟒 𝒘𝒘
𝟐𝟐 −𝟏𝟏𝟖𝟖 𝒘𝒘
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 193
𝑑𝑑3𝑦𝑦𝑑𝑑𝑡𝑡3
+ 4𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡
= 𝑡𝑡2 + 2𝑡𝑡 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 + 𝑐𝑐2 𝑐𝑐𝑓𝑓𝑠𝑠 2𝑡𝑡 + 𝑐𝑐3 𝑠𝑠𝑠𝑠𝑛𝑛 2𝑡𝑡
𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑡𝑡3 + 𝐵𝐵𝑡𝑡2 + 𝐶𝐶𝑡𝑡
𝐴𝐴 =1
12
We have: (1) ≡ 𝐷𝐷𝑥𝑥 − 4𝑥𝑥 + 𝐷𝐷2𝑦𝑦 = 𝑡𝑡2 ⇒ 𝐷𝐷 − 4 𝑥𝑥 + 𝐷𝐷2𝑦𝑦 = 𝑡𝑡2 2 ≡ 𝐷𝐷𝑥𝑥 + 𝑥𝑥 + 𝐷𝐷𝑦𝑦 = 0 ⇒ 𝐷𝐷 + 1 𝑥𝑥 + 𝐷𝐷𝑦𝑦 = 0
Solving for 𝑥𝑥 now: 1 ⇒ 𝐷𝐷 − 4 𝑥𝑥 + 𝐷𝐷2𝑦𝑦 = 𝑡𝑡2
2 ∗ 𝐷𝐷 ⇒ 𝐷𝐷 𝐷𝐷 + 1 𝑥𝑥 + 𝐷𝐷2𝑦𝑦 = 0
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 194
𝑫𝑫− 𝟒𝟒 − 𝑫𝑫 𝑫𝑫 + 𝟏𝟏 𝟐𝟐 = 𝒘𝒘𝟐𝟐 (−) (−) (−)
⇒ 𝐷𝐷 − 4 − 𝐷𝐷 𝐷𝐷 + 1 𝑥𝑥 = 𝑡𝑡2 ⇒ 𝐷𝐷 − 4 − 𝐷𝐷2 − 𝐷𝐷 𝑥𝑥 = 𝑡𝑡2 ⇒ − 4 + 𝐷𝐷2 = 𝑡𝑡2 ⇒ 𝐷𝐷2 + 4 𝑥𝑥 = −𝑡𝑡2 Aux. Equation: 𝑚𝑚2 + 4 = 0 ⇒ 𝑚𝑚1 = 2𝑗𝑗;𝑚𝑚2 = −2𝑗𝑗
𝑥𝑥𝑐𝑐 = 𝑐𝑐4 cos2𝑡𝑡 + 𝑐𝑐5 sin2𝑡𝑡
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 195
Particular Solution 𝟐𝟐: use undetermined coefficient 𝐴𝐴𝑠𝑠𝑠𝑠𝑜𝑜𝑚𝑚𝑒𝑒 𝑥𝑥𝑝𝑝 = 𝐴𝐴𝑡𝑡2 + 𝐵𝐵𝑡𝑡 + 𝐶𝐶 (Table 3.4.1) ⇒ 𝑥𝑥𝑝𝑝′ = 2𝐴𝐴𝑡𝑡 + 𝐵𝐵; 𝑥𝑥𝑝𝑝′′ = 2𝐴𝐴
𝐷𝐷2𝑥𝑥 + 4𝑥𝑥 = −𝑡𝑡2 →𝑑𝑑2𝑥𝑥𝑝𝑝𝑑𝑑𝑡𝑡2 + 4𝑥𝑥𝑝𝑝 = −𝑡𝑡2
⇒ 2𝐴𝐴 + 4 𝐴𝐴𝑡𝑡2 + 𝐵𝐵𝑡𝑡 + 𝐶𝐶 = −𝑡𝑡2 ⇒ 2𝐴𝐴 + 4𝐴𝐴𝑡𝑡2 + 4𝐵𝐵𝑡𝑡 + 4𝐶𝐶 = −𝑡𝑡2 4𝐴𝐴𝑡𝑡2 + 4𝐵𝐵𝑡𝑡 + 2𝐴𝐴 + 4𝐶𝐶 = −𝑡𝑡2
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 196
𝐷𝐷2 + 4 𝑥𝑥 = −𝑡𝑡2 𝑥𝑥𝑐𝑐 = 𝑐𝑐4 cos2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡
4𝐴𝐴𝑡𝑡2 + 4𝐵𝐵𝑡𝑡 + 2𝐴𝐴 + 4𝐶𝐶 = −𝑡𝑡2
4𝐴𝐴 = −1 ⇒ 𝐴𝐴 = −14
4𝐵𝐵 = 0 ⇒ 𝐵𝐵 = 0
2𝐴𝐴 + 4𝐶𝐶 = 0 ⇒ 𝐶𝐶 = −12𝐴𝐴 = −
12
−14
=18⇒ 𝐶𝐶 =
18
𝑥𝑥𝑝𝑝 = −14𝑡𝑡2 +
18
𝑥𝑥 = 𝑥𝑥𝑐𝑐 + 𝑥𝑥𝑝𝑝
⇒ 𝟐𝟐 = 𝒄𝒄𝟒𝟒 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐𝒘𝒘 + 𝒄𝒄𝟓𝟓 𝐜𝐜𝐬𝐬𝐥𝐥𝟐𝟐𝒘𝒘 −𝟏𝟏𝟒𝟒𝒘𝒘𝟐𝟐 +
𝟏𝟏𝟖𝟖
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 197
𝐷𝐷2 + 4 𝑥𝑥 = −𝑡𝑡2 𝑥𝑥𝑐𝑐 = 𝑐𝑐4 cos2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡
𝑥𝑥 = 𝑐𝑐4 cos2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡 −14𝑡𝑡2 +
18
𝑦𝑦 = 𝑐𝑐1 + 𝑐𝑐2 cos2𝑡𝑡 + 𝑐𝑐3 sin2𝑡𝑡 +1
12𝑡𝑡3 +
14𝑡𝑡2 −
18𝑡𝑡
Re-substituting 𝑥𝑥, 𝑦𝑦 in 𝟐𝟐′ + 𝟐𝟐 + 𝒚𝒚′ = 𝟎𝟎
⇒ −2𝑐𝑐4 cos2𝑡𝑡 + 2𝑐𝑐5 cos2𝑡𝑡 −12𝑡𝑡 + 𝑐𝑐4 cos2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡 −
14𝑡𝑡2 +
18
+ −2𝑐𝑐2 cos2𝑡𝑡 + 2𝑐𝑐3 cos2𝑡𝑡 +14𝑡𝑡3 +
12𝑡𝑡 −
18
= 0
⇒ sin 2𝑡𝑡 −2 𝑐𝑐4 + 𝑐𝑐5 − 2𝑐𝑐2 + cos2𝑡𝑡 2𝑐𝑐5 + 𝑐𝑐4 + 2𝑐𝑐3 = 0 ⇒ −2 𝑐𝑐4 + 𝑐𝑐5 − 2𝑐𝑐2 = 0 & 2𝑐𝑐5 + 𝑐𝑐4 + 2𝑐𝑐3 = 0
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 198
⇒ 𝑐𝑐5 − 2𝑐𝑐4 = 2𝑐𝑐2 ⇒ 2𝑐𝑐5 + 𝑐𝑐4 = −2𝑐𝑐3
⇒ 𝑐𝑐4 = − 15
4𝑐𝑐2 + 2𝑐𝑐3
⇒ 𝑐𝑐5 =15 2𝑐𝑐2 − 4𝑐𝑐3
⇒ 𝟐𝟐 = − 𝟏𝟏𝟓𝟓𝟒𝟒𝒄𝒄𝟐𝟐 + 𝟐𝟐𝒄𝒄𝟑𝟑 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐𝒘𝒘 + 𝟏𝟏
𝟓𝟓𝟐𝟐𝒄𝒄𝟐𝟐 − 𝟒𝟒𝒄𝒄𝟑𝟑 𝐜𝐜𝐬𝐬𝐥𝐥𝟐𝟐𝒘𝒘 − 𝟏𝟏
𝟒𝟒𝒘𝒘𝟐𝟐 + 𝟏𝟏
𝟖𝟖
⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 + 𝒄𝒄𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜𝟐𝟐𝒘𝒘 + 𝒄𝒄𝟑𝟑 𝐜𝐜𝐬𝐬𝐥𝐥𝟐𝟐𝒘𝒘 + 𝟏𝟏𝟏𝟏𝟐𝟐𝒘𝒘𝟑𝟑 + 𝟏𝟏
𝟒𝟒𝒘𝒘𝟐𝟐 − 𝟏𝟏
𝟖𝟖𝒘𝒘
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 199
𝑥𝑥 = 𝑐𝑐4 𝑐𝑐𝑓𝑓𝑠𝑠 2𝑡𝑡 + 𝑐𝑐5 𝑠𝑠𝑠𝑠𝑛𝑛 2𝑡𝑡 −14𝑡𝑡2 +
18
𝑦𝑦 = 𝑐𝑐1 + 𝑐𝑐2 𝑐𝑐𝑓𝑓𝑠𝑠 2𝑡𝑡 + 𝑐𝑐3 𝑠𝑠𝑠𝑠𝑛𝑛 2𝑡𝑡 +1
12𝑡𝑡3 +
14𝑡𝑡2 −
18𝑡𝑡
9/30/2014 Dr. Eli Saber 200
End of Chapter 3