Chapter 3-Higher Order Differential Equations

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

[email protected]

Chapter 3 Higher Order Differential

Equations

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Section 3.1 Theory of Linear Equations


Theory of Linear Equations

Objective: Investigate Differential Equations of Order 2++



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


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.




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.




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




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


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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


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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.


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E.g. 𝑥𝑥2𝑦𝑦′′ − 2𝑥𝑥𝑦𝑦′ + 2𝑦𝑦 = 6 𝑦𝑦 0 = 3; 𝑦𝑦′ 0 = 1 Solution: 𝑦𝑦 = 𝑐𝑐𝑥𝑥2 + 𝑥𝑥 + 3 in interval 𝐼𝐼 ≡ (−∞,∞)


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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


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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?


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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


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• 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


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Note: Even when the conditions for Unique Solution theorem are met, a BVP may have: 1) Many solutions 2) Unique Solution 3) No solution


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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


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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


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2) 𝑥𝑥 0 = 0; 𝑥𝑥𝜋𝜋8

= 0

• 𝑥𝑥 0 = 0 ⇒ 0 = 𝑐𝑐1 1 + 𝑐𝑐2 0 ⇒ 𝒄𝒄𝟏𝟏 = 𝟎𝟎


= 0 ⇒ 0 = 𝑐𝑐1 cos 𝜋𝜋2

+ 𝑐𝑐2 sin 𝜋𝜋2⇒ 0 = 𝑐𝑐1 0 + 𝑐𝑐2 1

– But 𝑐𝑐1 = 0


– Implies 𝒄𝒄𝟐𝟐 = 𝟎𝟎

– 𝟐𝟐 = 𝟎𝟎 is the solution of this new boundary problem

• Unique solution ≡ 𝟐𝟐 = 𝟎𝟎

𝑥𝑥 = 𝑐𝑐1 cos4𝑡𝑡 + 𝑐𝑐2 sin4𝑡𝑡


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3) 𝑥𝑥 0 = 0;𝑥𝑥𝜋𝜋2

= 1

• 𝑥𝑥 0 = 0 ⇒ 0 = 𝑐𝑐1 1 + 𝑐𝑐2 0 ⇒ 𝒄𝒄𝟏𝟏 = 𝟎𝟎


= 1 ⇒ 1 = 𝑐𝑐1 cos 4𝜋𝜋 ∙ 𝜋𝜋2

+ 𝑐𝑐2 sin 4𝜋𝜋 ∙ 𝜋𝜋2⇒ 1 = 𝑐𝑐1 cos(2𝜋𝜋) + 𝑐𝑐2 sin(2𝜋𝜋)

– 1 = 0 1 + 𝑐𝑐2 0


– Implies 𝒄𝒄𝟐𝟐 = 𝟏𝟏𝟎𝟎

= 𝑵𝑵.𝑫𝑫.

– Not possible to find 𝑐𝑐2

• No solution for BVP

𝑥𝑥 = 𝑐𝑐1 cos4𝑡𝑡 + 𝑐𝑐2 sin4𝑡𝑡


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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


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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


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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



can be written as 𝐿𝐿 𝑦𝑦 = 𝑔𝑔(𝑥𝑥)


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Differential Equations Definition: 𝑛𝑛𝑡𝑡𝑡order differential operator is:

𝐿𝐿 = 𝑎𝑎𝑛𝑛 𝑥𝑥 𝐷𝐷𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝐷𝐷𝑛𝑛−1 + ⋯+ 𝑎𝑎1 𝑥𝑥 𝐷𝐷 + 𝑎𝑎0(𝑥𝑥)


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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


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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


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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


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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


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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


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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)


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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


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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


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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


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Non Homogeneous Equations

𝑎𝑎𝑛𝑛 𝑥𝑥 𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛

+ 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑑𝑑𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1



,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


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E.g. 𝑦𝑦′′′ − 6𝑦𝑦′′ + 11𝑦𝑦′ − 6𝑦𝑦 = 3𝑥𝑥 non-homogeneous equation

Let 𝒚𝒚𝒑𝒑 = −𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐− 𝟏𝟏

𝟐𝟐𝟐𝟐. Is it a solution?


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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


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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

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Reduction of Order

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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

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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

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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

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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

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− 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

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⇒𝑑𝑑𝑤𝑤𝑑𝑑𝑥𝑥

= −2𝑤𝑤

⇒𝑑𝑑𝑤𝑤𝑤𝑤

= −2 𝑑𝑑𝑥𝑥

⇒ �𝑑𝑑𝑤𝑤𝑤𝑤 = �−2𝑑𝑑𝑥𝑥

⇒ ln 𝑤𝑤 = −2𝑥𝑥 + 𝑐𝑐 ⇒ 𝑤𝑤 = 𝑒𝑒−2𝑥𝑥+𝑐𝑐 = 𝑒𝑒−2𝑥𝑥 𝑒𝑒𝑐𝑐 = 𝑒𝑒−2𝑥𝑥 𝑐𝑐1 ⇒ 𝑤𝑤 = 𝑐𝑐1𝑒𝑒−2𝑥𝑥

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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

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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

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Check: 𝑦𝑦 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥


= 𝛼𝛼1𝑒𝑒𝑥𝑥 − 𝛼𝛼2𝑒𝑒−𝑥𝑥

⇒𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥

Hence, 𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2 − 𝑦𝑦 = 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥 − 𝛼𝛼1𝑒𝑒𝑥𝑥 + 𝛼𝛼2𝑒𝑒−𝑥𝑥 = 𝟎𝟎

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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)

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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

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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


= −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

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�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

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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

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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)

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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

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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

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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

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Homogeneous Linear Eq. with Constant Coefficients

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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


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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


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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


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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


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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


2𝑎𝑎

P(x) Q(x)

(Note: 𝑚𝑚1 = − 𝑏𝑏2𝑎𝑎⇒ − 𝑏𝑏

𝑎𝑎= 2𝑚𝑚1)


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Introduction

𝑦𝑦2 𝑥𝑥 = 𝑒𝑒𝑚𝑚1𝑥𝑥 �𝑒𝑒∫ 2𝑚𝑚1 𝑑𝑑𝑥𝑥

𝑒𝑒2𝑚𝑚1𝑥𝑥𝑑𝑑𝑥𝑥 = 𝑒𝑒𝑚𝑚1𝑥𝑥 �

𝑒𝑒2𝑚𝑚1𝑥𝑥

𝑒𝑒2𝑚𝑚1𝑥𝑥𝑑𝑑𝑥𝑥

= 𝑒𝑒𝑚𝑚1𝑥𝑥 �𝑑𝑑𝑥𝑥 = 𝑥𝑥𝑒𝑒𝑚𝑚1𝑥𝑥

𝑦𝑦2 𝑥𝑥 = 𝑥𝑥𝑒𝑒𝑚𝑚1𝑥𝑥 General solution: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝑚𝑚1𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒𝑚𝑚1𝑥𝑥


2𝑎𝑎


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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


2𝑎𝑎

Since 𝑦𝑦 = 𝑐𝑐1𝑒𝑒 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥 + 𝑐𝑐2𝑒𝑒 𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥 is a solution ∀𝑐𝑐1 &∀𝑐𝑐2


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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


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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


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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


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b) 𝑦𝑦′′ − 10𝑦𝑦′ + 25𝑦𝑦 = 0 𝑚𝑚2 − 10𝑚𝑚 + 25 = 0 ⇒ 𝑚𝑚 − 5 2 = 0 ⇒ 𝑚𝑚1 = 𝑚𝑚2 = 5 General solution: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒5𝑥𝑥 + 𝑐𝑐2𝑥𝑥𝑒𝑒5𝑥𝑥


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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


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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.


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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



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𝑦𝑦′′ + 𝐾𝐾2𝑦𝑦 = 0 & 𝑦𝑦′′ − 𝐾𝐾2𝑦𝑦 = 0 𝐾𝐾:real 𝑚𝑚2 − 𝐾𝐾2 = 0 ⇒ 𝑚𝑚2 = 𝐾𝐾2 ⇒ 𝑚𝑚 = ±𝐾𝐾 Which results in: 𝑦𝑦 = 𝑐𝑐1𝑒𝑒𝐾𝐾𝑥𝑥 + 𝑐𝑐2𝑒𝑒−𝐾𝐾𝑥𝑥



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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



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𝑎𝑎𝑛𝑛 𝑥𝑥 𝑑𝑑𝑛𝑛𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛

+ 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑑𝑑𝑛𝑛−1𝑦𝑦𝑑𝑑𝑥𝑥𝑛𝑛−1


+ 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


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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



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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


Section 3.4 Undetermined Coefficients

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Undetermined Coefficients

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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


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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, … …


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Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.


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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


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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


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⇒ 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


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⇒ 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


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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


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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


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⇒ 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


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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)


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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


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⇒ 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)


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


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


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


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


9/30/2014 Dr. Eli Saber 96



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.


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


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


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.


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


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


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


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


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



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



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



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


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


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.


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)


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


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)


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


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


′ + 𝑜𝑜2′𝑦𝑦2


⟹𝑑𝑑𝑑𝑑𝑥𝑥 𝑜𝑜1


+ 𝑃𝑃[𝑜𝑜1′𝑦𝑦1 + 𝑜𝑜2

′𝑦𝑦2] + 𝑜𝑜1′𝑦𝑦1

′ + 𝑜𝑜2′𝑦𝑦2


• Have two unknown functions 𝑜𝑜1 & 𝑜𝑜2 ⟹ Need two equations ⟹ make further assumption that 𝒖𝒖𝟏𝟏’𝒚𝒚𝟏𝟏 + 𝒖𝒖𝟐𝟐’𝒚𝒚𝟐𝟐 = 𝟎𝟎 ⟹ 𝒖𝒖𝟏𝟏’𝒚𝒚𝟏𝟏’ + 𝒖𝒖𝟐𝟐’𝒚𝒚𝟐𝟐’ = 𝒇𝒇(𝟐𝟐)


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


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


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)


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


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


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


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


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


𝑊𝑊 = 𝑒𝑒4𝑥𝑥 𝑊𝑊2 = 𝑥𝑥 + 1 𝑒𝑒4𝑥𝑥

𝑜𝑜1 = −13𝑥𝑥3 −

𝑥𝑥2

2


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


𝑊𝑊 = 𝑒𝑒4𝑥𝑥 𝑊𝑊2 = 𝑥𝑥 + 1 𝑒𝑒4𝑥𝑥

𝑜𝑜1 = −13𝑥𝑥3 −

𝑥𝑥2

2

𝑜𝑜2 =𝑥𝑥2

2+ 𝑥𝑥


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)


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


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


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


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)


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)


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


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


9/30/2014 Dr. Eli Saber 136

Try a solution of the form 𝑦𝑦 = 𝑥𝑥𝑚𝑚


= 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


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


9/30/2014 Dr. Eli Saber 138



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



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



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


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


𝑷𝑷(𝟐𝟐) Q(𝟐𝟐)


9/30/2014 Dr. Eli Saber 142


+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


𝑦𝑦1 = 𝑥𝑥𝑚𝑚1 𝑦𝑦 = 𝑜𝑜𝑦𝑦1

𝑦𝑦′ = 𝑜𝑜𝑦𝑦1′ + 𝑜𝑜′𝑦𝑦1 𝑦𝑦′′ = 𝑜𝑜𝑦𝑦′′ + 2𝑜𝑜′𝑦𝑦1′ + 𝑜𝑜′′𝑦𝑦1

=0 since 𝒚𝒚𝟏𝟏 = 𝟐𝟐𝒎𝒎 is a solution


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



9/30/2014 Dr. Eli Saber 144


= −2𝑚𝑚1𝑥𝑥𝑚𝑚1−1

𝑥𝑥𝑚𝑚1 𝑑𝑑𝑥𝑥 −


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


𝑑𝑑𝑤𝑤𝑤𝑤

= −2𝑦𝑦1′

𝑦𝑦1 𝑑𝑑𝑥𝑥 −


𝑑𝑑𝑥𝑥

𝑦𝑦1 = 𝑥𝑥𝑚𝑚1

⇒ 𝑦𝑦1′ = 𝑚𝑚1𝑥𝑥𝑚𝑚1−1


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


𝑚𝑚1 = −𝑏𝑏 − 𝑎𝑎2𝑎𝑎

⇒ 2𝑚𝑚1 = −𝑏𝑏 − 𝑎𝑎𝑎𝑎


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



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


𝑦𝑦 = 𝑐𝑐1𝑥𝑥𝑚𝑚1 + 𝑐𝑐2𝑥𝑥𝑚𝑚1 𝑔𝑔𝑛𝑛 𝑥𝑥

Repeated roots


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



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


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



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



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



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



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



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



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)


Auxiliary Equation

𝑚𝑚2 + 4 = 0 ⇒ 𝑚𝑚2 = −4 ⇒ 𝑚𝑚2 = 4𝑗𝑗2 ⇒ 𝑚𝑚 = ±2𝑗𝑗


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



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


Auxiliary Equation

𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥


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


𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2𝑥𝑥3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥3

𝑷𝑷(𝟐𝟐) 𝑸𝑸(𝟐𝟐) 𝒇𝒇(𝟐𝟐)


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 = −


2𝑥𝑥3 = −𝑥𝑥2𝑒𝑒𝑥𝑥

⇒ 𝑜𝑜2′ =𝑊𝑊2𝑊𝑊 =


2𝑥𝑥3 = 𝑒𝑒𝑥𝑥


𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2𝑥𝑥3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥2𝑒𝑒𝑥𝑥


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


𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2𝑥𝑥3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥2𝑒𝑒𝑥𝑥 𝑜𝑜1′ = −𝑥𝑥2𝑒𝑒𝑥𝑥

𝑜𝑜2′ = 𝑒𝑒𝑥𝑥

Integration by parts


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


𝑥𝑥2𝑦𝑦′′ − 3𝑥𝑥𝑦𝑦′ + 3𝑦𝑦 = 2𝑥𝑥4𝑒𝑒𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑥𝑥 + 𝑐𝑐2𝑥𝑥3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥2𝑒𝑒𝑥𝑥 𝑜𝑜1′ = −𝑥𝑥2𝑒𝑒𝑥𝑥

𝑜𝑜2 = 𝑒𝑒𝑥𝑥


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


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


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


= 𝟎𝟎 critically damped


< 𝟎𝟎 under damped

Now,


𝐶𝐶2𝐿𝐿 = −

𝑅𝑅2𝐿𝐿 ±

𝑅𝑅2 − 4𝐿𝐿𝐶𝐶

2𝐿𝐿


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


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]


𝑑𝑑 𝑡𝑡 = 𝑒𝑒−20𝑡𝑡 𝑐𝑐1 cos60𝑡𝑡 + 𝑐𝑐2 sin 60𝑡𝑡

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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


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


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


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


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


+𝑑𝑑𝐶𝐶

= 𝐸𝐸 𝑡𝑡

⇒ 1𝑑𝑑2𝑑𝑑𝑑𝑑𝑡𝑡2

+ 2𝑑𝑑𝑑𝑑𝑑𝑑𝑡𝑡

+𝑑𝑑

0.25= 50 cos 𝑡𝑡

Linear Models

9/30/2014 Dr. Eli Saber 176


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


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


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


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


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


+𝑑𝑑𝐶𝐶

= 𝐸𝐸 𝑡𝑡

Auxiliary Eqn: 𝐿𝐿𝑚𝑚2 + 𝑅𝑅𝑚𝑚 + 1𝐶𝐶

= 0


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


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.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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

Chapter 3-Higher Order Differential Equations

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Transcript of Chapter 3-Higher Order Differential Equations