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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

9/30/2014

Dr. Eli Saber

2

Section 3.1 Theory of Linear Equations

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Dr. Eli Saber

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Theory of Linear Equations Objective: Investigate Differential Equations of Order 2++

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Dr. Eli Saber

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Theory of Linear Equations Initial Value and Boundary Value Problems IVP: Initial Value Problem BVP: Boundary Value Problem •

IVP:

Solve:

𝑑𝑑 𝑛𝑛 𝑦𝑦 𝑑𝑑 𝑛𝑛−1 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑎𝑎𝑛𝑛 𝑥𝑥 + 𝑎𝑎 𝑥𝑥 + … + 𝑎𝑎 𝑥𝑥 + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) 𝑛𝑛−1 1 𝑑𝑑𝑥𝑥 𝑛𝑛 𝑑𝑑𝑥𝑥 𝑛𝑛−1 𝑑𝑑𝑑𝑑

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

Theory of Linear Equations 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 & 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 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.

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Dr. Eli Saber

6

Theory of Linear Equations Initial Value and Boundary Value Problems Theorem: Existence of a Unique Solution (for nth order D.E.) 𝑑𝑑 𝑛𝑛 𝑦𝑦 𝑑𝑑 𝑛𝑛−1 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑎𝑎𝑛𝑛 𝑥𝑥 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 + … + 𝑎𝑎1 𝑥𝑥 + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑛𝑛 𝑑𝑑𝑥𝑥 𝑛𝑛−1

Let 𝑎𝑎𝑛𝑛 𝑥𝑥 , 𝑎𝑎𝑛𝑛−1 𝑥𝑥 , … , 𝑎𝑎1 𝑥𝑥 , 𝑎𝑎0 𝑥𝑥 & 𝑔𝑔 𝑥𝑥 be continuous on an interval 𝐼𝐼, and let 𝑎𝑎𝑛𝑛 𝑥𝑥 ≠ 0 ∀ 𝑥𝑥 𝜖𝜖𝜖𝜖 If 𝑥𝑥 = 𝑥𝑥0 is any point in 𝐼𝐼,

a solution 𝑦𝑦(𝑥𝑥) of the IVP exists on the interval and is unique.

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Dr. Eli Saber

7

Theory of Linear Equations 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 9/30/2014

Dr. Eli Saber

8

Theory of Linear Equations Initial Value and Boundary Value Problems E.g. 𝑦𝑦 ′′ − 4𝑦𝑦 = 12𝑥𝑥

𝑦𝑦 0 = 4; 𝑦𝑦 ′ 0 = 1 Solution:

𝒚𝒚 = 𝟑𝟑𝒆𝒆𝟐𝟐𝟐𝟐 + 𝒆𝒆−𝟐𝟐𝟐𝟐 − 𝟑𝟑𝟑𝟑

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 − 𝟑𝟑𝟑𝟑 is the unique solution on interval 𝑰𝑰 9/30/2014

Dr. Eli Saber

9

Theory of Linear Equations Initial Value and Boundary Value Problems Check: 𝑦𝑦 ′′ − 4𝑦𝑦 = 12𝑥𝑥;

Solution: 𝑦𝑦 = 3𝑒𝑒 2𝑥𝑥 + 𝑒𝑒 −2𝑥𝑥 − 3𝑥𝑥

Now, 𝑦𝑦 ′ = 6𝑒𝑒 2𝑥𝑥 − 2𝑒𝑒 −2𝑥𝑥 − 3

And, 𝑦𝑦 ′′ = 12𝑒𝑒 2𝑥𝑥 + 4𝑒𝑒 −2𝑥𝑥

𝑦𝑦 ′′ − 4𝑦𝑦 = 12𝑒𝑒 2𝑥𝑥 + 4𝑒𝑒 −2𝑥𝑥 − 4 3𝑒𝑒 2𝑥𝑥 + 𝑒𝑒 −2𝑥𝑥 − 3𝑥𝑥

= 12𝑒𝑒 2𝑥𝑥 + 4𝑒𝑒 −2𝑥𝑥 − 12𝑒𝑒 2𝑥𝑥 − 4𝑒𝑒 −2𝑥𝑥 + 12𝑥𝑥

9/30/2014

Dr. Eli Saber

10

Theory of Linear Equations Initial Value and Boundary Value Problems Check: 𝑦𝑦 ′′ − 4𝑦𝑦 = 12𝑥𝑥; 𝑦𝑦 = 3𝑒𝑒 2𝑥𝑥 + 𝑒𝑒 −2𝑥𝑥 − 3𝑥𝑥 Now, 𝑦𝑦 ′ = 6𝑒𝑒 2𝑥𝑥 − 2𝑒𝑒 −2𝑥𝑥 − 3

And, 𝑦𝑦 ′′ = 12𝑒𝑒 2𝑥𝑥 + 4𝑒𝑒 −2𝑥𝑥

𝑦𝑦 ′′ − 4𝑦𝑦 = 12𝑒𝑒 2𝑥𝑥 + 4𝑒𝑒 −2𝑥𝑥 − 4 3𝑒𝑒 2𝑥𝑥 + 𝑒𝑒 −2𝑥𝑥 − 3𝑥𝑥

= 12𝑒𝑒 2𝑥𝑥 + 4𝑒𝑒 −2𝑥𝑥 − 12𝑒𝑒 2𝑥𝑥 − 4𝑒𝑒 −2𝑥𝑥 + 12𝑥𝑥

Verified. 9/30/2014

= 12𝑥𝑥

Dr. Eli Saber

11

Theory of Linear Equations Initial Value and Boundary Value Problems E.g. 𝑥𝑥 2 𝑦𝑦 ′′ − 2𝑥𝑥𝑦𝑦 ′ + 2𝑦𝑦 = 6 𝑦𝑦 0 = 3; 𝑦𝑦 ′ 0 = 1 Solution:

𝑦𝑦 = 𝑐𝑐𝑥𝑥 2 + 𝑥𝑥 + 3

9/30/2014

in interval 𝐼𝐼 ≡ (−∞, ∞)

Dr. Eli Saber

12

Theory of Linear Equations Initial Value and Boundary Value Problems Check: 𝑦𝑦 = 𝑐𝑐𝑥𝑥 2 + 𝑥𝑥 + 3 ⇒ 𝑦𝑦 ′ = 2𝑐𝑐𝑐𝑐 + 1; 𝑦𝑦 ′′ = 2𝑐𝑐

𝑥𝑥 2 𝑦𝑦 ′′ − 2𝑥𝑥𝑦𝑦 ′ + 2𝑦𝑦 = 𝑥𝑥 2 2𝑐𝑐 − 2𝑥𝑥 2𝑐𝑐𝑐𝑐 + 1 + 2 𝑐𝑐𝑥𝑥 2 + 𝑥𝑥 + 3 = 2𝑐𝑐𝑥𝑥 2 − 4𝑐𝑐𝑥𝑥 2 − 2𝑥𝑥 + 2𝑐𝑐𝑥𝑥 2 + 2𝑥𝑥 + 6

= 2𝑐𝑐𝑥𝑥 2 − 4𝑐𝑐𝑥𝑥 2 − 2𝑥𝑥 + 2𝑐𝑐𝑥𝑥 2 + 2𝑥𝑥 + 6

𝑥𝑥 2 𝑦𝑦 ′′ − 2𝑥𝑥𝑦𝑦 ′ + 2𝑦𝑦 = 6

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Dr. Eli Saber

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Theory of Linear Equations Initial Value and Boundary Value Problems IVP Check: 𝑦𝑦 = 𝑐𝑐𝑥𝑥 2 + 𝑥𝑥 + 3 • •

•

𝑦𝑦 0 = 3 ⇒ 3 = 𝑐𝑐 0

2

+ 0 +3⇒3=3

𝑦𝑦 ′ 0 = 1  𝑦𝑦 = 𝑐𝑐𝑥𝑥 2 + 𝑥𝑥 + 3 ⇒ 𝑦𝑦 ′ = 2𝑐𝑐𝑐𝑐 + 1  1 = 2𝑐𝑐 0 + 1 ⇒ 1 = 1

Note: For 𝑦𝑦 = 𝑐𝑐𝑥𝑥 2 + 𝑥𝑥 + 3 , the initial conditions of 𝑦𝑦 0 = 3 & 𝑦𝑦 ′ 0 = 1 did not provide a unique value for 𝑐𝑐

Hence: 𝑦𝑦 = 𝑐𝑐𝑥𝑥 2 + 𝑥𝑥 + 3 is a solution for the D.E. 𝑥𝑥 2 𝑦𝑦 ′′ − 2𝑥𝑥𝑦𝑦 ′ + 2𝑦𝑦 = 6 ∀𝑐𝑐 i.e. there is no unique solution But what w.r.t. unique solution theorem? 9/30/2014

Dr. Eli Saber

14

Theory of Linear Equations Initial Value and Boundary Value Problems Let us apply the theorem towards this example. 𝑥𝑥 2 𝑦𝑦 ′′ − 2𝑥𝑥 𝑦𝑦 ′ + 2 𝑦𝑦 = 6

𝑎𝑎2 𝑥𝑥 = 𝑥𝑥 2

𝑎𝑎1 𝑥𝑥 = −2𝑥𝑥

𝑎𝑎0 𝑥𝑥 = 2

𝑔𝑔 𝑥𝑥 = 6

Note: 𝑎𝑎2 𝑥𝑥 = 𝑥𝑥 2 = 0 𝑓𝑓𝑓𝑓𝑓𝑓 𝑥𝑥 = 0 and 𝑥𝑥 ∈ 𝐼𝐼 = (−∞, ∞)

 𝒂𝒂𝟐𝟐 𝒙𝒙 ≠ 𝟎𝟎∀ 𝒙𝒙 = 𝒙𝒙𝟎𝟎 ∈ 𝑰𝑰  this condition is NOT satisfied 9/30/2014

Dr. Eli Saber

15

Theory of Linear Equations Initial Value and Boundary Value Problems •

BVP (Boundary Value Problem): 𝑑𝑑 2 𝑦𝑦 𝑑𝑑𝑑𝑑 𝐷𝐷. 𝐸𝐸. ∶ 𝑎𝑎2 𝑥𝑥 + 𝑎𝑎1 𝑥𝑥 + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔 𝑥𝑥 𝑑𝑑𝑥𝑥 2 𝑑𝑑𝑑𝑑

Boundary conditions

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:

𝑨𝑨𝟏𝟏 𝒚𝒚 𝒂𝒂 + 𝑩𝑩𝟏𝟏 𝒚𝒚′ 𝒂𝒂 = 𝑪𝑪𝟏𝟏 𝑨𝑨𝟐𝟐 𝒚𝒚 𝒃𝒃 + 𝑩𝑩𝟐𝟐 𝒚𝒚′ 𝒃𝒃 = 𝑪𝑪𝟐𝟐

9/30/2014

Dr. Eli Saber

16

Theory of Linear Equations 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

9/30/2014

Dr. Eli Saber

17

Theory of Linear Equations Initial Value and Boundary Value Problems E.g. 𝑑𝑑 2 𝑥𝑥 + 16𝑥𝑥 = 0 𝑑𝑑𝑡𝑡 2

Solution: 𝑥𝑥 = 𝑐𝑐1 cos 4𝑡𝑡 + 𝑐𝑐2 sin 4𝑡𝑡 Check:

𝑑𝑑𝑑𝑑 = −4𝑐𝑐1 sin 4𝑡𝑡 + 4𝑐𝑐2 cos 4𝑡𝑡 𝑑𝑑𝑑𝑑

𝑑𝑑 2 𝑥𝑥 = −16𝑐𝑐1 cos 4𝑡𝑡 − 16𝑐𝑐2 sin 4𝑡𝑡 𝑑𝑑𝑡𝑡 2

𝑑𝑑 2 𝑥𝑥 + 16𝑥𝑥 = −16𝑐𝑐1 cos 4𝑡𝑡 − 16𝑐𝑐2 sin 4𝑡𝑡 + 16 𝑐𝑐1 cos 4𝑡𝑡 + 𝑐𝑐2 sin 4𝑡𝑡 = 0 𝑑𝑑𝑡𝑡 2 9/30/2014

Dr. Eli Saber

18

Theory of Linear Equations Initial Value and Boundary Value Problems Now, let us consider these different sets of BV Conditions: 𝑥𝑥 = 𝑐𝑐1 cos 4𝑡𝑡 + 𝑐𝑐2 sin 4𝑡𝑡 𝜋𝜋 1) 𝑥𝑥 0 = 0; 𝑥𝑥 =0 2 • •

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

𝜋𝜋 2

= 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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Dr. Eli Saber

19

Theory of Linear Equations Initial Value and Boundary Value Problems 2) 𝑥𝑥 0 = 0; 𝑥𝑥 • •

𝜋𝜋 =0 8

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

𝜋𝜋 8

= 0 ⇒ 0 = 𝑐𝑐1 cos

– But 𝑐𝑐1 = 0

𝜋𝜋 2

+ 𝑐𝑐2 sin

𝜋𝜋 2

𝑥𝑥 = 𝑐𝑐1 cos 4𝑡𝑡 + 𝑐𝑐2 sin 4𝑡𝑡

⇒ 0 = 𝑐𝑐1 0 + 𝑐𝑐2 1

– That means, 𝑐𝑐2 1 = 0 – Implies 𝒄𝒄𝟐𝟐 = 𝟎𝟎 •

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

Unique solution ≡ 𝒙𝒙 = 𝟎𝟎

9/30/2014

Dr. Eli Saber

20

Theory of Linear Equations Initial Value and Boundary Value Problems 3) 𝑥𝑥 0 = 0; 𝑥𝑥 • •

𝜋𝜋 =1 2

𝑥𝑥 0 = 0 ⇒ 0 = 𝑐𝑐1 1 + 𝑐𝑐2 0 ⇒ 𝒄𝒄𝒄𝒄 = 𝟎𝟎 𝑥𝑥

𝜋𝜋 2

= 1 ⇒ 1 = 𝑐𝑐1 cos 4𝜋𝜋 ∙

– 1 = 0 1 + 𝑐𝑐2 0

𝜋𝜋 2

+ 𝑐𝑐2 sin 4𝜋𝜋 ∙

𝑥𝑥 = 𝑐𝑐1 cos 4𝑡𝑡 + 𝑐𝑐2 sin 4𝑡𝑡 𝜋𝜋 2

⇒ 1 = 𝑐𝑐1 cos(2𝜋𝜋) + 𝑐𝑐2 sin(2𝜋𝜋)

– That means, 𝑐𝑐2 0 = 1 𝟏𝟏 𝟎𝟎

– Implies 𝒄𝒄𝟐𝟐 = = 𝑵𝑵. 𝑫𝑫. •

– Not possible to find 𝑐𝑐2

No solution for BVP

9/30/2014

Dr. Eli Saber

21

Theory of Linear Equations Differential Operators “D” E.g.

𝑑𝑑𝑑𝑑 = 𝐷𝐷𝐷𝐷 𝑑𝑑𝑑𝑑

𝑑𝑑𝑦𝑦 2 𝑑𝑑 𝑑𝑑𝑑𝑑 = = 𝐷𝐷 𝐷𝐷𝐷𝐷 = 𝐷𝐷2 𝑦𝑦 2 𝑑𝑑𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 i.e.

𝑑𝑑 𝑑𝑑𝑑𝑑

cos 4𝑥𝑥 = −4 sin 4𝑥𝑥 ⇒ 𝐷𝐷 cos 4𝑥𝑥 = −4 sin 4𝑥𝑥

𝒅𝒅𝒏𝒏 𝒚𝒚 ⇒ 𝐼𝐼𝐼𝐼 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔: = 𝑫𝑫𝒏𝒏 𝒚𝒚 𝒏𝒏 𝒅𝒅𝒙𝒙

9/30/2014

Dr. Eli Saber

22

Theory of Linear Equations Differential Equations Note: • • •

𝐷𝐷 𝑐𝑐 𝑓𝑓 𝑥𝑥

= 𝑐𝑐 𝐷𝐷𝐷𝐷(𝑥𝑥)

𝐷𝐷 𝑓𝑓 𝑥𝑥 + 𝑔𝑔 𝑥𝑥

= 𝐷𝐷𝐷𝐷 𝑥𝑥 + 𝐷𝐷𝐷𝐷(𝑥𝑥)

𝐷𝐷 𝛼𝛼 𝑓𝑓 𝑥𝑥 + 𝛽𝛽 𝑔𝑔 𝑥𝑥

= 𝛼𝛼 𝐷𝐷 𝑓𝑓 𝑥𝑥

– 𝛼𝛼, 𝛽𝛽 are constants

9/30/2014

Linear

+ 𝛽𝛽 𝐷𝐷 𝑔𝑔 𝑥𝑥

Dr. Eli Saber

23

Theory of Linear Equations Differential Equations  Let 𝑦𝑦 ′′ + 5𝑦𝑦 ′ + 6𝑦𝑦 = 5𝑥𝑥 − 3 This can be written was

𝑑𝑑2 𝑦𝑦 𝑑𝑑𝑥𝑥 2

+5

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

+ 6𝑦𝑦 = 5𝑥𝑥 − 3

Which can also be written as: 𝐷𝐷2 𝑦𝑦 + 5𝐷𝐷𝐷𝐷 + 6𝑦𝑦 = 5𝑥𝑥 − 3  Similarly, 𝑎𝑎𝑛𝑛 𝑥𝑥

𝑑𝑑𝑛𝑛 𝑦𝑦 𝑑𝑑𝑥𝑥 𝑛𝑛

+ 𝑎𝑎𝑛𝑛−1 𝑥𝑥

can be written as 𝐿𝐿 𝑦𝑦 = 0  And, 𝑎𝑎𝑛𝑛 𝑥𝑥

𝑑𝑑𝑛𝑛 𝑦𝑦 𝑑𝑑𝑥𝑥 𝑛𝑛

+ 𝑎𝑎𝑛𝑛−1 𝑥𝑥

can be written as 𝐿𝐿 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) 9/30/2014

𝑑𝑑𝑛𝑛−1 𝑦𝑦 𝑑𝑑𝑥𝑥 𝑛𝑛−1

𝑑𝑑𝑛𝑛−1 𝑦𝑦 𝑑𝑑𝑥𝑥 𝑛𝑛−1

+ … + 𝑎𝑎1 𝑥𝑥

+ … + 𝑎𝑎1 𝑥𝑥 Dr. Eli Saber

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

+ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 0

+ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) 24

Theory of Linear Equations Definition:

Differential Equations

𝑛𝑛𝑡𝑡𝑡 order differential operator is:

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

9/30/2014

Dr. Eli Saber

25

Theory of Linear Equations 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 = 𝑐𝑐1 𝑦𝑦1 𝑥𝑥 + 𝑐𝑐2 𝑦𝑦2 𝑥𝑥 + ⋯ + 𝑐𝑐𝑘𝑘 𝑦𝑦𝑘𝑘 𝑥𝑥

,where 𝑐𝑐1 , 𝑐𝑐2 , … , 𝑐𝑐𝑘𝑘 as are arbitrary constants, is also a solution on 𝐼𝐼

Corollaries: • A constant multiple 𝑦𝑦 = 𝑐𝑐1 𝑦𝑦1 (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

9/30/2014

Dr. Eli Saber

26

Theory of Linear Equations Superposition Principle E.g.

3

𝑑𝑑 𝑦𝑦 𝑥𝑥 3 3 𝑑𝑑𝑥𝑥

− 2𝑥𝑥

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

+ 4𝑦𝑦 = 0

And 𝑦𝑦1 = 𝑥𝑥 2 & 𝑦𝑦2 = 𝑥𝑥 2 ln 𝑥𝑥 are both solutions Check:

First solution: 𝑦𝑦1 = 𝑥𝑥 2

2 𝑦𝑦 3 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑦𝑦 = 𝑥𝑥 2 ⇒ = 2𝑥𝑥 ⇒ 2 = 2 & 3 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥

Implies, 𝑥𝑥 3 0 − 2𝑥𝑥 2𝑥𝑥 + 4 𝑥𝑥 2 = −4𝑥𝑥 2 + 4𝑥𝑥 2 = 0 9/30/2014

Dr. Eli Saber

27

Theory of Linear Equations Superposition Principle E.g. 𝑥𝑥 3

𝑑𝑑3 𝑦𝑦 𝑑𝑑𝑥𝑥 3

− 2𝑥𝑥

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

+ 4𝑦𝑦 = 0

And 𝑦𝑦1 = 𝑥𝑥 2 & 𝑦𝑦2 = 𝑥𝑥 2 ln 𝑥𝑥 are both solutions Check:

Second solution: 𝑦𝑦1 = 𝑥𝑥 2 ln 𝑥𝑥 𝑦𝑦 = 𝑥𝑥 2 ln 𝑥𝑥 ⇒

𝑑𝑑𝑑𝑑 1 = 2𝑥𝑥 ln 𝑥𝑥 + 𝑥𝑥 2 = 2𝑥𝑥 ln 𝑥𝑥 + 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑥𝑥

𝑥𝑥 𝑑𝑑 3 𝑦𝑦 2 𝑑𝑑 2 𝑦𝑦 = 2 ln 𝑥𝑥 + + 1 = 2 ln 𝑥𝑥 + 3 ⇒ 3 = 𝑥𝑥 𝑥𝑥 𝑑𝑑𝑥𝑥 2 𝑑𝑑𝑥𝑥 Implies, 𝑥𝑥 3

2 𝑥𝑥

− 2𝑥𝑥 2𝑥𝑥 ln 𝑥𝑥 + 𝑥𝑥 + 4 𝑥𝑥 2 ln 𝑥𝑥 = 2𝑥𝑥 2 − 4𝑥𝑥 2 ln 𝑥𝑥 − 2𝑥𝑥 2 + 4𝑥𝑥 2 ln 𝑥𝑥 = 0

By superposition  𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙𝟐𝟐 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝟐𝟐 𝐥𝐥𝐥𝐥 𝒙𝒙 is also a solution 9/30/2014

Dr. Eli Saber

28

Theory of Linear Equations 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: 𝒄𝒄𝟏𝟏 𝒇𝒇𝟏𝟏 𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒇𝒇𝟐𝟐 𝒙𝒙 + ⋯ + 𝒄𝒄𝒏𝒏 𝒇𝒇𝒏𝒏 𝒙𝒙 = 𝟎𝟎 ∀𝒙𝒙 ∈ 𝑰𝑰

• A set of functions is linearly independent on an interval if the only constants for which 𝒄𝒄𝟏𝟏 𝒇𝒇𝟏𝟏 𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒇𝒇𝟐𝟐 𝒙𝒙 + ⋯ + 𝒄𝒄𝒏𝒏 𝒇𝒇𝒏𝒏 𝒙𝒙 = 𝟎𝟎 ∀𝒙𝒙 ∈ 𝑰𝑰 are 𝒄𝒄𝟏𝟏 = 𝒄𝒄𝟐𝟐 = 𝒄𝒄𝟑𝟑 = ⋯ = 𝒄𝒄𝒏𝒏 = 𝟎𝟎 9/30/2014

Dr. Eli Saber

29

Theory of Linear Equations 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: 𝒄𝒄𝟏𝟏 𝒇𝒇𝟏𝟏 𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒇𝒇𝟐𝟐 𝒙𝒙 + ⋯ + 𝒄𝒄𝒏𝒏 𝒇𝒇𝒏𝒏 𝒙𝒙 = 𝟎𝟎 ∀𝒙𝒙 ∈ 𝑰𝑰

• A set of functions is linearly independent on an interval if the only constants for which 𝒄𝒄𝟏𝟏 𝒇𝒇𝟏𝟏 𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒇𝒇𝟐𝟐 𝒙𝒙 + ⋯ + 𝒄𝒄𝒏𝒏 𝒇𝒇𝒏𝒏 𝒙𝒙 = 𝟎𝟎 ∀𝒙𝒙 ∈ 𝑰𝑰 are 𝒄𝒄𝟏𝟏 = 𝒄𝒄𝟐𝟐 = 𝒄𝒄𝟑𝟑 = ⋯ = 𝒄𝒄𝒏𝒏 = 𝟎𝟎 9/30/2014

Dr. Eli Saber

30

Theory of Linear Equations Wronskian

Definition:

Suppose each of the functions 𝑓𝑓1 𝑥𝑥 , 𝑓𝑓2 𝑥𝑥 , … , 𝑓𝑓𝑛𝑛 (𝑥𝑥) possesses at least 𝑛𝑛 − 1 derivatives Then

9/30/2014

𝑊𝑊 𝑓𝑓1 , 𝑓𝑓2 , … , 𝑓𝑓𝑛𝑛

𝑓𝑓1 𝑓𝑓2 … ⋯ 𝑓𝑓𝑛𝑛 𝑓𝑓1′ 𝑓𝑓2′ … … 𝑓𝑓𝑛𝑛′ = ⋮ (𝑛𝑛−1) (𝑛𝑛−1) (𝑛𝑛−1) 𝑓𝑓1 𝑓𝑓2 … 𝑓𝑓𝑛𝑛 Dr. Eli Saber

31

Theory of Linear Equations 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 ∀𝒙𝒙 ∈ 𝑰𝑰

9/30/2014

Dr. Eli Saber

32

Theory of Linear Equations Wronskian E.g.

𝑾𝑾 𝒇𝒇𝟏𝟏 , 𝒇𝒇𝟐𝟐 , … , 𝒇𝒇𝒏𝒏 ≠ 𝟎𝟎 ∀𝒙𝒙 ∈ 𝑰𝑰

𝑦𝑦1 = 𝑒𝑒 3𝑥𝑥 and 𝑦𝑦2 = 𝑒𝑒 −3𝑥𝑥 are both the solutions of the homogeneous linear equation 𝑦𝑦 ′′ − 9𝑦𝑦 = 0; 𝐼𝐼 = (−∞, ∞) Check: 𝑊𝑊

𝑒𝑒 3𝑥𝑥 , 𝑒𝑒 −3𝑥𝑥

3𝑥𝑥 𝑒𝑒 = 3𝑒𝑒 3𝑥𝑥

𝑒𝑒 −3𝑥𝑥 −3𝑒𝑒 −3𝑥𝑥

= 𝑒𝑒 3𝑥𝑥 −3𝑒𝑒 −3𝑥𝑥 − 𝑒𝑒 −3𝑥𝑥 3𝑒𝑒 3𝑥𝑥 = −3 − 3 = −6 ≠ 0

Thus, 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆𝟑𝟑𝟑𝟑 + 𝒄𝒄𝟐𝟐 𝒆𝒆−𝟑𝟑𝟑𝟑 is the general solution 9/30/2014

Dr. Eli Saber

33

Theory of Linear Equations Wronskian E.g.

𝑾𝑾 𝒇𝒇𝟏𝟏 , 𝒇𝒇𝟐𝟐 , … , 𝒇𝒇𝒏𝒏 ≠ 𝟎𝟎 ∀𝒙𝒙 ∈ 𝑰𝑰

𝑦𝑦 ′′′ − 6𝑦𝑦 ′′ + 11𝑦𝑦 ′ − 6𝑦𝑦 = 0

The functions 𝑦𝑦1 = 𝑒𝑒 𝑥𝑥 ; 𝑦𝑦2 = 𝑒𝑒 2𝑥𝑥 & 𝑦𝑦3 = 𝑒𝑒 3𝑥𝑥 satisfy the D.E. above Check:

𝑊𝑊 𝑒𝑒 𝑥𝑥 , 𝑒𝑒 2𝑥𝑥 , 𝑒𝑒 3𝑥𝑥 2𝑥𝑥 = 𝑒𝑒 𝑥𝑥 2𝑒𝑒 2𝑥𝑥 4𝑒𝑒

= 2𝑒𝑒 6𝑥𝑥 ≠ 0

𝑒𝑒 𝑥𝑥 = 𝑒𝑒 𝑥𝑥 𝑒𝑒 𝑥𝑥

𝑒𝑒 2𝑥𝑥 2𝑒𝑒 2𝑥𝑥 4𝑒𝑒 2𝑥𝑥

3𝑒𝑒 3𝑥𝑥 − 𝑒𝑒 2𝑥𝑥 𝑒𝑒 𝑥𝑥 9𝑒𝑒 3𝑥𝑥 𝑒𝑒 𝑥𝑥

𝑒𝑒 3𝑥𝑥 3𝑒𝑒 3𝑥𝑥 9𝑒𝑒 3𝑥𝑥

3𝑒𝑒 3𝑥𝑥 + 𝑒𝑒 3𝑥𝑥 𝑒𝑒 𝑥𝑥 9𝑒𝑒 3𝑥𝑥 𝑒𝑒 𝑥𝑥

2𝑒𝑒 2𝑥𝑥 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 4𝑒𝑒 2𝑥𝑥

Hence, 𝒆𝒆𝒙𝒙 , 𝒆𝒆𝟐𝟐𝟐𝟐 , 𝒆𝒆𝟑𝟑𝟑𝟑 form a fundamental set & 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒆𝒆𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟑𝟑 𝒆𝒆𝟑𝟑𝟑𝟑 is the general solution 9/30/2014

Dr. Eli Saber

34

Theory of Linear Equations Non Homogeneous Equations 𝑑𝑑 𝑛𝑛 𝑦𝑦 𝑑𝑑 𝑛𝑛−1 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑎𝑎𝑛𝑛 𝑥𝑥 + 𝑎𝑎 𝑥𝑥 + … + 𝑎𝑎 𝑥𝑥 + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) 𝑛𝑛−1 1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑛𝑛 𝑑𝑑𝑥𝑥 𝑛𝑛−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 •

a

If 𝑦𝑦1 , 𝑦𝑦2 , … , 𝑦𝑦𝑛𝑛 are solutions of Homogeneous equations and 𝑦𝑦𝑝𝑝 is any particular solution, 𝑦𝑦 = 𝑐𝑐1 𝑦𝑦1 𝑥𝑥 + 𝑐𝑐2 𝑦𝑦2 𝑥𝑥 + ⋯ + 𝑐𝑐𝑛𝑛 𝑦𝑦𝑛𝑛 𝑥𝑥 + 𝑦𝑦𝑝𝑝

General solution

9/30/2014

Complementary S𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝒚𝒚𝒄𝒄 Dr. Eli Saber

Particular Solution 𝒚𝒚𝒑𝒑

35

Theory of Linear Equations Non Homogeneous Equations E.g. 𝑦𝑦 ′′′ − 6𝑦𝑦 ′′ + 11𝑦𝑦 ′ − 6𝑦𝑦 = 3𝑥𝑥  non-homogeneous equation Let 𝒚𝒚𝒑𝒑 = −

9/30/2014

𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏

𝟏𝟏 𝟐𝟐

− 𝒙𝒙. Is it a solution?

Dr. Eli Saber

36

Theory of Linear Equations Non Homogeneous Equations E.g.

1 𝑦𝑦𝑝𝑝′ = − ; 𝑦𝑦𝑝𝑝′′ = 0; 𝑦𝑦𝑝𝑝′′′ = 0 2 ⇒ 𝑦𝑦

′′′

𝑦𝑦 ′′′ − 6𝑦𝑦 ′′ + 11𝑦𝑦 ′ − 6𝑦𝑦 = 3𝑥𝑥 ; 𝑦𝑦𝑝𝑝 = −

11 12

1

− 𝑥𝑥 2

1 11 1 − 6𝑦𝑦 + 11𝑦𝑦 − 6𝑦𝑦 = 0 − 6 0 + 11 − −6 − − 𝑥𝑥 2 12 2 ′′

′

11 11 =− + + 3𝑥𝑥 = 𝟑𝟑𝟑𝟑 2 2 Verified.

9/30/2014

Dr. Eli Saber

37

Theory of Linear Equations Non Homogeneous Equations Homogeneous Equation:

𝑦𝑦 ′′′ − 6𝑦𝑦 ′′ + 11𝑦𝑦 ′ − 6𝑦𝑦 = 3𝑥𝑥 ; 𝑦𝑦𝑝𝑝 = −

𝑦𝑦 ′′′ − 6𝑦𝑦 ′′ + 11𝑦𝑦 ′ − 6𝑦𝑦 = 0

11 12

1

− 𝑥𝑥 2

Let 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 2𝑥𝑥 + 𝑐𝑐3 𝑒𝑒 3𝑥𝑥 be a complimentary solution Hence, the general solution is given by: 𝑦𝑦 = 𝑐𝑐1

𝑒𝑒 𝑥𝑥

9/30/2014

+ 𝑐𝑐2

𝑒𝑒 2𝑥𝑥 𝒚𝒚𝒄𝒄

+ 𝑐𝑐3

𝑒𝑒 3𝑥𝑥

11 1 + (− − 𝑥𝑥) 12 2 𝒚𝒚𝒑𝒑

Dr. Eli Saber

38

Section 3.2 Reduction of Order

9/30/2014

Dr. Eli Saber

39

Reduction of Order Introduction 2nd 𝑜𝑜𝑟𝑟𝑑𝑑𝑒𝑒𝑟𝑟 𝐻𝐻𝑜𝑜𝑚𝑚𝑜𝑜𝑔𝑔𝑒𝑒𝑛𝑛𝑒𝑒𝑜𝑜𝑢𝑢𝑠𝑠 𝐷𝐷.𝐸𝐸.: 𝑎𝑎2 𝑥𝑥 𝑦𝑦 ′′ + 𝑎𝑎1 𝑥𝑥 𝑦𝑦 ′ + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 0 Solution: 𝑦𝑦 = 𝑐𝑐1 𝑦𝑦1 + 𝑐𝑐2 𝑦𝑦2

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

9/30/2014

Dr. Eli Saber

40

Reduction of Order Introduction Approach: •

𝑦𝑦

Recall if 𝑦𝑦1 𝑥𝑥 & 𝑦𝑦2 (𝑥𝑥) are L.I. => 𝑦𝑦2 is non-constant 1



𝑦𝑦2 𝑦𝑦1

= 𝑢𝑢 𝑥𝑥 𝑜𝑜𝑜𝑜 𝑦𝑦2 𝑥𝑥 = 𝑢𝑢 𝑥𝑥 𝑦𝑦1 (𝑥𝑥)

Seek to find 𝑢𝑢(𝑥𝑥) in order to find 𝒚𝒚𝟐𝟐 𝒙𝒙 = 𝒖𝒖 𝒙𝒙 𝒚𝒚𝟏𝟏 (𝒙𝒙)

9/30/2014

Dr. Eli Saber

41

Reduction of Order E.g. Given

𝑑𝑑2 𝑦𝑦 𝑑𝑑𝑥𝑥 2

− 𝑦𝑦 = 0; 𝐼𝐼 = (−∞, ∞) and assume that 𝑦𝑦1 = 𝑒𝑒 𝑥𝑥 is a solution. Find

a second solution 𝑦𝑦2 Check:

2 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑦𝑦 𝑥𝑥 𝑥𝑥 𝑦𝑦 = 𝑒𝑒 ⇒ = 𝑒𝑒 ⇒ 2 = 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥

And substituting back in the equation,

9/30/2014

𝑑𝑑2 𝑦𝑦 𝑑𝑑𝑥𝑥 2

− 𝑦𝑦 = 𝑒𝑒 𝑥𝑥 − 𝑒𝑒 𝑥𝑥 = 𝟎𝟎

Dr. Eli Saber

42

Reduction of Order Let 𝑦𝑦 𝑥𝑥 = 𝑢𝑢 𝑥𝑥 𝑦𝑦1 𝑥𝑥 = 𝑢𝑢 𝑥𝑥 𝑒𝑒 𝑥𝑥 ⇒

𝑑𝑑𝑑𝑑 = 𝑢𝑢 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 𝑢𝑢′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑

𝑑𝑑 2 𝑦𝑦 ⇒ 2 = 𝑢𝑢 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 𝑢𝑢′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 𝑢𝑢′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 𝑢𝑢′′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑥𝑥 𝑑𝑑 2 𝑦𝑦 ⇒ 2 = 𝑢𝑢 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 2𝑢𝑢′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 𝑢𝑢′′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑥𝑥

Hence, 𝑑𝑑 2 𝑦𝑦 − 𝑦𝑦 = 0 ⇒ 𝑢𝑢 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 2𝑢𝑢′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 𝑢𝑢′′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 − 𝑢𝑢 𝑥𝑥 𝑒𝑒 𝑥𝑥 = 0 2 𝑑𝑑𝑥𝑥 9/30/2014

Dr. Eli Saber

43

Reduction of Order 𝑑𝑑 2 𝑦𝑦 − 𝑦𝑦 = 0 ⇒ 2𝑢𝑢′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 + 𝑢𝑢′′ 𝑥𝑥 𝑒𝑒 𝑥𝑥 = 0 2 𝑑𝑑𝑥𝑥 ⇒ 𝑒𝑒 𝑥𝑥 𝑢𝑢′′ + 2𝑢𝑢′ = 0 But 𝑒𝑒 𝑥𝑥 ≠ 0.

⇒ 𝑢𝑢′′ + 2𝑢𝑢′ = 0

Let 𝑤𝑤 = 𝑢𝑢𝑢  change of variable ⇒ 𝑤𝑤 ′ + 2𝑤𝑤 = 0

(Linear First Order D.E.)

𝑑𝑑𝑑𝑑 ⇒ + 2𝑤𝑤 = 0 𝑑𝑑𝑑𝑑 ⇒

𝑑𝑑𝑑𝑑 = −2𝑤𝑤 𝑑𝑑𝑑𝑑

9/30/2014

Dr. Eli Saber

44

Reduction of Order ⇒

𝑑𝑑𝑑𝑑 = −2𝑤𝑤 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 ⇒ = −2 𝑑𝑑𝑑𝑑 𝑤𝑤 ⇒�

𝑑𝑑𝑑𝑑 = � −2 𝑑𝑑𝑑𝑑 𝑤𝑤

⇒ ln 𝑤𝑤 = −2𝑥𝑥 + 𝑐𝑐

⇒ 𝑤𝑤 = 𝑒𝑒 −2𝑥𝑥+𝑐𝑐 = 𝑒𝑒 −2𝑥𝑥 𝑒𝑒 𝑐𝑐 = 𝑒𝑒 −2𝑥𝑥 𝑐𝑐1 ⇒ 𝑤𝑤 = 𝑐𝑐1 𝑒𝑒 −2𝑥𝑥

9/30/2014

Dr. Eli Saber

45

Reduction of Order Introduction 𝑤𝑤 = 𝑐𝑐1 𝑒𝑒 −2𝑥𝑥

But 𝑤𝑤 = 𝑢𝑢′ ⇒ 𝑢𝑢′ = 𝑐𝑐1 𝑒𝑒 −2𝑥𝑥 ⇒ Hence, ∫ 𝑑𝑑𝑑𝑑 = ∫ 𝑐𝑐1 𝑒𝑒 −2𝑥𝑥 𝑑𝑑𝑑𝑑 1 ⇒ 𝑢𝑢 = − 𝑐𝑐1 𝑒𝑒 −2𝑥𝑥 + 𝑐𝑐2 2

Hence, 𝑦𝑦 = 𝑢𝑢 𝑥𝑥 𝑒𝑒 𝑥𝑥 = − ⇒ 𝑦𝑦 = − 9/30/2014

𝑐𝑐1 −𝑥𝑥 𝑒𝑒 + 𝑐𝑐2 𝑒𝑒 𝑥𝑥 2

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑐𝑐1 −2𝑥𝑥 𝑒𝑒 2

= 𝑐𝑐1 𝑒𝑒 −2𝑥𝑥

+ 𝑐𝑐2 𝑒𝑒 𝑥𝑥

Dr. Eli Saber

46

Reduction of Order 𝑦𝑦 = −

𝑐𝑐1 −𝑥𝑥 𝑒𝑒 + 𝑐𝑐2 𝑒𝑒 𝑥𝑥 2

Let, 𝑐𝑐1 = −2 & 𝑐𝑐2 = 0 ⇒ 𝑦𝑦2 𝑥𝑥 = 𝑒𝑒 −𝑥𝑥

Let us check for independence in the two solutions 𝑊𝑊

𝑒𝑒 𝑥𝑥 , 𝑒𝑒 −𝑥𝑥

𝑒𝑒 𝑥𝑥 = 𝑥𝑥 𝑒𝑒

𝑒𝑒 −𝑥𝑥 = −𝑒𝑒 𝑥𝑥 𝑒𝑒 −𝑥𝑥 − 𝑒𝑒 𝑥𝑥 𝑒𝑒 −𝑥𝑥 = −1 − 1 = −2 ≠ 0 −𝑥𝑥 −𝑒𝑒

 𝑒𝑒 𝑥𝑥 & 𝑒𝑒 −𝑥𝑥 are independent

 General solution: 𝑦𝑦 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 −𝑥𝑥 Wronskian

9/30/2014

Dr. Eli Saber

47

Reduction of Order Check: 𝑦𝑦 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 −𝑥𝑥

𝑑𝑑𝑦𝑦 ⇒ = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 − 𝛼𝛼2 𝑒𝑒 −𝑥𝑥 𝑑𝑑𝑑𝑑

𝑑𝑑 2 𝑦𝑦 ⇒ 2 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 −𝑥𝑥 𝑑𝑑𝑥𝑥 Hence,

𝑑𝑑 2 𝑦𝑦 − 𝑦𝑦 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 −𝑥𝑥 − 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 −𝑥𝑥 = 𝟎𝟎 2 𝑑𝑑𝑥𝑥 9/30/2014

Dr. Eli Saber

48

Reduction of Order General case: 𝑎𝑎2 𝑥𝑥 𝑦𝑦 ′′ + 𝑎𝑎1 𝑥𝑥 𝑦𝑦𝑦 + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 0 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑏𝑏𝑏𝑏 𝑎𝑎2 𝑥𝑥 ⇒ 𝑦𝑦 ′′ +

𝑎𝑎1 𝑥𝑥 ′ 𝑎𝑎0 𝑥𝑥 𝑦𝑦 + 𝑦𝑦 = 0 𝑎𝑎2 𝑥𝑥 𝑎𝑎2 𝑥𝑥 P(x)

Q(x)

⇒ 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 9/30/2014

Dr. Eli Saber

49

Reduction of Order Introduction Let 𝑦𝑦 𝑥𝑥 = 𝑢𝑢 𝑥𝑥 𝑦𝑦1 𝑥𝑥

⇒ 𝑦𝑦 ′ 𝑥𝑥 = 𝑢𝑢 𝑥𝑥 𝑦𝑦1′ 𝑥𝑥 + 𝑢𝑢′ 𝑥𝑥 𝑦𝑦1 𝑥𝑥 ⇒ 𝒚𝒚′ = 𝒖𝒖𝒚𝒚′𝟏𝟏 + 𝒖𝒖′ 𝒚𝒚𝟏𝟏

⇒ 𝑦𝑦 ′′ = 𝑢𝑢𝑦𝑦1′′ + 𝑢𝑢′ 𝑦𝑦1′ + 𝑢𝑢′ 𝑦𝑦1′ + 𝑢𝑢′′ 𝑦𝑦1 ′ ′ ′′ ⇒ 𝒚𝒚′′ = 𝒖𝒖𝒚𝒚′′ 𝟏𝟏 + 𝟐𝟐𝒖𝒖 𝒚𝒚𝟏𝟏 + 𝒖𝒖 𝒚𝒚𝟏𝟏

Now, 𝑦𝑦 ′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦 ′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦 = 0 Replacing,

𝑢𝑢𝑦𝑦1′′ + 2𝑢𝑢′ 𝑦𝑦1′ + 𝑢𝑢′′ 𝑦𝑦1 + 𝑃𝑃 𝑥𝑥 𝑢𝑢𝑦𝑦1′ + 𝑢𝑢′ 𝑦𝑦1 + 𝑄𝑄 𝑥𝑥 𝑢𝑢𝑦𝑦1 = 0

9/30/2014

Dr. Eli Saber

50

Reduction of Order Rearranging terms,

𝑢𝑢𝑦𝑦1′′ + 2𝑢𝑢′ 𝑦𝑦1′ + 𝑢𝑢′′ 𝑦𝑦1 + 𝑃𝑃 𝑥𝑥 𝑢𝑢𝑦𝑦1′ + 𝑢𝑢′ 𝑦𝑦1 + 𝑄𝑄 𝑥𝑥 𝑢𝑢𝑦𝑦1 = 0

⇒ 𝑢𝑢 𝑦𝑦1′′ + 𝑃𝑃𝑦𝑦1′ + 𝑄𝑄𝑦𝑦1 + 𝑦𝑦1 𝑢𝑢′′ + 2𝑦𝑦1′ + 𝑃𝑃𝑦𝑦1 𝑢𝑢′ = 0 =0 since 𝑦𝑦1 is a solution

⇒ 𝑦𝑦1 𝑢𝑢′′ + 2𝑦𝑦1′ + 𝑃𝑃𝑦𝑦1 𝑢𝑢′ = 0

Let 𝑤𝑤 = 𝑢𝑢𝑢  change of variables  𝑤𝑤 ′ = 𝑢𝑢𝑢𝑢

 𝑦𝑦1 𝑤𝑤 ′ + 2𝑦𝑦1′ + 𝑃𝑃𝑦𝑦1 𝑤𝑤 = 0  linear and separable ⇒ 𝑦𝑦1 𝑤𝑤 ′ = − 2𝑦𝑦1′ + 𝑃𝑃𝑦𝑦1 𝑤𝑤 𝑜𝑜𝑜𝑜 𝑦𝑦1 ⇒

𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑦𝑦1 =− 2 + 𝑃𝑃 𝑦𝑦1 𝑑𝑑𝑑𝑑 𝑤𝑤 𝑦𝑦1 𝑑𝑑𝑑𝑑

9/30/2014

𝑑𝑑𝑦𝑦1 𝑑𝑑𝑑𝑑 =− 2 + 𝑃𝑃 𝑦𝑦1 𝑤𝑤 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

Dr. Eli Saber

51

Reduction of Order �

𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑦𝑦1 = �− 2 + 𝑃𝑃 𝑦𝑦1 𝑑𝑑𝑑𝑑 𝑤𝑤 𝑦𝑦1 𝑑𝑑𝑑𝑑

⇒�

𝑑𝑑𝑑𝑑 𝑑𝑑𝑦𝑦1 = � −2 − � 𝑃𝑃𝑑𝑑𝑑𝑑 𝑤𝑤 𝑦𝑦1

⇒ ln 𝑤𝑤 = −2 ln 𝑦𝑦1 − � 𝑃𝑃𝑃𝑃𝑃𝑃 + 𝑐𝑐 ⇒ ln 𝑤𝑤 + 2 ln 𝑦𝑦1 = − � 𝑃𝑃𝑃𝑃𝑃𝑃 + 𝑐𝑐 ⇒ ln 𝑤𝑤 + ln 𝑦𝑦12 = − � 𝑃𝑃𝑃𝑃𝑃𝑃 + 𝑐𝑐 ⇒ ln 𝑤𝑤𝑦𝑦12 = − � 𝑃𝑃𝑃𝑃𝑃𝑃 + 𝑐𝑐 ⇒ 𝑤𝑤𝑦𝑦12 = 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 +𝑐𝑐 = 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 𝑒𝑒 𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 9/30/2014

Dr. Eli Saber

52

Reduction of Order 𝑤𝑤𝑦𝑦12 = 𝑐𝑐1 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 ⇒ 𝑤𝑤 = 𝑐𝑐1 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 /𝑦𝑦12 But,

𝑤𝑤 =

𝑢𝑢′

⇒ 𝑤𝑤 =

𝑢𝑢′

𝑑𝑑𝑑𝑑 𝑐𝑐1 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 = = 𝑑𝑑𝑑𝑑 𝑦𝑦12

𝑐𝑐1 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 𝑐𝑐1 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 ⇒ 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 ⇒ � 𝑑𝑑𝑑𝑑 = � 𝑑𝑑𝑑𝑑 𝑦𝑦12 𝑦𝑦12 𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 ⇒ 𝑢𝑢 = 𝑐𝑐1 � 𝑑𝑑𝑑𝑑 + 𝑐𝑐2 𝑦𝑦12

Let 𝑐𝑐1 = 1 & 𝑐𝑐2 = 0 & note: 𝑦𝑦2 𝑥𝑥 = 𝑢𝑢 𝑥𝑥 𝑦𝑦1 (𝑥𝑥)

𝒆𝒆− ∫ 𝑷𝑷𝑷𝑷𝑷𝑷 𝒂𝒂𝟏𝟏 𝒙𝒙 ⇒ 𝒚𝒚𝟐𝟐 𝒙𝒙 = 𝒚𝒚𝟏𝟏 𝒙𝒙 � 𝟐𝟐 𝒅𝒅𝒅𝒅 ; 𝑷𝑷 𝒙𝒙 = 𝒂𝒂𝟐𝟐 𝒙𝒙 𝒚𝒚𝟏𝟏 (𝒙𝒙) 9/30/2014

Dr. Eli Saber

53

Reduction of Order E.g. 𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 4𝑦𝑦 = 0; 𝐼𝐼 ≡ 0, ∞

Let 𝑦𝑦1 𝑥𝑥 = 𝑥𝑥 2 be a solution. Find a 2nd solution 𝑦𝑦2 (𝑥𝑥) and the general solution 𝑦𝑦(𝑥𝑥) Solution:

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 4𝑦𝑦 = 0 ⇒ 𝑦𝑦 ′′ −

3𝑥𝑥 ′ 4 𝑦𝑦 + 2 𝑦𝑦 = 0 𝑥𝑥 2 𝑥𝑥

⇒ 𝑦𝑦 ′′ + −

3 4 𝑦𝑦 ′ + 2 𝑦𝑦 = 0 𝑥𝑥 𝑥𝑥

P(x)

9/30/2014

Q(x)

Dr. Eli Saber

54

Reduction of Order According to our derivation, 𝑦𝑦2 𝑥𝑥 = 𝑦𝑦1 𝑥𝑥 ∫ = 𝑥𝑥 2 � = 𝑥𝑥 2 �

𝑒𝑒

3 − ∫ − 𝑑𝑑𝑑𝑑 𝑥𝑥

𝑒𝑒 ∫

𝑥𝑥 2

2

3 𝑑𝑑𝑑𝑑 𝑥𝑥

𝑥𝑥 4

𝑒𝑒 − ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 𝑦𝑦12 (𝑥𝑥)

𝑑𝑑𝑑𝑑

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 4𝑦𝑦 = 0

𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 3

𝑒𝑒 3 ln 𝑥𝑥 𝑒𝑒 ln 𝑥𝑥 𝑥𝑥 3 2 2 2 = 𝑥𝑥 � 4 𝑑𝑑𝑑𝑑 = 𝑥𝑥 � 4 𝑑𝑑𝑑𝑑 = 𝑥𝑥 � 4 𝑑𝑑𝑑𝑑 𝑥𝑥 𝑥𝑥 𝑥𝑥 1 = 𝑥𝑥 2 � 𝑑𝑑𝑑𝑑 = 𝑥𝑥 2 ln 𝑥𝑥 𝑥𝑥 ⇒ 𝒚𝒚𝟐𝟐 𝒙𝒙 = 𝒙𝒙𝟐𝟐 𝐥𝐥𝐥𝐥 𝒙𝒙 9/30/2014

General solution: 𝒚𝒚 𝒙𝒙 = 𝒄𝒄𝟏𝟏 𝒙𝒙𝟐𝟐 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝟐𝟐 𝐥𝐥𝐥𝐥 𝒙𝒙 Dr. Eli Saber

55

Reduction of Order 𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 4𝑦𝑦 = 0

Check: 𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 2 + 𝑐𝑐2 𝑥𝑥 2 ln 𝑥𝑥 ⇒ 𝑦𝑦 ′ = 2 𝑐𝑐1 𝑥𝑥 + 𝑐𝑐2

𝑥𝑥 2 2𝑥𝑥 ln 𝑥𝑥 + 𝑥𝑥

⇒ 𝑦𝑦 ′ = 2𝑐𝑐1 𝑥𝑥 + 2𝑐𝑐2 𝑥𝑥 ln 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥 𝑦𝑦 ′′ = 2𝑐𝑐1 + 2𝑐𝑐2 ln 𝑥𝑥 +

𝑥𝑥 + 𝑐𝑐2 𝑥𝑥

⇒ 𝑦𝑦 ′′ = 2𝑐𝑐1 + 2𝑐𝑐2 ln 𝑥𝑥 + 2𝑐𝑐2 + 𝑐𝑐2 = 2𝑐𝑐1 + 3𝑐𝑐2 + 2𝑐𝑐2 ln 𝑥𝑥

9/30/2014

Dr. Eli Saber

56

Reduction of Order 𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 4𝑦𝑦 = 0

We know,

𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 2 + 𝑐𝑐2 𝑥𝑥 2 ln 𝑥𝑥

𝑦𝑦 ′ = 2𝑐𝑐1 𝑥𝑥 + 2𝑐𝑐2 𝑥𝑥 ln 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥 & 𝑦𝑦 ′′ = 2𝑐𝑐1 + 2𝑐𝑐2 ln 𝑥𝑥 + 2𝑐𝑐2 + 𝑐𝑐2 = 2𝑐𝑐1 + 3𝑐𝑐2 + 2𝑐𝑐2 ln 𝑥𝑥 Replace in D.E.:

𝑥𝑥 2 2𝑐𝑐1 + 3𝑐𝑐2 + 2𝑐𝑐2 ln 𝑥𝑥 − 3x 2𝑐𝑐1 𝑥𝑥 + 2𝑐𝑐2 𝑥𝑥 ln 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥 + 4 𝑐𝑐1 𝑥𝑥 2 + 𝑐𝑐2 𝑥𝑥 2 ln 𝑥𝑥 = 2𝑐𝑐1 𝑥𝑥 2 + 3𝑐𝑐2 𝑥𝑥 2 + 2𝑐𝑐2 ln 𝑥𝑥 𝑥𝑥 2 − 6𝑐𝑐1 𝑥𝑥 2 − 6𝑐𝑐2 𝑥𝑥 2 ln 𝑥𝑥 − 3𝑐𝑐2 𝑥𝑥 2 + 4𝑐𝑐1 𝑥𝑥 2 + 4𝑐𝑐2 𝑥𝑥 2 ln 𝑥𝑥

= 𝟎𝟎

9/30/2014

Dr. Eli Saber

57

Section 3.3 Homogeneous Linear Eq. with Constant Coefficients

9/30/2014

Dr. Eli Saber

58

Homogeneous Linear Eq. with Constant Coefficients

Introduction 𝒂𝒂𝒏𝒏 𝒚𝒚

•

𝒏𝒏

+ 𝒂𝒂𝒏𝒏−𝟏𝟏 𝒚𝒚(𝒏𝒏−𝟏𝟏) + ⋯ + 𝒂𝒂𝟏𝟏 𝒚𝒚′ + 𝒂𝒂𝟎𝟎 𝒚𝒚 = 𝟎𝟎

𝑛𝑛𝑡𝑡𝑡 order Linear Constant Coefficients Differential Equation

𝑎𝑎𝑖𝑖 ; 𝑖𝑖 = 0,1, … , 𝑛𝑛 are real constant coefficients and 𝑎𝑎𝑛𝑛 ≠ 0

Objective: To find a solution to the above homogeneous solution

9/30/2014

Dr. Eli Saber

59

Homogeneous Linear Eq. with Constant Coefficients

Auxiliary Equation Consider the special Case ( 2nd order LCCDE) given as: 𝑎𝑎𝑦𝑦 ′′ + 𝑏𝑏𝑦𝑦 ′ + 𝑐𝑐𝑐𝑐 = 0

Try a solution of the form 𝑦𝑦 = 𝑒𝑒 𝑚𝑚𝑚𝑚 ⇒ 𝑦𝑦 ′ = 𝑚𝑚𝑒𝑒 𝑚𝑚𝑚𝑚

⇒ 𝑦𝑦 ′′ = 𝑚𝑚2 𝑒𝑒 𝑚𝑚𝑚𝑚

Substituting back in the given D.E., 𝑎𝑎 𝑚𝑚2 𝑒𝑒 𝑚𝑚𝑚𝑚 + 𝑏𝑏 𝑚𝑚𝑒𝑒 𝑚𝑚𝑚𝑚 + 𝑐𝑐 𝑒𝑒 𝑚𝑚𝑚𝑚 = 0 ⇒ 𝑎𝑎𝑚𝑚2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 𝑒𝑒 𝑚𝑚𝑚𝑚 = 0 Now, 𝒆𝒆𝒎𝒎𝒎𝒎 ≠ 𝟎𝟎 ∀𝒙𝒙𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓

⇒

𝒂𝒂𝒎𝒎𝟐𝟐

9/30/2014

+ 𝒃𝒃𝒃𝒃 + 𝒄𝒄 = 𝟎𝟎

Auxiliary Eqn. of the LCCDE Dr. Eli Saber

60

Homogeneous Linear Eq. with Constant Coefficients

Introduction 𝒂𝒂𝒎𝒎𝟐𝟐 + 𝒃𝒃𝒃𝒃 + 𝒄𝒄 = 𝟎𝟎  Auxiliary Eqn. of the LCCDE

The only way that 𝑦𝑦 = 𝑒𝑒 𝑚𝑚𝑚𝑚 can satisfy the D.E. is if 𝑎𝑎𝑚𝑚2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0 Hence, choose 𝒎𝒎 as the root of the equation to solve the problem ⇒ 𝑚𝑚1,2

−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 = 2𝑎𝑎

The 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 leads to 3 cases:

1) 2) 3)

𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 > 0 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 = 0 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 < 0

9/30/2014

Dr. Eli Saber

61

Homogeneous Linear Eq. with Constant Coefficients

Introduction Case 1: 𝒃𝒃𝟐𝟐 −𝟒𝟒𝟒𝟒𝟒𝟒 > 𝟎𝟎

𝑚𝑚1,2

Here, 𝑚𝑚1 & 𝑚𝑚2 are real and distinct

−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 = 2𝑎𝑎

 2 solutions: 𝑦𝑦1 = 𝑒𝑒 𝑚𝑚1𝑥𝑥 & 𝑦𝑦2 = 𝑒𝑒 𝑚𝑚2𝑥𝑥

 𝑦𝑦1 &𝑦𝑦2 are linearly independent

 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 𝑚𝑚1𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 𝑚𝑚2𝑥𝑥 is the general solution

9/30/2014

Dr. Eli Saber

62

Homogeneous Linear Eq. with Constant Coefficients

Introduction Case 2: 𝒃𝒃𝟐𝟐 −𝟒𝟒𝟒𝟒𝟒𝟒 = 𝟎𝟎 𝑚𝑚1 = 𝑚𝑚2 = − Digression: 𝑎𝑎𝑦𝑦 ′′

+

𝑏𝑏𝑦𝑦 ′

𝑏𝑏 ⇒ 𝑦𝑦1 = 𝑒𝑒 𝑚𝑚1𝑥𝑥 & 𝑦𝑦2 = 𝑥𝑥𝑒𝑒 𝑚𝑚1𝑥𝑥 2𝑎𝑎

+ 𝑐𝑐𝑐𝑐 = 0 ⇒

⇒ 𝑦𝑦2 𝑥𝑥 = 𝑦𝑦1 𝑥𝑥 � 9/30/2014

𝑒𝑒

𝑦𝑦 ′′

𝑏𝑏 ′ 𝑐𝑐 + 𝑦𝑦 + 𝑦𝑦 = 0 𝑎𝑎 𝑎𝑎 P(x)

− ∫ 𝑃𝑃 𝑥𝑥 𝑑𝑑𝑑𝑑

𝑦𝑦1 𝑥𝑥

𝑚𝑚1,2

−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 = 2𝑎𝑎

2

Q(x)

𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑚𝑚1𝑥𝑥 �

𝑏𝑏 − ∫ 𝑑𝑑𝑑𝑑 𝑎𝑎 𝑒𝑒

𝑒𝑒 2𝑚𝑚1𝑥𝑥

Dr. Eli Saber

∫ 2𝑚𝑚1 𝑑𝑑𝑑𝑑 𝑒𝑒 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑚𝑚1𝑥𝑥 � 2𝑚𝑚 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑒𝑒 1

(Note: 𝑚𝑚1 = −

𝑏𝑏 2𝑎𝑎

𝑏𝑏 𝑎𝑎

⇒ − = 2𝑚𝑚1 )

63

Homogeneous Linear Eq. with Constant Coefficients

Introduction ∫ 2𝑚𝑚1 𝑑𝑑𝑑𝑑 𝑒𝑒 𝑒𝑒 2𝑚𝑚1𝑥𝑥 𝑚𝑚1 𝑥𝑥 𝑚𝑚1 𝑥𝑥 𝑦𝑦2 𝑥𝑥 = 𝑒𝑒 � 2𝑚𝑚 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑒𝑒 � 2𝑚𝑚 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑒𝑒 1 𝑒𝑒 1

𝑚𝑚1,2

−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 = 2𝑎𝑎

= 𝑒𝑒 𝑚𝑚1𝑥𝑥 � 𝑑𝑑𝑑𝑑 = 𝑥𝑥𝑒𝑒 𝑚𝑚1𝑥𝑥 𝑦𝑦2 𝑥𝑥 = 𝑥𝑥𝑒𝑒 𝑚𝑚1𝑥𝑥

 General solution: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 𝑚𝑚1𝑥𝑥 + 𝑐𝑐2 𝑥𝑥𝑒𝑒 𝑚𝑚1𝑥𝑥

9/30/2014

Dr. Eli Saber

64

Homogeneous Linear Eq. with Constant Coefficients

Introduction Case 3: 𝒃𝒃𝟐𝟐 −𝟒𝟒𝟒𝟒𝟒𝟒 < 𝟎𝟎

 𝑚𝑚1 & 𝑚𝑚2 are complex conjugate numbers

𝑚𝑚1,2

−𝑏𝑏 ± 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 = 2𝑎𝑎

 𝑚𝑚1 = 𝛼𝛼 + 𝑗𝑗𝑗𝑗 & 𝑚𝑚2 = 𝛼𝛼 − 𝑗𝑗𝑗𝑗 • 𝛼𝛼, 𝛽𝛽 > 0 and are real • 𝑗𝑗 2 = −1

 General solution: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 𝑚𝑚1𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 𝑚𝑚2𝑥𝑥  𝑦𝑦 = 𝑐𝑐1 𝑒𝑒

9/30/2014

𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥

+ 𝑐𝑐2 𝑒𝑒

𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥

Dr. Eli Saber

65

Homogeneous Linear Eq. with Constant Coefficients

Introduction Since 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 𝑦𝑦1 = 𝑐𝑐1 𝑒𝑒 = 𝑒𝑒

𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥

+ 𝑐𝑐2 𝑒𝑒

𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥

Choose 𝑐𝑐1 = 𝑐𝑐2 = 1 𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥

𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥

+ 𝑒𝑒

+ 𝑐𝑐2 𝑒𝑒

𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥

= 𝑒𝑒 𝛼𝛼𝑥𝑥 𝑒𝑒 𝑗𝑗𝑗𝑗𝑥𝑥 + 𝑒𝑒 −𝑗𝑗𝑗𝑗𝑥𝑥 = 𝑒𝑒 𝛼𝛼𝛼𝛼 2 cos 𝛽𝛽𝑥𝑥

𝑦𝑦1 = 2𝑒𝑒 𝛼𝛼𝛼𝛼 cos 𝛽𝛽𝛽𝛽

𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥

is a solution ∀𝑐𝑐1 &∀𝑐𝑐2

Choose 𝑐𝑐1 = 1 & 𝑐𝑐2 = −1

𝑦𝑦2 = 𝑐𝑐1 𝑒𝑒 = 𝑒𝑒

𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥

− 𝑒𝑒

+ 𝑐𝑐2 𝑒𝑒

𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥

𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥

= 𝑒𝑒 𝛼𝛼𝛼𝛼 𝑒𝑒 𝑗𝑗𝑗𝑗𝑗𝑗 − 𝑒𝑒 −𝑗𝑗𝑗𝑗𝑗𝑗 = 𝑒𝑒 𝛼𝛼𝛼𝛼 2𝑗𝑗 sin 𝛽𝛽𝛽𝛽

𝑦𝑦2 = 2𝑗𝑗𝑒𝑒 𝛼𝛼𝛼𝛼 sin 𝛽𝛽𝛽𝛽

General solution: 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆𝜶𝜶𝜶𝜶 𝐜𝐜𝐜𝐜𝐜𝐜 𝜷𝜷𝜷𝜷 + 𝒄𝒄𝟐𝟐 𝒆𝒆𝜶𝜶𝜶𝜶 𝐬𝐬𝐬𝐬𝐬𝐬 𝜷𝜷𝜷𝜷

9/30/2014

𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥

Dr. Eli Saber

66

Homogeneous Linear Eq. with Constant Coefficients

Alternate Derivation: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒

𝛼𝛼+𝑗𝑗𝑗𝑗 𝑥𝑥

+ 𝑐𝑐2 𝑒𝑒

𝛼𝛼−𝑗𝑗𝑗𝑗 𝑥𝑥

= 𝑐𝑐1 𝑒𝑒 𝛼𝛼𝛼𝛼 𝑒𝑒 𝑗𝑗𝑗𝑗𝑗𝑗 + 𝑐𝑐2 𝑒𝑒 𝛼𝛼𝛼𝛼 𝑒𝑒 −𝑗𝑗𝑗𝑗𝑗𝑗

= 𝑐𝑐1 𝑒𝑒 𝛼𝛼𝛼𝛼 cos 𝛽𝛽𝛽𝛽 + 𝑗𝑗 sin 𝛽𝛽𝛽𝛽 + 𝑐𝑐2 𝑒𝑒 𝛼𝛼𝛼𝛼 cos 𝛽𝛽𝛽𝛽 − 𝑗𝑗 sin 𝛽𝛽𝛽𝛽

= 𝑐𝑐1 𝑒𝑒 𝛼𝛼𝛼𝛼 cos 𝛽𝛽𝛽𝛽 + 𝑗𝑗𝑐𝑐1 𝑒𝑒 𝛼𝛼𝛼𝛼 sin 𝛽𝛽𝛽𝛽 + 𝑐𝑐2 𝑒𝑒 𝛼𝛼𝛼𝛼 cos 𝛽𝛽𝛽𝛽 − 𝑗𝑗𝑐𝑐2 𝑒𝑒 𝛼𝛼𝛼𝛼 sin 𝛽𝛽𝛽𝛽 = 𝑒𝑒 𝛼𝛼𝛼𝛼 𝑐𝑐1 + 𝑐𝑐2 cos 𝛽𝛽𝛽𝛽 + 𝑒𝑒 𝛼𝛼𝛼𝛼 𝑗𝑗𝑐𝑐1 − 𝑗𝑗𝑐𝑐2 sin 𝛽𝛽𝛽𝛽 Hence,

∝1

∝2

𝒚𝒚 = ∝𝟏𝟏 𝒆𝒆𝜶𝜶𝜶𝜶 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷𝜷𝜷 +∝𝟐𝟐 𝒆𝒆𝜶𝜶𝜶𝜶 𝒔𝒔𝒔𝒔𝒔𝒔 𝜷𝜷𝜷𝜷 9/30/2014

Dr. Eli Saber

67

Homogeneous Linear Eq. with Constant Coefficients

Example: a) 2𝑦𝑦 ′′ − 5𝑦𝑦 ′ − 3𝑦𝑦 = 0

Now, 2𝑚𝑚2 − 5𝑚𝑚 − 3 = 0

⇒ 2𝑚𝑚 + 1 𝑚𝑚 − 3 = 0 ⇒ 𝑚𝑚1 = −

1 ; 𝑚𝑚 = 3 2 2

General solution: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒

9/30/2014

1

− 2𝑥𝑥

+ 𝑐𝑐2 𝑒𝑒 3𝑥𝑥

Dr. Eli Saber

68

Homogeneous Linear Eq. with Constant Coefficients

b) 𝑦𝑦 ′′ − 10𝑦𝑦 ′ + 25𝑦𝑦 = 0

𝑚𝑚2 − 10𝑚𝑚 + 25 = 0 ⇒ 𝑚𝑚 − 5

2

=0

⇒ 𝑚𝑚1 = 𝑚𝑚2 = 5

General solution: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 5𝑥𝑥 + 𝑐𝑐2 𝑥𝑥𝑒𝑒 5𝑥𝑥

9/30/2014

Dr. Eli Saber

69

Homogeneous Linear Eq. with Constant Coefficients

c) 𝑦𝑦 ′′ + 4𝑦𝑦 ′ + 7𝑦𝑦 = 0

⇒ 𝑚𝑚2 + 4𝑚𝑚 + 7 = 0 ⇒ 𝑚𝑚 = ⇒ 𝑚𝑚 = ⇒ 𝑚𝑚 =

−4 ±

4 2 − 4(1)(7) −4 ± 16 − 28 = 2(1) 2

−4 ± −12 −4 ± 12 −1 −4 ± 𝑗𝑗 12 = = 2 2 2

−4 ± 𝑗𝑗 2 3 = −2 ± 𝑗𝑗 3 2

General solution: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 9/30/2014

−2+𝑗𝑗 3 𝑥𝑥

+ 𝑐𝑐2 𝑒𝑒

−2−𝑗𝑗 3 𝑥𝑥

or 𝑦𝑦 = 𝑒𝑒 −2𝑥𝑥 𝑐𝑐1 cos 3𝑥𝑥 + 𝑐𝑐2 sin 3𝑥𝑥 Dr. Eli Saber

70

Homogeneous Linear Eq. with Constant Coefficients

Two important Equations 𝑦𝑦 ′′ + 𝐾𝐾 2 𝑦𝑦 = 0 & 𝑦𝑦 ′′ − 𝐾𝐾 2 𝑦𝑦 = 0

𝐾𝐾:real

Where do we see these equations??

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

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

D.E. Free of Undamped Motion:

𝑑𝑑2 𝑥𝑥 𝑑𝑑𝑡𝑡 2

+ 𝜔𝜔2 𝑥𝑥 = 0

With the solution: 𝒙𝒙 = 𝒄𝒄𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜 𝒘𝒘𝒘𝒘 + 𝒄𝒄𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬 𝒘𝒘𝒘𝒘 9/30/2014

Dr. Eli Saber

HOW? 71

Homogeneous Linear Eq. with Constant Coefficients

Two important Equations 𝑦𝑦 ′′ + 𝐾𝐾 2 𝑦𝑦 = 0 & 𝑦𝑦 ′′ − 𝐾𝐾 2 𝑦𝑦 = 0

𝐾𝐾:real

𝑚𝑚2 + 𝐾𝐾 2 = 0

⇒ 𝑚𝑚2 = −𝐾𝐾 2 = 𝐾𝐾 2 𝑗𝑗 2

⇒ 𝑚𝑚 = ±𝐾𝐾𝐾𝐾

Which results in: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 𝐾𝐾𝐾𝐾𝐾𝐾 + 𝑐𝑐2 𝑒𝑒 −𝐾𝐾𝐾𝐾𝐾𝐾 or 𝑦𝑦 = 𝑐𝑐1 cos 𝐾𝐾𝐾𝐾 + 𝑐𝑐2 sin 𝐾𝐾𝐾𝐾

9/30/2014

Dr. Eli Saber

72

Homogeneous Linear Eq. with Constant Coefficients

Two important Equations 𝑦𝑦 ′′ + 𝐾𝐾 2 𝑦𝑦 = 0 & 𝑦𝑦 ′′ − 𝐾𝐾 2 𝑦𝑦 = 0

𝐾𝐾:real

𝑚𝑚2 − 𝐾𝐾 2 = 0 ⇒ 𝑚𝑚2 = 𝐾𝐾 2

⇒ 𝑚𝑚 = ±𝐾𝐾

Which results in: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 𝐾𝐾𝐾𝐾 + 𝑐𝑐2 𝑒𝑒 −𝐾𝐾𝐾𝐾

9/30/2014

Dr. Eli Saber

73

Homogeneous Linear Eq. with Constant Coefficients

Two important Equations Note: 𝑦𝑦 ′′ − 𝐾𝐾 2 𝑦𝑦 = 0  𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆𝑲𝑲𝑲𝑲 + 𝒄𝒄𝟐𝟐 𝒆𝒆−𝑲𝑲𝑲𝑲 • • •

1 2

If 𝑐𝑐1 = 𝑐𝑐2 = ⇒ 𝑦𝑦 = If 𝑐𝑐1 =

1 2

1 2

1 2

𝑒𝑒 𝐾𝐾𝐾𝐾 +

& 𝑐𝑐2 = − ⇒ 𝑦𝑦 =

1 2

1 2

𝑒𝑒 −𝐾𝐾𝐾𝐾 = cosh 𝐾𝐾𝐾𝐾

𝑒𝑒 𝐾𝐾𝐾𝐾 −

1 2

𝑒𝑒 −𝐾𝐾𝐾𝐾 = sinh 𝐾𝐾𝐾𝐾

Since cosh 𝐾𝐾𝐾𝐾 & sinh 𝐾𝐾𝐾𝐾 are linearly independent

– Alternate solution of 𝑦𝑦 ′′ − 𝐾𝐾 2 𝑦𝑦 = 0 is 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜 𝑲𝑲𝑲𝑲 + 𝒄𝒄𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 𝑲𝑲𝑲𝑲

9/30/2014

Dr. Eli Saber

74

Homogeneous Linear Eq. with Constant Coefficients

Higher Order Equations 𝑑𝑑 𝑛𝑛 𝑦𝑦 𝑑𝑑 𝑛𝑛−1 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑎𝑎𝑛𝑛 𝑥𝑥 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 + … + 𝑎𝑎1 𝑥𝑥 + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 0 𝑑𝑑𝑥𝑥 𝑛𝑛 𝑑𝑑𝑥𝑥 𝑛𝑛−1 𝑑𝑑𝑑𝑑

,where 𝑎𝑎𝑖𝑖 , 𝑖𝑖 = 0,1, … , 𝑛𝑛 are real constants

Auxiliary Equation: 𝑎𝑎𝑛𝑛 𝑚𝑚𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑚𝑚𝑛𝑛−1 + ⋯ + 𝑎𝑎2 𝑚𝑚2 + 𝑎𝑎1 𝑚𝑚 + 𝑎𝑎0 𝑚𝑚0 = 0 Case 1:

If all roots are distinct – general solution is given by: 𝑦𝑦 = 𝑐𝑐1 𝑒𝑒 𝑚𝑚1𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 𝑚𝑚2𝑥𝑥 + ⋯ + 𝑐𝑐𝑛𝑛 𝑒𝑒 𝑚𝑚𝑛𝑛𝑥𝑥

(similar to a 2nd order D.E.) 9/30/2014

Dr. Eli Saber

75

Homogeneous Linear Eq. with Constant Coefficients

Higher Order Equations 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: 𝑒𝑒 𝑚𝑚1𝑥𝑥 , 𝑥𝑥𝑒𝑒 𝑚𝑚1𝑥𝑥 , 𝑥𝑥 2 𝑒𝑒 𝑚𝑚1𝑥𝑥 ,…, 𝑥𝑥 𝑘𝑘−1 𝑒𝑒 𝑚𝑚1𝑥𝑥

Case 3:

Complex roots appear in conjugate pairs when the coefficients of the D.E. are real

9/30/2014

Dr. Eli Saber

76

Homogeneous Linear Eq. with Constant Coefficients

Higher Order Equations 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 = 𝒄𝒄𝟏𝟏 𝒆𝒆𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒆𝒆−𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟑𝟑 𝒙𝒙𝒆𝒆−𝟐𝟐𝟐𝟐 9/30/2014

Dr. Eli Saber

77

Section 3.4 Undetermined Coefficients

9/30/2014

Dr. Eli Saber

78

Undetermined Coefficients Solve a non-homogeneous Linear Differential Equation: 𝑎𝑎𝑛𝑛 𝑦𝑦 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑦𝑦 𝑛𝑛−1 + ⋯ + 𝑎𝑎1 𝑦𝑦1 + 𝑎𝑎0 𝑦𝑦 0 = 𝑔𝑔 𝑥𝑥 By: 1. Finding a complementary solution 𝑦𝑦𝑐𝑐 for the homogeneous equation. 2. Finding a particular solution 𝑦𝑦𝑝𝑝 .

⇒ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

9/30/2014

Dr. Eli Saber

79

Undetermined Coefficients Method of undetermined coefficient 𝒖𝒖𝒖𝒖 “Educated guess about the form of 𝒚𝒚𝒑𝒑 ”

Method is limited to non-homogeneous linear D.E. such that: 1. The coefficient 𝑎𝑎𝑖𝑖 , 𝑖𝑖 = 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 = 𝑥𝑥 2 − 5𝑥𝑥, … …

9/30/2014

Dr. Eli Saber

80

Undetermined Coefficients

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

9/30/2014

Dr. Eli Saber

81

Undetermined Coefficients E.g. 1:

𝑦𝑦 ′′ + 4𝑦𝑦 ′ − 2𝑦𝑦 = 2𝑥𝑥 2 − 3𝑥𝑥 + 6

𝑔𝑔 𝑥𝑥 Step 1: Solve the associated Homogeneous equation. 𝑦𝑦 ′′ + 4𝑦𝑦 ′ − 2𝑦𝑦 = 0 𝑚𝑚2 + 4𝑚𝑚 − 2 = 0 ⇒ 𝑚𝑚 = ⇒ 𝑚𝑚 =

−4 ± 16 − 4 1 −2 2

−4 ± 24 −4 ± 2 6 = 2 2

⇒ 𝑚𝑚 = −2 ± 6 ⇒ 𝑚𝑚1 = −2 − 6 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚2 = −2 + 6 9/30/2014

Dr. Eli Saber

82

Undetermined Coefficients 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 𝑚𝑚1𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 𝑚𝑚2𝑥𝑥

⇒ 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒

−2− 6 𝑥𝑥

𝑚𝑚1 = −2 − 6 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚2 = −2 + 6

+ 𝑐𝑐2 𝑒𝑒

−2+ 6 𝑥𝑥

Step 2: Note 𝑔𝑔(𝑥𝑥) is a quadratic ⇒ assume a particular solution of quadratic form.

(See Table 3.4.1)

⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥 2 + 𝐵𝐵𝐵𝐵 + 𝐶𝐶

⇒ 𝑦𝑦𝑝𝑝′ = 2𝐴𝐴𝐴𝐴 + 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑝𝑝′′ = 2𝐴𝐴 Substitute into D.E.

𝑦𝑦𝑝𝑝′′ + 4𝑦𝑦𝑝𝑝′ − 2𝑦𝑦𝑝𝑝 = 3𝑥𝑥 2 − 3𝑥𝑥 + 6

⇒ 2𝐴𝐴 + 4 2𝐴𝐴𝐴𝐴 + 𝐵𝐵 − 2 𝐴𝐴𝑥𝑥 2 + 𝐵𝐵𝐵𝐵 + 𝐶𝐶 = 2𝑥𝑥 2 − 3𝑥𝑥 + 6 9/30/2014

Dr. Eli Saber

83

Undetermined Coefficients ⇒ 2𝐴𝐴 + 8𝐴𝐴𝐴𝐴 + 4𝐵𝐵 − 2𝐴𝐴𝑥𝑥 2 − 2𝐵𝐵𝐵𝐵 − 2𝐶𝐶 = 2𝑥𝑥 2 − 3𝑥𝑥 + 6

⇒ −𝟐𝟐𝟐𝟐𝑥𝑥 2 + 𝟖𝟖𝟖𝟖 − 𝟐𝟐𝑩𝑩 𝑥𝑥 + 𝟐𝟐𝟐𝟐 + 𝟒𝟒𝟒𝟒 − 𝟐𝟐𝟐𝟐 = 2𝑥𝑥 2 − 3𝑥𝑥 + 6 • •

−2𝐴𝐴 = 2 ⇒ 𝐴𝐴 = −1

8𝐴𝐴 − 2𝐵𝐵 = −3 ⇒ 2𝐵𝐵 = 8𝐴𝐴 + 3 = 8 −1 + 3 = −5

−5 ⇒ 2𝐵𝐵 = −5 ⇒ 𝐵𝐵 = 2 •

2𝐴𝐴 + 4𝐵𝐵 − 2𝐶𝐶 = 6 ⇒ 2𝐶𝐶 = 2𝐴𝐴 + 4𝐵𝐵 − 6 = 2 −1 + 4

2𝐶𝐶 = −2 − 10 − 6 ⇒ 2𝐶𝐶 = −18 ⇒ 𝐶𝐶 = −9 9/30/2014

Dr. Eli Saber

−5 2

−6 84

Undetermined Coefficients 5 ⇒ 𝑦𝑦𝑝𝑝 = −𝑥𝑥 2 − 𝑥𝑥 − 9 2 ⇒ 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒

−2− 6 𝑥𝑥

+ 𝑐𝑐2 𝑒𝑒

−2+ 6 𝑥𝑥

⇒ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝 ⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆−

9/30/2014

𝟐𝟐+ 𝟔𝟔 𝒙𝒙

+ 𝒄𝒄𝟐𝟐 𝒆𝒆

−𝟐𝟐+ 𝟔𝟔 𝒙𝒙

Dr. Eli Saber

𝟓𝟓 − 𝒙𝒙𝟐𝟐 − 𝒙𝒙 − 𝟗𝟗 𝟐𝟐

85

Undetermined Coefficients E.g. 2: 𝑦𝑦 ′′ − 𝑦𝑦 ′ + 𝑦𝑦 = 2 sin 3𝑥𝑥

Step 1: Find 𝑦𝑦𝑐𝑐 𝑓𝑓𝑓𝑓𝑓𝑓 𝑦𝑦 ′′ − 𝑦𝑦 ′ + 𝑦𝑦 = 0 𝑚𝑚2 − 𝑚𝑚 + 1 = 0 ⇒ 𝑚𝑚 =

1 ± 3𝑗𝑗 2 1 3 ⇒ 𝑚𝑚 = ⇒ 𝑚𝑚 = ± 𝑗𝑗 2 2 2 ⇒ 𝑚𝑚1 =

3 3 1 1 + 𝑗𝑗 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚2 = − 𝑗𝑗 2 2 2 2

⇒ 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 9/30/2014

1± 1−4 1 1 2

1 3 +𝑗𝑗 𝑥𝑥 2 2

+ 𝑐𝑐2 𝑒𝑒

1 3 − 𝑗𝑗 𝑥𝑥 2 2

Dr. Eli Saber

86

Undetermined Coefficients Step 2: 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑦𝑦𝑝𝑝 . 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑦𝑦𝑝𝑝 = 𝐴𝐴 cos 3𝑥𝑥 + 𝐵𝐵 sin 3𝑥𝑥 (see Table 3.4.1)

⇒ 𝑦𝑦𝑝𝑝′ = −3𝐴𝐴 sin 3𝑥𝑥 + 3𝐵𝐵 cos 3𝑥𝑥 𝑦𝑦𝑝𝑝′′ = −9𝐴𝐴 cos 3𝑥𝑥 − 9𝐵𝐵 sin 3𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝′′ − 𝑦𝑦𝑝𝑝′ + 𝑦𝑦 = 2 sin 3𝑥𝑥

⇒ −9𝐴𝐴 cos 3𝑥𝑥 − 9𝐵𝐵 sin 3𝑥𝑥 − −3𝐴𝐴 sin 3𝑥𝑥 + 3𝐵𝐵 cos 3𝑥𝑥 + 𝐴𝐴 cos 3𝑥𝑥 + 𝐵𝐵 sin 3𝑥𝑥 = 2 sin 3𝑥𝑥

⇒ −9𝐴𝐴 cos 3𝑥𝑥 − 9𝐵𝐵 sin 3𝑥𝑥 + 3𝐴𝐴 sin 3𝑥𝑥 − 3𝐵𝐵 cos 3𝑥𝑥 + 𝐴𝐴 cos 3𝑥𝑥 + 𝐵𝐵 sin 3𝑥𝑥 = 2 sin 𝑥𝑥

⇒ −9𝐴𝐴 cos 3𝑥𝑥 − 3𝐵𝐵 cos 3𝑥𝑥 + 𝐴𝐴 cos 3𝑥𝑥 − 9𝐵𝐵 sin 3𝑥𝑥 + 3𝐴𝐴 sin 3𝑥𝑥 + 𝐵𝐵 sin 3𝑥𝑥 = 2 sin 𝑥𝑥 ⇒ 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑 −𝟖𝟖𝟖𝟖 − 𝟑𝟑𝟑𝟑 + 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 −𝟖𝟖𝟖𝟖 + 𝟑𝟑𝟑𝟑 = 2 sin 𝑥𝑥 ⇒ 3𝐴𝐴 − 8𝐵𝐵 = 2 𝑎𝑎𝑎𝑎𝑎𝑎 − 8𝐴𝐴 − 3𝐵𝐵 = 0 9/30/2014

Dr. Eli Saber

87

Undetermined Coefficients

⇒ 𝐴𝐴 =

6 16 𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵 = − 73 73

6 16 ⇒ 𝑦𝑦𝑝𝑝 = cos 3𝑥𝑥 − sin 3𝑥𝑥 73 73

⇒ 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝 ⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆

9/30/2014

𝟏𝟏 𝟑𝟑 +𝒋𝒋 𝒙𝒙 𝟐𝟐 𝟐𝟐

+ 𝒄𝒄𝟐𝟐 𝒆𝒆

𝟏𝟏 𝟑𝟑 −𝒋𝒋 𝒙𝒙 𝟐𝟐 𝟐𝟐

+

Dr. Eli Saber

𝟔𝟔 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝐜𝐜 𝟑𝟑𝟑𝟑 − 𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 𝟕𝟕𝟕𝟕 𝟕𝟕𝟕𝟕

88

Undetermined Coefficients E.g. 3: Using superposition 𝑦𝑦 ′′ − 2𝑦𝑦 ′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥 Given:

𝒈𝒈𝟏𝟏 (𝒙𝒙)

𝒈𝒈𝟐𝟐 (𝒙𝒙)

𝑦𝑦 ′′ − 2𝑦𝑦 ′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥

polynomial exponential

Step 1: 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 −𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 3𝑥𝑥 Step 2: Find 𝑦𝑦𝑝𝑝

9/30/2014

Dr. Eli Saber

89

Undetermined Coefficients

𝑦𝑦 ′′ − 2𝑦𝑦 ′ − 3𝑦𝑦 = 0

𝑦𝑦 ′′ − 2𝑦𝑦 ′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥

𝑚𝑚2 − 2𝑚𝑚 − 3 = 0

⇒ (𝑚𝑚 − 3)(𝑚𝑚 + 1) = 0 ⇒ 𝑚𝑚1 = 3, 𝑚𝑚2 = −1

Complimentary Solution: 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 −𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 3𝑥𝑥

9/30/2014

Dr. Eli Saber

90

Undetermined Coefficients ⇒ 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑝𝑝 = 𝑦𝑦𝑝𝑝1 + 𝑦𝑦𝑝𝑝2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶𝐶𝐶𝑒𝑒 2𝑥𝑥 + 𝐷𝐷𝑒𝑒 2𝑥𝑥 𝒚𝒚𝒑𝒑𝟏𝟏 :for 𝒈𝒈𝟏𝟏 (𝒙𝒙)

𝑦𝑦 ′′ − 2𝑦𝑦 ′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥

𝒚𝒚𝒑𝒑𝟐𝟐 :for 𝒈𝒈𝟐𝟐 (𝒙𝒙)

⇒ 𝑦𝑦𝑝𝑝′ = 𝐴𝐴 + 𝐶𝐶 𝑒𝑒 2𝑥𝑥 + 2𝑥𝑥𝑒𝑒 2𝑥𝑥 + 2𝐷𝐷𝑒𝑒 2𝑥𝑥

⇒ 𝑦𝑦𝑝𝑝′′ = 2𝐶𝐶𝑒𝑒 2𝑥𝑥 + 2𝐶𝐶 𝑒𝑒 2𝑥𝑥 + 2𝑥𝑥𝑒𝑒 2𝑥𝑥 + 4𝐷𝐷𝑒𝑒 2𝑥𝑥 ⇒ 𝑦𝑦 ′′ − 2𝑦𝑦 ′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥 Substitute 𝑦𝑦 ′′ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦 ′ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦:

⇒ 2𝐶𝐶𝑒𝑒 2𝑥𝑥 + 2𝐶𝐶𝑒𝑒 2𝑥𝑥 + 4𝐶𝐶𝐶𝐶𝑒𝑒 2𝑥𝑥 + 4𝐷𝐷𝑒𝑒 2𝑥𝑥 − 2𝐴𝐴 − 2𝐶𝐶𝑒𝑒 2𝑥𝑥 − 4𝐶𝐶𝐶𝐶𝑒𝑒 2𝑥𝑥 − 4𝐷𝐷𝑒𝑒 2𝑥𝑥 − 3𝐴𝐴𝐴𝐴 − 3𝐵𝐵 − 3𝐶𝐶𝐶𝐶𝑒𝑒 2𝑥𝑥 − 3𝐷𝐷𝑒𝑒 2𝑥𝑥 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥 9/30/2014

Dr. Eli Saber

91

Undetermined Coefficients 𝑦𝑦 ′′ − 2𝑦𝑦 ′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥

⇒ −3𝐴𝐴𝐴𝐴 − 2𝐴𝐴 − 3𝐵𝐵 − 3𝐶𝐶𝐶𝐶𝑒𝑒 2𝑥𝑥 + 𝑒𝑒 2𝑥𝑥 2𝐶𝐶 − 3𝐷𝐷 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥 4 3

•

−3𝐴𝐴 = 4 ⇒ 𝐴𝐴 = −

•

−2𝐴𝐴 − 3𝐵𝐵 = −5 ⇒ −3𝐵𝐵 = −5 + 2𝐴𝐴 = −5 + 2 −

⇒ −3𝐵𝐵 = −5 − • •

8 15 8 23 23 =− − =− ⇒ 𝐵𝐵 = 3 3 3 3 9

−3𝐶𝐶 = 6 ⇒ 𝐶𝐶 = −2

4 3

2𝐶𝐶 − 3𝐷𝐷 = 0 ⇒ 3𝐷𝐷 = 2𝐶𝐶 = 2 −2 = −4 ⇒ D = −

9/30/2014

Dr. Eli Saber

4 3 92

Undetermined Coefficients 4 23 4 𝑦𝑦𝑝𝑝 = − 𝑥𝑥 + − 2𝑥𝑥𝑒𝑒 2𝑥𝑥 − 𝑒𝑒 2𝑥𝑥 3 9 3

𝑦𝑦 ′′ − 2𝑦𝑦 ′ − 3𝑦𝑦 = 4𝑥𝑥 − 5 + 6𝑥𝑥𝑒𝑒 2𝑥𝑥

𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

𝟒𝟒 𝟐𝟐𝟐𝟐 𝟒𝟒 ⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆−𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒆𝒆𝟑𝟑𝟑𝟑 − 𝒙𝒙 + − 𝟐𝟐𝟐𝟐𝒆𝒆𝟐𝟐𝟐𝟐 − 𝒆𝒆𝟐𝟐𝟐𝟐 𝟑𝟑 𝟗𝟗 𝟑𝟑

9/30/2014

Dr. Eli Saber

93

Undetermined Coefficients E.g. 4: 𝑦𝑦 ′′ − 5𝑦𝑦 ′ + 4𝑦𝑦 = 8𝑒𝑒 𝑥𝑥

Step 1: Find 𝑦𝑦𝑐𝑐 → 𝑦𝑦𝑐𝑐 = 𝐶𝐶1 𝑒𝑒 𝑥𝑥 + 𝐶𝐶2 𝑒𝑒 4𝑥𝑥 Step 2: Find 𝑦𝑦𝑝𝑝 → 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑒𝑒 𝑥𝑥 ⇒ 𝑦𝑦𝑝𝑝′ = 𝐴𝐴𝑒𝑒 𝑥𝑥 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑝𝑝′′ = 𝐴𝐴𝑒𝑒 𝑥𝑥

Re-substituting back ⇒ 𝐴𝐴𝑒𝑒 𝑥𝑥 − 5𝐴𝐴𝑒𝑒 𝑥𝑥 + 4𝐴𝐴𝑒𝑒 𝑥𝑥 = 8𝑒𝑒 𝑥𝑥 ⇒ 0𝐴𝐴𝑒𝑒 𝑥𝑥 = 8𝑒𝑒 𝑥𝑥 ⇒ 𝟎𝟎 = 𝟖𝟖𝒆𝒆𝒙𝒙 − 𝒏𝒏𝒏𝒏𝒏𝒏 𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑 Note: 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 4𝑥𝑥 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑒𝑒 𝑥𝑥

Not Independent

• 𝑒𝑒 𝑥𝑥 is already present in 𝑦𝑦𝑐𝑐 ⇒ 𝑒𝑒 𝑥𝑥 is a solution of the homogeneous equation. ⇒ 𝐴𝐴𝑒𝑒 𝑥𝑥 when substituted into the D.E. produces zero ⇒(see case II in section 3.3) 9/30/2014

Dr. Eli Saber

94

Undetermined Coefficients 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥

𝑦𝑦𝑝𝑝′ = 𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑝𝑝′′ = 𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 = 2𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥

𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 4𝑥𝑥

⇒ 2𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 − 5 𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 + 4𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 = 8𝑒𝑒 𝑥𝑥 ⇒ 2𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 − 5𝐴𝐴𝑒𝑒 𝑥𝑥 − 5𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 + 4𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 = 8𝑒𝑒 𝑥𝑥 ⇒ −3𝐴𝐴𝑒𝑒 𝑥𝑥 = 8𝑒𝑒 𝑥𝑥 ⇒ −3𝐴𝐴 = 8 ⇒ 𝐴𝐴 = − 8 ⇒ 𝑦𝑦𝑝𝑝 = − 𝑥𝑥𝑒𝑒 𝑥𝑥 3

Now, 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

9/30/2014

8 3

𝟖𝟖 ⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒆𝒆𝟒𝟒𝟒𝟒 − 𝒙𝒙𝒆𝒆𝒙𝒙 𝟑𝟑 Dr. Eli Saber

95

Undetermined Coefficients

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

9/30/2014

Dr. Eli Saber

96

Undetermined Coefficients Case I: No function in the assumed particular solution 𝑦𝑦𝑝𝑝 is a solution of the associated Homogeneous Differential Equation. E.g. 5.1 𝑦𝑦 ′′ − 8𝑦𝑦 ′ + 25𝑦𝑦 = 5𝑥𝑥 3 𝑒𝑒 −𝑥𝑥 − 7𝑒𝑒 −𝑥𝑥

⇒ 𝑦𝑦 ′′ − 8𝑦𝑦 ′ + 25𝑦𝑦 = 5𝑥𝑥 3 − 7 𝑒𝑒 −𝑥𝑥

Homogeneous solution: 𝑦𝑦𝑐𝑐 = 𝑒𝑒 4𝑥𝑥 𝑐𝑐1 cos 3𝑥𝑥 + 𝑐𝑐2 sin 3𝑥𝑥 Assume 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥 3 + 𝐵𝐵𝑥𝑥 2 + 𝐶𝐶𝐶𝐶 + 𝐸𝐸 𝑒𝑒 −𝑥𝑥

Note no duplication of terms between 𝑦𝑦𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑐𝑐 9/30/2014

Dr. Eli Saber

97

Undetermined Coefficients E.g. 5.2 𝑦𝑦 ′′ + 4𝑦𝑦 = 𝑥𝑥 cos 𝑥𝑥

𝑦𝑦𝑐𝑐 = 𝑐𝑐1 cos 2𝑥𝑥 + 𝑐𝑐2 sin 2𝑥𝑥

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝐴𝐴 + 𝐵𝐵 cos 𝑥𝑥 + 𝐶𝐶𝐶𝐶 + 𝐸𝐸 sin 𝑥𝑥 No duplication of terms between 𝑦𝑦𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑐𝑐

9/30/2014

Dr. Eli Saber

98

Undetermined Coefficients E.g. 6: 𝑦𝑦 ′′ − 9𝑦𝑦 ′ + 14𝑦𝑦 = 3𝑥𝑥 2 − 5 sin 2𝑥𝑥 + 7𝑥𝑥𝑒𝑒 6𝑥𝑥 Given 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 2𝑥𝑥 + 𝑐𝑐2 𝑒𝑒 7𝑥𝑥

(computer earlier)

Since 𝑔𝑔 𝑥𝑥 has various terms, form 𝑦𝑦𝑝𝑝 by superposition 3𝑥𝑥 2

→ 𝑦𝑦𝑝𝑝1 = 𝐴𝐴𝑥𝑥 2 + 𝐵𝐵𝐵𝐵 + 𝐶𝐶

7𝑥𝑥𝑒𝑒 6𝑥𝑥

→ 𝑦𝑦𝑝𝑝3 = 𝐺𝐺𝐺𝐺 + 𝐻𝐻 𝑒𝑒 6𝑥𝑥

−5 sin 2𝑥𝑥 → 𝑦𝑦𝑝𝑝2 = 𝐸𝐸 cos 2𝑥𝑥 + 𝐹𝐹 sin 2𝑥𝑥

⇒ 𝑦𝑦𝑝𝑝 = 𝑦𝑦𝑝𝑝1 + 𝑦𝑦𝑝𝑝2 + 𝑦𝑦𝑝𝑝3

⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥 2 + 𝐵𝐵𝐵𝐵 + 𝐶𝐶 + 𝐸𝐸 cos 2𝑥𝑥 + 𝐹𝐹 sin 2𝑥𝑥 + 𝐺𝐺𝐺𝐺 + 𝐻𝐻 𝑒𝑒 6𝑥𝑥 Note: No duplication of terms between 𝑦𝑦𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑐𝑐 9/30/2014

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99

Undetermined Coefficients Case II: A function in the potential particular solution is also a solution of the associated Homogeneous Differential Equation. E.g. 7: 𝑦𝑦 ′′ − 2𝑦𝑦 ′ + 𝑦𝑦 = 𝑒𝑒 𝑥𝑥 With 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥𝑒𝑒 𝑥𝑥

(computed earlier)

What do we assume for 𝑦𝑦𝑝𝑝 ?

• •

𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑒𝑒 𝑥𝑥 → will fail since 𝑒𝑒 𝑥𝑥 is part of 𝑦𝑦𝑐𝑐

𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥𝑥𝑥 𝑥𝑥 → will fail since 𝑥𝑥𝑥𝑥 𝑥𝑥 is part of 𝑦𝑦𝑐𝑐

⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑥𝑥 2 𝑒𝑒 𝑥𝑥

⇒ 𝑦𝑦𝑝𝑝′ = 2𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝑥𝑥 2 𝑒𝑒 𝑥𝑥 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑝𝑝′′ = 2𝐴𝐴𝑒𝑒 𝑥𝑥 + 2𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 + 2𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝑥𝑥 2 𝑒𝑒 𝑥𝑥

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Undetermined Coefficients ⇒ 2𝐴𝐴𝑒𝑒 𝑥𝑥 + 4𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝑥𝑥 2 𝑒𝑒 𝑥𝑥 − 4𝐴𝐴𝐴𝐴𝑒𝑒 𝑥𝑥 − 2𝐴𝐴𝑥𝑥 2 𝑒𝑒 𝑥𝑥 + 𝐴𝐴𝑥𝑥 2 𝑒𝑒 𝑥𝑥 = 𝑒𝑒 𝑥𝑥 1 ⇒ 2𝐴𝐴𝑒𝑒 = 𝑒𝑒 ⇒ 2𝐴𝐴 = 1 ⇒ 𝐴𝐴 = 2 𝑥𝑥

⇒ 𝒚𝒚𝒑𝒑 =

𝑥𝑥

𝟏𝟏 𝟐𝟐 𝒙𝒙 𝒙𝒙 𝒆𝒆 𝟐𝟐

𝟏𝟏 ⇒ 𝒚𝒚𝒄𝒄 = 𝒄𝒄𝟏𝟏 𝒆𝒆𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝒆𝒆𝒙𝒙 + 𝒙𝒙𝟐𝟐 𝒆𝒆𝒙𝒙 𝟐𝟐

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Undetermined Coefficients Hence if 𝑔𝑔 𝑥𝑥 consists of no terms similar to Table 3.4.1 and that: 𝑦𝑦𝑝𝑝 = 𝑦𝑦𝑝𝑝1 + 𝑦𝑦𝑝𝑝2 + ⋯ + 𝑦𝑦𝑝𝑝𝑚𝑚

(assumption)

Where 𝑦𝑦𝑝𝑝𝑖𝑖 , 𝑖𝑖 = 1, 2, 3, … … , 𝑚𝑚 are potential particular solution

Multiplication rule: If any 𝒚𝒚𝒑𝒑𝒊𝒊 contains terms that duplicate terms in 𝒚𝒚𝒄𝒄 , then that 𝒚𝒚𝒑𝒑𝒊𝒊 must be multiplied by 𝒙𝒙𝒏𝒏 , where n is the smallest positive integer that eliminates that duplication

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102

Undetermined Coefficients E.g. 8: 𝑦𝑦 ′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥 Initial conditions: 𝑦𝑦 𝜋𝜋 = 0; 𝑦𝑦 ′ 𝜋𝜋 = 2 Step 1: 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑦𝑦 ′′ + 𝑦𝑦 = 0

𝑚𝑚2 + 1 = 0 ⇒ 𝑚𝑚2 = −1 = 𝑗𝑗 2 ⇒ 𝑚𝑚 = ±𝑗𝑗

⇒ 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 𝑗𝑗𝑗𝑗 + 𝑐𝑐2 𝑒𝑒 −𝑗𝑗𝑗𝑗 = 𝑐𝑐1 cos 𝑥𝑥 + 𝑗𝑗𝑐𝑐1 sin 𝑥𝑥 + 𝑐𝑐2 cos 𝑥𝑥 − 𝑗𝑗𝑐𝑐2 sin 𝑥𝑥 = 𝑐𝑐1 + 𝑐𝑐2 cos 𝑥𝑥 + 𝑗𝑗 𝑐𝑐1 − 𝑐𝑐2 sin 𝑥𝑥 ⇒ 𝑦𝑦𝑐𝑐 = 𝛼𝛼1 cos 𝑥𝑥 + 𝛼𝛼2 sin 𝑥𝑥

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103

Undetermined Coefficients 𝑦𝑦 ′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥

𝑔𝑔 𝑥𝑥 = 4𝑥𝑥 + 10 sin 𝑥𝑥 • •

𝑦𝑦

𝑐𝑐 4𝑥𝑥 → 𝐴𝐴𝐴𝐴+𝐵𝐵

10 sin 𝑥𝑥 → 𝐶𝐶 cos 𝑥𝑥 + 𝐸𝐸 sin 𝑥𝑥 (but these are part of 𝑦𝑦𝑐𝑐 ) = 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 + 𝐸𝐸𝐸𝐸 sin 𝑥𝑥

⇒ 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 + 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 •

𝑦𝑦𝑝𝑝′ = 𝐴𝐴 + 𝐶𝐶 cos 𝑥𝑥 + 𝑥𝑥 − sin 𝑥𝑥

+ 𝐸𝐸 sin 𝑥𝑥 + 𝑥𝑥 cos 𝑥𝑥

= 𝐴𝐴 + 𝐶𝐶 cos 𝑥𝑥 − 𝐶𝐶𝐶𝐶 sin 𝑥𝑥 + 𝐸𝐸 sin 𝑥𝑥 + 𝐸𝐸𝐸𝐸 cos 𝑥𝑥 9/30/2014

Dr. Eli Saber

104

Undetermined Coefficients

•

𝑦𝑦 ′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥 𝑦𝑦𝑝𝑝′′ = −𝐶𝐶 sin 𝑥𝑥 − 𝐶𝐶 sin 𝑥𝑥 + 𝑥𝑥 cos 𝑥𝑥 + 𝐸𝐸 cos 𝑥𝑥 + 𝐸𝐸 cos 𝑥𝑥 − 𝑥𝑥 sin 𝑥𝑥

= −𝐶𝐶 sin 𝑥𝑥 − 𝐶𝐶 sin 𝑥𝑥 − 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 + 𝐸𝐸 cos 𝑥𝑥 + 𝐸𝐸 cos 𝑥𝑥 − 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 = −2𝐶𝐶 sin 𝑥𝑥 + 2𝐸𝐸 cos 𝑥𝑥 − 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 − 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 We have: 𝑦𝑦 ′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥

⇒ −2𝐶𝐶 sin 𝑥𝑥 + 2𝐸𝐸 cos 𝑥𝑥 − 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 − 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 + 𝐴𝐴𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 + 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 = 4𝑥𝑥 + 10 sin 𝑥𝑥

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Dr. Eli Saber

105

Undetermined Coefficients 𝑦𝑦 ′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥

⇒ −2𝐶𝐶 sin 𝑥𝑥 + 2𝐸𝐸 cos 𝑥𝑥 − 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 − 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 + 𝐴𝐴𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 + 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 = 4𝑥𝑥 + 10 sin 𝑥𝑥 ⇒ 𝐴𝐴𝐴𝐴 + 𝐵𝐵 − 2𝐶𝐶 sin 𝑥𝑥 − 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 + 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 + 2𝐸𝐸 cos 𝑥𝑥 − 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 + 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 = 4𝑥𝑥 + 10 sin 𝑥𝑥

⇒ 𝑨𝑨𝑨𝑨 + 𝑩𝑩 + −𝟐𝟐𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙 + 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 = 𝟒𝟒𝟒𝟒 + 𝟎𝟎 + 𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙 + 𝟎𝟎 ⇒ 𝐴𝐴𝐴𝐴 = 4𝑥𝑥 ⇒ 𝐴𝐴 = 4

⇒ 𝐵𝐵 = 0

⇒ −2𝐶𝐶 sin 𝑥𝑥 = 10 sin 𝑥𝑥 ⇒ −2𝐶𝐶 = 10 ⇒ 𝐶𝐶 = −5 ⇒ 2𝐸𝐸 cos 𝑥𝑥 = 0 ⇒ 𝐸𝐸 = 0

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Undetermined Coefficients WE know, 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶𝐶𝐶 cos 𝑥𝑥 + 𝐸𝐸𝐸𝐸 sin 𝑥𝑥 & 𝐴𝐴 = 4, 𝐵𝐵 = 0, 𝐶𝐶 = −5, 𝐷𝐷 = 0

𝑦𝑦 ′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥

⇒ 𝒚𝒚𝒑𝒑 = 𝟒𝟒𝟒𝟒 − 𝟓𝟓𝟓𝟓 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 We know: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

⇒ 𝑦𝑦 = 𝛼𝛼1 cos 𝑥𝑥 + 𝛼𝛼2 sin 𝑥𝑥 + 4𝑥𝑥 − 5𝑥𝑥 cos 𝑥𝑥

Initial conditions: 𝑦𝑦 𝑥𝑥 = 0 𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦 ′ 𝑥𝑥 = 2 •

𝑦𝑦 𝜋𝜋 = 0 ⇒ 0 = 𝛼𝛼1 cos 𝜋𝜋 + 𝛼𝛼2 sin 𝜋𝜋 + 4𝜋𝜋 − 5𝜋𝜋 cos 𝜋𝜋

⇒ 0 = −𝛼𝛼1 + 0 + 4𝜋𝜋 + 5𝜋𝜋 ⇒ 𝛼𝛼1 = 9𝜋𝜋 9/30/2014

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Undetermined Coefficients •

𝑦𝑦 ′ 𝜋𝜋 = 2

⇒ 𝑦𝑦 ′ = −𝛼𝛼1 sin 𝑥𝑥 + 𝛼𝛼2 cos 𝑥𝑥 + 4 − 5 cos 𝑥𝑥 − 𝑥𝑥 sin 𝑥𝑥

𝑦𝑦 ′′ + 𝑦𝑦 = 4𝑥𝑥 + 10 sin 𝑥𝑥 𝑦𝑦𝑝𝑝 = 4𝑥𝑥 − 5𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥

⇒ 2 = −9𝜋𝜋 sin 𝜋𝜋 + 𝛼𝛼2 cos 𝜋𝜋 + 4 − 5 cos 𝜋𝜋 + 5𝜋𝜋 sin 𝜋𝜋 ⇒ 2 = −𝛼𝛼2 + 4 + 5 ⇒ 𝛼𝛼2 = 9 − 2 ⇒ 𝛼𝛼2 = 7

Therefore:

𝒚𝒚 = 𝟗𝟗𝟗𝟗 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙 + 𝟕𝟕 𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙 + 𝟒𝟒𝟒𝟒 − 𝟓𝟓𝟓𝟓 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙

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Undetermined Coefficients Summary •

To solve a non-Homogeneous D.E. 𝑎𝑎𝑛𝑛 𝑦𝑦

•

•

𝑛𝑛

+ 𝑎𝑎𝑛𝑛−1 𝑦𝑦 𝑛𝑛−1 + ⋯ + 𝑎𝑎1 𝑦𝑦1 + 𝑎𝑎0 𝑦𝑦 0 = 𝑔𝑔 𝑥𝑥

Step1: Finding a complementary solution 𝑦𝑦𝑐𝑐 by equating it to 0. Step2: Finding a particular solution 𝑦𝑦𝑝𝑝 . Step3: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

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

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 𝒚𝒚𝒑𝒑𝒊𝒊 contains terms that duplicate terms in 𝒚𝒚𝒄𝒄 , then that 𝒚𝒚𝒑𝒑𝒊𝒊 must be multiplied by 𝒙𝒙𝒏𝒏 , where n is the smallest positive integer that eliminates that duplication

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Section 3.5 Variation of Parameters

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Variation of Parameters • 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.

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Variation of Parameters Given

Divide by 𝑎𝑎2 𝑥𝑥

𝑎𝑎2 𝑥𝑥 𝑦𝑦 ′′ + 𝑎𝑎1 𝑥𝑥 𝑦𝑦 ′ + 𝑎𝑎0 𝑥𝑥 𝑦𝑦 = 𝑔𝑔 𝑥𝑥 ⟹ 𝑦𝑦 ′′ +

𝑎𝑎1 𝑎𝑎0 𝑔𝑔 𝑥𝑥 𝑥𝑥 𝑦𝑦 ′ + 𝑥𝑥 𝑦𝑦 = 𝑎𝑎2 𝑎𝑎2 𝑎𝑎2 𝑷𝑷(𝒙𝒙)

𝑸𝑸(𝒙𝒙)

𝒇𝒇(𝒙𝒙)

⟹ 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 9/30/2014

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Variation of Parameters Method: • For first order differential equation 𝑦𝑦 ′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦 = 𝑓𝑓 𝑥𝑥 , seek a solution

𝒚𝒚𝒑𝒑 = 𝝁𝝁𝟏𝟏 (𝒙𝒙)𝒚𝒚𝟏𝟏 (𝒙𝒙)

𝒚𝒚𝟏𝟏 (𝒙𝒙): 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 𝒙𝒙 𝒚𝒚𝟏𝟏 𝒙𝒙 + 𝝁𝝁𝟐𝟐 (𝒙𝒙)𝒚𝒚𝟐𝟐 (𝒙𝒙)

𝒚𝒚𝟏𝟏 𝒙𝒙 , 𝒚𝒚𝟐𝟐 (𝒙𝒙): fundamental solution for homogeneous D.E 9/30/2014

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Variation of Parameters 𝑦𝑦𝑝𝑝 = 𝜇𝜇1 𝑦𝑦1 + 𝜇𝜇2 𝑦𝑦2

⟹ 𝑦𝑦𝑝𝑝′ = 𝜇𝜇1 𝑦𝑦1 ′ + 𝜇𝜇1′ 𝑦𝑦1 + 𝜇𝜇2 𝑦𝑦2 ′ + 𝜇𝜇2′ 𝑦𝑦2

⇒ 𝑦𝑦𝑝𝑝 ′′ = 𝜇𝜇1 𝑦𝑦1 ′′ + 𝜇𝜇1′ 𝑦𝑦1′ + 𝜇𝜇1′ 𝑦𝑦1′ + 𝜇𝜇1′′ 𝑦𝑦1 + 𝜇𝜇2 𝑦𝑦2 ′′ + 𝜇𝜇2′ 𝑦𝑦2′ + 𝜇𝜇2′ 𝑦𝑦2′ + 𝜇𝜇2′′ 𝑦𝑦2 Substitute into D.E:

𝑦𝑦𝑝𝑝 ′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦𝑝𝑝 ′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦𝑝𝑝 = 𝑓𝑓 𝑥𝑥

⟹ 𝜇𝜇1 𝑦𝑦1 ′′ + 𝜇𝜇1′ 𝑦𝑦1′ + 𝜇𝜇1′ 𝑦𝑦1′ + 𝜇𝜇1′′ 𝑦𝑦1 + 𝜇𝜇2 𝑦𝑦2 ′′ + 𝜇𝜇2′ 𝑦𝑦2′ + 𝜇𝜇2′ 𝑦𝑦2′ + 𝜇𝜇2′′ 𝑦𝑦2 𝑃𝑃 𝑥𝑥 𝜇𝜇1 𝑦𝑦1 ′ + 𝜇𝜇1′ 𝑦𝑦1 + 𝜇𝜇2 𝑦𝑦2 ′ + 𝜇𝜇2′ 𝑦𝑦2 +𝑄𝑄 𝑥𝑥 𝜇𝜇1 𝑦𝑦1 + 𝜇𝜇2 𝑦𝑦2 = 𝑓𝑓(𝑥𝑥) 9/30/2014

Dr. Eli Saber

114

Variation of Parameters Rearranging the equations,

= 𝟎𝟎

Since 𝑦𝑦1 & 𝑦𝑦2 are the solutions to the homogeneous equation = 𝟎𝟎

𝑢𝑢1 𝑦𝑦1′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦1′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦1 + 𝑢𝑢2 𝑦𝑦2′′ + 𝑃𝑃 𝑥𝑥 𝑦𝑦2′ + 𝑄𝑄 𝑥𝑥 𝑦𝑦2

+𝑢𝑢1′′ 𝑦𝑦1 + 𝑢𝑢1′ 𝑦𝑦1′ + 𝑢𝑢2′′ 𝑦𝑦2 + 𝑢𝑢2′ 𝑦𝑦2′ + 𝑃𝑃 𝑢𝑢1′ 𝑦𝑦1 + 𝑢𝑢2′ 𝑦𝑦2 + 𝑢𝑢1′ 𝑦𝑦1′ + 𝑢𝑢2′ 𝑦𝑦2′ = 𝑓𝑓(𝑥𝑥)

9/30/2014

Dr. Eli Saber

115

Variation of Parameters 𝑢𝑢1′′ 𝑦𝑦1 + 𝑢𝑢1′ 𝑦𝑦1′ + 𝑢𝑢2′′ 𝑦𝑦2 + 𝑢𝑢 2′ 𝑦𝑦2′ + 𝑃𝑃 𝑢𝑢1′ 𝑦𝑦1 + 𝑢𝑢 2′ 𝑦𝑦2 + 𝑢𝑢1′ 𝑦𝑦1′ + 𝑢𝑢2′ 𝑦𝑦2′ = 𝑓𝑓(𝑥𝑥) ⟹

𝒅𝒅 𝒅𝒅 𝒖𝒖𝟏𝟏′ 𝒚𝒚𝟏𝟏 + 𝒖𝒖𝟐𝟐′ 𝒚𝒚𝟐𝟐 + 𝑃𝑃 𝑢𝑢1′ 𝑦𝑦1 + 𝑢𝑢2′ 𝑦𝑦2 + 𝑢𝑢1′ 𝑦𝑦1′ + 𝑢𝑢 2′ 𝑦𝑦2′ = 𝑓𝑓(𝑥𝑥) 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅

𝑑𝑑 ⟹ 𝑢𝑢1′ 𝑦𝑦1 + 𝑢𝑢 2′ 𝑦𝑦2 + 𝑃𝑃[𝑢𝑢1′ 𝑦𝑦1 + 𝑢𝑢2′ 𝑦𝑦2 ] + 𝑢𝑢1′ 𝑦𝑦1′ + 𝑢𝑢2′ 𝑦𝑦2′ = 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑

• Have two unknown functions 𝑢𝑢1 & 𝑢𝑢2 ⟹ Need two equations ⟹ make further assumption that 𝒖𝒖𝟏𝟏’𝒚𝒚𝟏𝟏 + 𝒖𝒖𝟐𝟐’𝒚𝒚𝟐𝟐 = 𝟎𝟎 ⟹ 𝒖𝒖𝟏𝟏’𝒚𝒚𝟏𝟏’ + 𝒖𝒖𝟐𝟐’𝒚𝒚𝟐𝟐’ = 𝒇𝒇(𝒙𝒙) 9/30/2014

Dr. Eli Saber

116

Variation of Parameters •

Hence, we have two equations with two unknowns: 𝑦𝑦1𝑢𝑢1’ + 𝑦𝑦2𝑢𝑢2’ = 0

• •

𝑦𝑦1 ′𝑢𝑢1’ + 𝑦𝑦2′ 𝑢𝑢2’ = 𝑓𝑓(𝑥𝑥)

(1) (2)

Solve for 𝑢𝑢1′ & 𝑢𝑢2′ & then integrate to get 𝑢𝑢1 & 𝑢𝑢2

Using Cramer’s rule:

9/30/2014

𝑦𝑦1 𝑦𝑦1′

𝑨𝑨

𝑦𝑦2 𝑢𝑢 ′ 0 1 = 𝑦𝑦2′ 𝑢𝑢2′ 𝑓𝑓(𝑥𝑥) 𝒖𝒖

Dr. Eli Saber

𝒃𝒃

117

Variation of Parameters We have,

Note: 𝑦𝑦1 W≡ 𝑦𝑦 ′ 1

𝑦𝑦1 𝑦𝑦1′

𝑦𝑦2 𝑢𝑢 ′ 0 1 = 𝑦𝑦2′ 𝑢𝑢2′ 𝑓𝑓(𝑥𝑥)

𝟎𝟎 𝒚𝒚𝟐𝟐 𝒚𝒚𝟏𝟏 𝟎𝟎 𝒇𝒇(𝒙𝒙) 𝒚𝒚𝟐𝟐′ 𝒚𝒚𝟏𝟏′ 𝒇𝒇(𝒙𝒙) ′ 𝒖𝒖′𝟏𝟏 = 𝒚𝒚 & 𝒖𝒖 = 𝟐𝟐 𝒚𝒚𝟐𝟐 𝒚𝒚𝟏𝟏 𝒚𝒚𝟐𝟐 𝟏𝟏 𝒚𝒚𝟏𝟏′ 𝒚𝒚𝟐𝟐′ 𝒚𝒚𝟏𝟏′ 𝒚𝒚𝟐𝟐′ 𝑦𝑦2 𝑦𝑦2′ ⇒ 𝑡𝑡𝑡𝑡𝑡 𝑾𝑾𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓 𝑜𝑜𝑜𝑜 𝑦𝑦1 & 𝑦𝑦2

Hence, Since y1 & y2 are independent ⟹ W≠ 𝟎𝟎 𝒇𝒇𝒇𝒇𝒇𝒇 ∀𝒙𝒙 ∈ 𝑰𝑰 9/30/2014

Dr. Eli Saber

118

Variation of Parameters Summary: Given

𝑎𝑎2 𝑦𝑦′′ + 𝑎𝑎1𝑦𝑦′ + 𝑎𝑎0𝑦𝑦 = 𝑔𝑔(𝑥𝑥)

1. Put Eq. into standard form by dividing throughout by a2(x) 𝑎𝑎1 ′ 𝑎𝑎0 𝑔𝑔 𝑥𝑥 𝑦𝑦 ′′ + 𝑦𝑦 + 𝑦𝑦 = 𝑎𝑎2 𝑎𝑎2 𝑎𝑎2 (𝑥𝑥) 𝑷𝑷(𝒙𝒙)

𝑸𝑸(𝒙𝒙)

𝒇𝒇(𝒙𝒙)

2. Find the complementary solution 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2 𝑦𝑦 3. Compute Wronskian of 𝑦𝑦1 & 𝑦𝑦2  𝑊𝑊 = 𝑦𝑦 1′ 1

9/30/2014

Dr. Eli Saber

𝑦𝑦2 𝑦𝑦2′

119

Variation of Parameters 4. Compute 𝑢𝑢1’ & 𝑢𝑢2’ using:

0 𝑦𝑦2 𝑦𝑦1 0 𝑦𝑦1′ 𝑓𝑓(𝑥𝑥) 𝑓𝑓(𝑥𝑥) 𝑦𝑦2′ ′ ; 𝑢𝑢 = 𝑢𝑢1′ = 𝑦𝑦 2 𝑦𝑦2 𝑦𝑦1 𝑦𝑦2 1 𝑦𝑦1′ 𝑦𝑦2′ 𝑦𝑦1′ 𝑦𝑦2′

5. Find 𝑢𝑢1 & 𝑢𝑢2 by integrating 𝑢𝑢1′ & 𝑢𝑢2′ respectively. 6. Form 𝑦𝑦𝑝𝑝 = 𝑢𝑢1(𝑥𝑥)𝑦𝑦1(𝑥𝑥) + 𝑢𝑢2(𝑥𝑥)𝑦𝑦2(𝑥𝑥)

7. General Solution: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

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Dr. Eli Saber

120

Variation of Parameters Note: When integrating 𝑢𝑢1’ & 𝑢𝑢2’, you don’t need to introduce any constants because: • 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2

� 𝑢𝑢1′ 𝑑𝑑𝑑𝑑 = 𝑢𝑢1 + 𝑎𝑎1 , � 𝑢𝑢2 ′ 𝑑𝑑𝑑𝑑 = 𝑢𝑢2 + 𝑎𝑎2 ⟹ 𝑦𝑦𝑝𝑝 = (𝑢𝑢1 + 𝑎𝑎1)𝑦𝑦1 + (𝑢𝑢2 + 𝑎𝑎2)𝑦𝑦2

𝑎𝑎1 , 𝑎𝑎2 are constants

⟹ 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝 = 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2 + 𝑢𝑢1𝑦𝑦1 + 𝑎𝑎1𝑦𝑦1 + 𝑢𝑢2𝑦𝑦2 + 𝑎𝑎2𝑦𝑦2

Rearranging, 𝑦𝑦 = (𝑐𝑐1 + 𝑎𝑎1)𝑦𝑦1 + (𝑐𝑐2 + 𝑎𝑎2)𝑦𝑦2 + 𝑢𝑢1𝑦𝑦1 + 𝑢𝑢2𝑦𝑦2 ⟹ 𝑦𝑦 = 𝜶𝜶𝟏𝟏𝑦𝑦1 + 𝜶𝜶𝟐𝟐𝑦𝑦2 + 𝑢𝑢1𝑦𝑦1 + 𝑢𝑢2𝑦𝑦2

Where, 𝛼𝛼1 &𝛼𝛼2 : constants computed using initial conditions or boundary conditions 9/30/2014

Dr. Eli Saber

121

Variation of Parameters E.g.1:

𝑦𝑦 ′′ − 4𝑦𝑦 ′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒 2𝑥𝑥

1. Equation is already in standard form: 𝑦𝑦 ′′ − 4𝑦𝑦 ′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒 2𝑥𝑥 2. Find 𝑦𝑦𝑐𝑐 : ⇒ 𝑚𝑚2 − 4𝑚𝑚 + 4 = 0 → 𝑚𝑚 − 2 ⇒ 𝑚𝑚1 = 𝑚𝑚2 = 2 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 2𝑥𝑥 + 𝑐𝑐2 𝑥𝑥𝑒𝑒 2𝑥𝑥

3. Compute 𝑊𝑊

𝒚𝒚𝟏𝟏

𝑷𝑷(𝒙𝒙) 2

=0

𝒚𝒚𝟐𝟐

2𝑥𝑥 𝑦𝑦1 𝑦𝑦2 𝑊𝑊 = 𝑦𝑦 ′ 𝑦𝑦 ′ = 𝑒𝑒 2𝑥𝑥 1 2 2𝑒𝑒

𝑸𝑸(𝒙𝒙)

𝒇𝒇(𝒙𝒙)

𝑥𝑥𝑒𝑒 2𝑥𝑥 𝑒𝑒 2𝑥𝑥 + 2𝑥𝑥 𝑒𝑒 2𝑥𝑥

= 𝑒𝑒 2𝑥𝑥 𝑒𝑒 2𝑥𝑥 + 2𝑥𝑥 𝑒𝑒 2𝑥𝑥 − 𝑥𝑥𝑒𝑒 2𝑥𝑥 2𝑒𝑒 2𝑥𝑥 = 𝑒𝑒 4𝑥𝑥 + 2𝑥𝑥 𝑒𝑒 4𝑥𝑥 − 2𝑥𝑥 𝑒𝑒 4𝑥𝑥 ⇒ 𝑊𝑊 = 𝑒𝑒 4𝑥𝑥 9/30/2014

Dr. Eli Saber

122

Variation of Parameters 4. Compute 𝑢𝑢1′ & 𝑢𝑢2 ′ 𝑊𝑊1 =

0 𝑥𝑥 + 1 𝑒𝑒 2𝑥𝑥

𝑒𝑒 2𝑥𝑥 𝑊𝑊2 = 2𝑒𝑒 2𝑥𝑥 Now,

𝑥𝑥𝑒𝑒 2𝑥𝑥 𝟒𝟒𝟒𝟒 2𝑥𝑥 2𝑥𝑥 = −𝒙𝒙 𝒙𝒙 + 𝟏𝟏 𝒆𝒆 𝑒𝑒 + 2𝑥𝑥𝑒𝑒

𝑦𝑦 ′′ − 4𝑦𝑦 ′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒 2𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 2𝑥𝑥 + 𝑐𝑐2 𝑥𝑥𝑒𝑒 2𝑥𝑥 𝑊𝑊 = 𝑒𝑒 4𝑥𝑥

0 = 𝒙𝒙 + 𝟏𝟏 𝒆𝒆𝟒𝟒𝟒𝟒 2𝑥𝑥 𝑥𝑥 + 1 𝑒𝑒

4𝑥𝑥 𝑊𝑊 𝑥𝑥 + 1 𝑥𝑥𝑒𝑒 1 =− = −𝑥𝑥 𝑥𝑥 + 1 ⇒ 𝑢𝑢1′ = −𝑥𝑥 2 − 𝑥𝑥 𝑢𝑢1′ = 4𝑥𝑥 𝑊𝑊 𝑒𝑒

𝑢𝑢1 = � −𝑥𝑥 2 − 𝑥𝑥 𝑑𝑑𝑑𝑑 1 3 𝑥𝑥 2 𝑢𝑢1 = − 𝑥𝑥 − 2 3 9/30/2014

Dr. Eli Saber

123

Variation of Parameters 𝑦𝑦 ′′ − 4𝑦𝑦 ′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒 2𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 2𝑥𝑥 + 𝑐𝑐2 𝑥𝑥𝑒𝑒 2𝑥𝑥 𝑊𝑊 = 𝑒𝑒 4𝑥𝑥 𝑊𝑊2 = 𝑥𝑥 + 1 𝑒𝑒 4𝑥𝑥 1 3 𝑥𝑥 2 𝑢𝑢1 = − 𝑥𝑥 − 2 3

Now, 𝑢𝑢2′

𝑊𝑊2 𝑥𝑥 + 1 𝑒𝑒 4𝑥𝑥 = = = 𝑥𝑥 + 1 → 𝑢𝑢2 ′ 𝑊𝑊 𝑒𝑒 4𝑥𝑥

𝑢𝑢2 = � 𝑥𝑥 + 1 𝑑𝑑𝑑𝑑 𝑥𝑥 2 𝑢𝑢2 = + 𝑥𝑥 2

Hence, 𝑦𝑦𝑝𝑝 = 𝑢𝑢1 𝑥𝑥 𝑦𝑦1 𝑥𝑥 + 𝑢𝑢2 𝑥𝑥 𝑦𝑦2 𝑥𝑥 1

𝑦𝑦𝑝𝑝 𝑥𝑥 = − 𝑥𝑥 3 − 3

9/30/2014

𝑥𝑥 2 2

𝑒𝑒 2𝑥𝑥 +

𝑥𝑥 2 𝑥𝑥

+ 𝑥𝑥 𝑥𝑥𝑒𝑒 2𝑥𝑥 Dr. Eli Saber

124

Variation of Parameters 1

𝑦𝑦𝑝𝑝 𝑥𝑥 = − 𝑥𝑥 3 − =−

3

𝑥𝑥 2

𝑒𝑒 2𝑥𝑥 +

2

𝑥𝑥 2 𝑥𝑥

+ 𝑥𝑥 𝑥𝑥𝑒𝑒 2𝑥𝑥

1 3 2𝑥𝑥 1 2 2𝑥𝑥 1 3 2𝑥𝑥 𝑥𝑥 𝑒𝑒 − 𝑥𝑥 𝑒𝑒 + 𝑥𝑥 𝑒𝑒 + 𝑥𝑥 2 𝑒𝑒 2𝑥𝑥 3 2 2

1 3 2𝑥𝑥 1 2 2𝑥𝑥 𝑦𝑦𝑝𝑝 = 𝑥𝑥 𝑒𝑒 + 𝑥𝑥 𝑒𝑒 6 2

And, 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝 ⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏

9/30/2014

𝒆𝒆𝟐𝟐𝟐𝟐

+ 𝒄𝒄𝟐𝟐

𝒙𝒙𝒆𝒆𝟐𝟐𝟐𝟐

𝑦𝑦 ′′ − 4𝑦𝑦 ′ + 4𝑦𝑦 = 𝑥𝑥 + 1 𝑒𝑒 2𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 2𝑥𝑥 + 𝑐𝑐2 𝑥𝑥𝑒𝑒 2𝑥𝑥 𝑊𝑊 = 𝑒𝑒 4𝑥𝑥 𝑊𝑊2 = 𝑥𝑥 + 1 𝑒𝑒 4𝑥𝑥 1 3 𝑥𝑥 2 𝑢𝑢1 = − 𝑥𝑥 − 2 3 2 𝑥𝑥 + 𝑥𝑥 𝑢𝑢2 = 2

𝟏𝟏 𝟑𝟑 𝟐𝟐𝟐𝟐 𝟏𝟏 𝟐𝟐 𝟐𝟐𝟐𝟐 + 𝒙𝒙 𝒆𝒆 + 𝒙𝒙 𝒆𝒆 𝟔𝟔 𝟐𝟐 Dr. Eli Saber

125

Variation of Parameters E.g.2:

𝑦𝑦 ′′

−

5𝑦𝑦 ′

+ 4𝑦𝑦 =

Sec. 3.4. E.g. 4 (slide 94-95)

8𝑒𝑒 𝑥𝑥

1. Equation is already in standard form: 𝑦𝑦 ′′ − 5𝑦𝑦 ′ + 4𝑦𝑦 = 8𝑒𝑒 𝑥𝑥 𝑷𝑷(𝒙𝒙)

𝑸𝑸(𝒙𝒙)

2. Find 𝑦𝑦𝑐𝑐 : ⇒ 𝑚𝑚2 − 5𝑚𝑚 + 4 = 0 → (𝑚𝑚 − 1)(𝑚𝑚 − 4) = 0 ⇒ 𝑚𝑚1 = 1&𝑚𝑚2 = 4

𝒇𝒇(𝒙𝒙)

𝑦𝑦𝑐𝑐 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 4𝑥𝑥

3. Compute 𝑊𝑊

𝒚𝒚𝟏𝟏

𝒚𝒚𝟐𝟐

𝑥𝑥 𝑦𝑦1 𝑦𝑦2 𝑊𝑊 = 𝑦𝑦 ′ 𝑦𝑦 ′ = 𝑒𝑒 𝑥𝑥 1 2 𝑒𝑒

= 𝑒𝑒 𝑥𝑥 4𝑒𝑒 4𝑥𝑥 − 𝑒𝑒 𝑥𝑥 𝑒𝑒 4𝑥𝑥 = 4𝑒𝑒 5𝑥𝑥 − 𝑒𝑒 5𝑥𝑥 ⇒ 𝑊𝑊 = 3𝑒𝑒 5𝑥𝑥 9/30/2014

Dr. Eli Saber

𝑒𝑒 4𝑥𝑥 4𝑒𝑒 4𝑥𝑥

126

Variation of Parameters 4. Compute 𝑢𝑢1′ & 𝑢𝑢2 ′

0 𝑦𝑦2 0 𝑒𝑒 4𝑥𝑥 𝑥𝑥 4𝑥𝑥 8𝑒𝑒 5𝑥𝑥 8 𝑓𝑓(𝑥𝑥) 𝑦𝑦2 ′ 8𝑒𝑒 4𝑒𝑒 ′ 𝑢𝑢1 = = = − = − 𝑊𝑊 3𝑒𝑒 5𝑥𝑥 3𝑒𝑒 5𝑥𝑥 3

𝑦𝑦 ′′ − 5𝑦𝑦 ′ + 4𝑦𝑦 = 8𝑒𝑒 𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 4𝑥𝑥 𝑊𝑊 = 3𝑒𝑒 5𝑥𝑥

𝑦𝑦1 0 𝑒𝑒 𝑥𝑥 0 2𝑥𝑥 𝑥𝑥 8𝑒𝑒 𝑥𝑥 8𝑒𝑒 8 −3𝑥𝑥 𝑦𝑦 ′ 𝑓𝑓(𝑥𝑥) 1 𝑒𝑒 𝑢𝑢2′ = = = = 𝑒𝑒 𝑊𝑊 3𝑒𝑒 5𝑥𝑥 3𝑒𝑒 5𝑥𝑥 3 5. Find 𝑢𝑢1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑢𝑢2

8 8 𝑢𝑢1 = � − 𝑑𝑑𝑑𝑑 = − 𝑥𝑥 3 3

8 −3𝑥𝑥 8 −3𝑥𝑥 𝑢𝑢2 = � 𝑒𝑒 𝑑𝑑𝑑𝑑 = − 𝑒𝑒 3 9 9/30/2014

Dr. Eli Saber

127

Variation of Parameters 𝑦𝑦 ′′ − 5𝑦𝑦 ′ + 4𝑦𝑦 = 8𝑒𝑒 𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 4𝑥𝑥 𝑊𝑊 = 3𝑒𝑒 5𝑥𝑥 8 8 𝑢𝑢1 = − 𝑥𝑥 & 𝑢𝑢2 = − 𝑒𝑒 −3𝑥𝑥 3 9

Hence, 𝑦𝑦𝑝𝑝 = 𝑢𝑢1 𝑥𝑥 𝑦𝑦1 𝑥𝑥 + 𝑢𝑢2 𝑥𝑥 𝑦𝑦2 𝑥𝑥

8 8 𝑦𝑦𝑝𝑝 = − 𝑥𝑥 𝑒𝑒 𝑥𝑥 + − 𝑒𝑒 −3𝑥𝑥 𝑒𝑒 4𝑥𝑥 3 9 8 8 𝑦𝑦𝑝𝑝 = − 𝑥𝑥 𝑒𝑒 𝑥𝑥 − 𝑒𝑒 𝑥𝑥 3 9

And, 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

𝟖𝟖 𝟖𝟖 ⇒ 𝒚𝒚 = 𝜶𝜶𝟏𝟏 𝒆𝒆𝒙𝒙 + 𝜶𝜶𝟐𝟐 𝒆𝒆𝟒𝟒𝟒𝟒 − 𝒙𝒙𝒆𝒆𝒙𝒙 − 𝒆𝒆𝒙𝒙 𝟑𝟑 𝟗𝟗 9/30/2014

Dr. Eli Saber

128

Variation of Parameters We got,

𝟖𝟖 𝟖𝟖 ⇒ 𝒚𝒚 = 𝜶𝜶𝟏𝟏 𝒆𝒆𝒙𝒙 + 𝜶𝜶𝟐𝟐 𝒆𝒆𝟒𝟒𝟒𝟒 − 𝒙𝒙𝒆𝒆𝒙𝒙 − 𝒆𝒆𝒙𝒙 𝟑𝟑 𝟗𝟗

From Sec. 3.4. E.g. 4 (slide 94-95), we have: 𝟖𝟖 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒆𝒆𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒆𝒆𝟒𝟒𝟒𝟒 − 𝒙𝒙𝒆𝒆𝒙𝒙 𝟑𝟑 as the solution. 8 3

8 9

8 9

8 3

Notice, 𝑦𝑦 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 4𝑥𝑥 − 𝑥𝑥𝑒𝑒 𝑥𝑥 − 𝑒𝑒 𝑥𝑥 = 𝛼𝛼1 𝑒𝑒 𝑥𝑥 − 𝑒𝑒 𝑥𝑥 + 𝛼𝛼2 𝑒𝑒 4𝑥𝑥 − 𝑥𝑥𝑒𝑒 𝑥𝑥 8 𝑥𝑥 8 𝑥𝑥 4𝑥𝑥 = 𝛼𝛼1 − 𝑒𝑒 + 𝛼𝛼2 𝑒𝑒 − 𝑥𝑥𝑒𝑒 9 3 = 𝒄𝒄𝟏𝟏

𝑒𝑒 𝑥𝑥

9/30/2014

+ 𝒄𝒄𝟐𝟐

𝑒𝑒 4𝑥𝑥

8 𝑥𝑥 − 𝑥𝑥𝑒𝑒 3

Dr. Eli Saber

129

Variation of Parameters • 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+. . . +𝑐𝑐𝑛𝑛 𝑦𝑦𝑛𝑛 is the complementary function , then a particular solution is: 𝑦𝑦𝑝𝑝 = 𝑢𝑢1(𝑥𝑥)𝑦𝑦1(𝑥𝑥) + 𝑢𝑢2(𝑥𝑥)𝑦𝑦2(𝑥𝑥)+. . . +𝑢𝑢𝑛𝑛 (𝑥𝑥)𝑦𝑦𝑛𝑛 (𝑥𝑥) ,where the 𝑢𝑢𝑘𝑘′ , 𝑘𝑘 = 1,2, … , 𝑛𝑛 are determined by the 𝑛𝑛 eqn.

9/30/2014

𝑦𝑦1 𝑢𝑢1′ + 𝑦𝑦2 𝑢𝑢2′ + ⋯ + 𝑦𝑦𝑛𝑛 𝑢𝑢𝑛𝑛′ = 0 𝑦𝑦1′ 𝑢𝑢1′ + 𝑦𝑦2′ 𝑢𝑢2′ + ⋯ + 𝑦𝑦𝑛𝑛′ 𝑢𝑢𝑛𝑛′ = 0 ⋮ ⋮ (𝑛𝑛−1) ′ (𝑛𝑛−1) ′ 𝑛𝑛−1 ′ 𝑦𝑦1 𝑢𝑢1 + 𝑦𝑦2 𝑢𝑢2 + ⋯ + 𝑦𝑦𝑛𝑛 𝑢𝑢𝑛𝑛 = 𝑓𝑓(𝑥𝑥) Dr. Eli Saber

130

Variation of Parameters

⇒

𝑦𝑦1 𝑦𝑦1 ′ ⋮

𝑦𝑦1

𝑛𝑛−1

And, 𝑢𝑢𝑘𝑘′ =

𝑦𝑦2 𝑦𝑦2′ ⋮

𝑦𝑦2

𝑊𝑊𝑘𝑘 ; 𝑊𝑊

𝑛𝑛−1

… … ⋮ …

𝑦𝑦𝑛𝑛 𝑦𝑦𝑛𝑛 ′ ⋮

𝑛𝑛−1

𝑦𝑦𝑛𝑛

𝑘𝑘 = 1,2, … , 𝑛𝑛

0 0 Where, 𝑊𝑊1 = ⋮ 𝑓𝑓(𝑥𝑥)

𝑦𝑦2 𝑦𝑦2′ ⋮

𝑦𝑦2 (𝑛𝑛−1)

𝑢𝑢1′ 0 0 𝑢𝑢2′ = ⋮ ⋮ 𝑓𝑓(𝑥𝑥) 𝑢𝑢𝑛𝑛′

⋯ 𝑦𝑦𝑛𝑛 ⋯ 𝑦𝑦𝑛𝑛′ ⋮ ⋮ (𝑛𝑛−1) ⋯ 𝑦𝑦𝑛𝑛

𝑢𝑢𝑘𝑘 can be computed by integrating 𝑢𝑢𝑘𝑘′ ; 𝑘𝑘 = 1,2, … , 𝑛𝑛

𝑦𝑦𝑝𝑝 = 𝑢𝑢1(𝑥𝑥)𝑦𝑦1(𝑥𝑥) + 𝑢𝑢2(𝑥𝑥)𝑦𝑦2(𝑥𝑥)+. . . +𝑢𝑢𝑛𝑛 (𝑥𝑥)𝑦𝑦𝑛𝑛 (𝑥𝑥) 9/30/2014

Dr. Eli Saber

131

Variation of Parameters Summary Given

𝑎𝑎2 𝑦𝑦𝑦𝑦 + 𝑎𝑎1𝑦𝑦𝑦 + 𝑎𝑎0𝑦𝑦 = 𝑔𝑔(𝑥𝑥)

1. Put Eq. into standard form by dividing throughout by a2(x) 𝑎𝑎1 ′ 𝑎𝑎0 𝑔𝑔 𝑥𝑥 𝑦𝑦 ′′ + 𝑦𝑦 + 𝑦𝑦 = 𝑎𝑎2 𝑎𝑎2 𝑎𝑎2 (𝑥𝑥) 2. Find 𝑦𝑦𝑐𝑐 = 𝑐𝑐1𝑦𝑦1 + 𝑐𝑐2𝑦𝑦2

Verify 𝑦𝑦𝑐𝑐 for the D.E. 𝑦𝑦 𝑦𝑦 3. Compute 𝑊𝑊 = 𝑦𝑦 1′ 𝑦𝑦 2′ 1 2

9/30/2014

4. Compute 𝑢𝑢1’ & 𝑢𝑢2’ using: 0 𝑦𝑦2 𝑦𝑦1 0 𝑦𝑦1′ 𝑓𝑓(𝑥𝑥) 𝑓𝑓(𝑥𝑥) 𝑦𝑦2′ ′ ; 𝑢𝑢 = 𝑢𝑢1′ = 𝑦𝑦 2 𝑦𝑦2 𝑦𝑦1 𝑦𝑦2 1 𝑦𝑦1′ 𝑦𝑦2′ 𝑦𝑦1′ 𝑦𝑦2′ 5. Find 𝑢𝑢1 & 𝑢𝑢2 by integrating 𝑢𝑢1′ & 𝑢𝑢2′ respectively.

6. Form 𝑦𝑦𝑝𝑝 = 𝑢𝑢1(𝑥𝑥)𝑦𝑦1(𝑥𝑥) + 𝑢𝑢2(𝑥𝑥)𝑦𝑦2(𝑥𝑥)

7. General Solution: 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝 verify the solution for the D.E.

Dr. Eli Saber

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Section 3.6 Cauchy-Euler Equations

9/30/2014

Dr. Eli Saber

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Cauchy-Euler Equations Any linear Differential Equation of the form: 𝑑𝑑 𝑛𝑛−1 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑛𝑛 𝑦𝑦 𝑛𝑛 𝑛𝑛−1 + 𝑎𝑎 𝑥𝑥 + … + 𝑎𝑎 𝑥𝑥 + 𝑎𝑎0 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑛𝑛−1 1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 𝑛𝑛 𝑑𝑑𝑥𝑥 𝑛𝑛−1 , where 𝑎𝑎𝑛𝑛 , 𝑎𝑎𝑛𝑛−1 , … , 𝑎𝑎1 , 𝑎𝑎0 are constants And the degree 𝑛𝑛 at

𝑥𝑥 𝑛𝑛

matches the order 𝑛𝑛 of the differentiation

is called a Cauchy-Euler Equation E.g.

same

1) 𝑥𝑥 2

𝑑𝑑𝑛𝑛 𝑦𝑦 𝑑𝑑𝑥𝑥 𝑛𝑛

same

𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑦𝑦 − 2𝑥𝑥 − 4𝑦𝑦 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 2

2 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑑𝑑 2) 𝑥𝑥 2 2 − 3𝑥𝑥 + 3𝑦𝑦 = 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 9/30/2014

Dr. Eli Saber

134

Cauchy-Euler Equations

General

2nd

order:

2

𝑑𝑑 𝑦𝑦 𝑎𝑎𝑥𝑥 2 2 𝑑𝑑𝑥𝑥

+ 𝑏𝑏𝑏𝑏

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

+ 𝑐𝑐𝑐𝑐 = 0

Homogeneous

 Proceed to develop solution for 2nd order and then generalize. 2 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑦𝑦 2 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐𝑐𝑐 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 2

Note: 𝑎𝑎𝑥𝑥 2 = 0 @ 𝑥𝑥 = 0  confine attention to interval 𝐼𝐼 ≡ 0, ∞ For (−∞, 0), let 𝑡𝑡 = −𝑥𝑥

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135

Cauchy-Euler Equations Try a solution of the form 𝑦𝑦 = 𝑥𝑥 𝑚𝑚

𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑦𝑦 𝑚𝑚−1 ⇒ = 𝑚𝑚𝑥𝑥 & 2 = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥

𝑑𝑑2 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐𝑐𝑐 = 0 𝑑𝑑𝑥𝑥 2 𝑑𝑑𝑑𝑑 2

⇒ 𝑎𝑎𝑥𝑥 2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 + 𝑏𝑏𝑏𝑏 𝑚𝑚𝑥𝑥 𝑚𝑚−1 + 𝑐𝑐𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑎𝑎𝑎𝑎 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚 + 𝑏𝑏𝑏𝑏𝑥𝑥 𝑚𝑚 + 𝑐𝑐𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑎𝑎𝑎𝑎 𝑚𝑚 − 1 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑎𝑎𝑚𝑚2 − 𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 𝑥𝑥 𝑚𝑚 = 0

⇒ 𝑎𝑎𝑚𝑚2 + 𝑏𝑏 − 𝑎𝑎 𝑚𝑚 + 𝑐𝑐 𝑥𝑥 𝑚𝑚 = 0

Thus, 𝒚𝒚 = 𝒙𝒙𝒎𝒎 is a solution of the D.E. whenever 𝑚𝑚 is a solution to the auxiliary equation 𝒂𝒂𝒎𝒎𝟐𝟐 + 𝒃𝒃 − 𝒂𝒂 𝒎𝒎 + 𝒄𝒄 = 𝟎𝟎 9/30/2014

Dr. Eli Saber

136

Cauchy-Euler Equations Case 1: Distinct Real Roots Case 1: Distinct Real Roots If 𝑚𝑚1 & 𝑚𝑚2 are the real roots of 𝑎𝑎𝑚𝑚2 + 𝑏𝑏 − 𝑎𝑎 𝑚𝑚 + 𝑐𝑐 = 0 with 𝑚𝑚1 ≠ 𝑚𝑚2 ⇒ 𝑦𝑦1 = 𝑥𝑥 𝑚𝑚1 & 𝑦𝑦2 = 𝑥𝑥 𝑚𝑚2 form a fundamental set of solutions and the general solution is 𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 𝑚𝑚1 + 𝑐𝑐2 𝑥𝑥 𝑚𝑚2

General case: 𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 𝑚𝑚1 + 𝑐𝑐2 𝑥𝑥 𝑚𝑚2 + ⋯ + 𝑐𝑐𝑛𝑛 𝑥𝑥 𝑚𝑚𝑛𝑛  𝑛𝑛𝑡𝑡𝑡 order

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Dr. Eli Saber

137

Cauchy-Euler Equations Case 1: Distinct Real Roots E.g.

2

𝑑𝑑 𝑦𝑦 𝑥𝑥 2 2 𝑑𝑑𝑥𝑥

− 2𝑥𝑥

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = 𝑚𝑚𝑥𝑥 𝑚𝑚−1 𝑑𝑑𝑑𝑑

− 4𝑦𝑦 = 0

𝑑𝑑 2 𝑦𝑦 = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 𝑑𝑑𝑥𝑥 2

Assume 𝑦𝑦 = 𝑥𝑥 𝑚𝑚 as the solution.

⇒ 𝑥𝑥 2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 − 2𝑥𝑥 𝑚𝑚𝑥𝑥 𝑚𝑚−1 − 4𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚 − 2𝑚𝑚 𝑥𝑥 𝑚𝑚 − 4𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑚𝑚 𝑚𝑚 − 1 − 2𝑚𝑚 − 4 𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑚𝑚2 − 𝑚𝑚 − 2𝑚𝑚 − 4 𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑚𝑚2 − 3𝑚𝑚 − 4 𝑥𝑥 𝑚𝑚 = 0 Auxiliary Equation

9/30/2014

Dr. Eli Saber

138

Cauchy-Euler Equations Case 1: Distinct Real Roots 𝑚𝑚2 − 3𝑚𝑚 − 4 = 0

⇒ 𝑚𝑚 + 1 𝑚𝑚 − 4 = 0 ⇒ 𝑚𝑚 = −1 𝑜𝑜𝑜𝑜 𝑚𝑚 = 4

Hence,

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𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙−𝟏𝟏 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝟒𝟒

Dr. Eli Saber

139

Cauchy-Euler Equations Case 2: Repeated Real Roots Case 2: Repeated Real Roots If the roots are repeated i.e. 𝑚𝑚1 = 𝑚𝑚2 , only one solution: 𝒚𝒚 = 𝒙𝒙𝒎𝒎𝟏𝟏 ⇒ 𝑎𝑎𝑚𝑚2 + 𝑏𝑏 − 𝑎𝑎 𝑚𝑚 + 𝑐𝑐 = 0 ⇒ 𝑚𝑚 =

− 𝑏𝑏 − 𝑎𝑎 ±

𝑏𝑏 − 𝑎𝑎 2𝑎𝑎

For 𝑚𝑚1 = 𝑚𝑚2 , ⇒ 𝑏𝑏 − 𝑎𝑎 ⇒ 𝑏𝑏 − 𝑎𝑎

2

= 4𝑎𝑎𝑎𝑎

Hence, 𝑚𝑚1 = 𝑚𝑚2 = − 9/30/2014

2

2

− 4𝑎𝑎𝑎𝑎

− 4𝑎𝑎𝑎𝑎 = 0

(𝑏𝑏−𝑎𝑎) 2𝑎𝑎 Dr. Eli Saber

140

Cauchy-Euler Equations Case 2: Repeated Real Roots Construct a second solution like Section 3.2. 𝑎𝑎𝑥𝑥 2

𝑑𝑑 2 𝑦𝑦 𝑑𝑑𝑑𝑑 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐𝑐𝑐 = 0 𝑑𝑑𝑥𝑥 2 𝑑𝑑𝑑𝑑

𝑑𝑑 2 𝑦𝑦 𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑 𝑐𝑐 ⇒ 2+ 2 + 𝑦𝑦 = 0 𝑑𝑑𝑥𝑥 𝑎𝑎𝑥𝑥 𝑑𝑑𝑑𝑑 𝑎𝑎𝑥𝑥 2 𝑑𝑑 2 𝑦𝑦 𝑏𝑏 𝑑𝑑𝑑𝑑 𝑐𝑐 ⇒ 2+ + 2 𝑦𝑦 = 0 𝑑𝑑𝑥𝑥 𝑎𝑎𝑥𝑥 𝑑𝑑𝑑𝑑 𝑎𝑎𝑥𝑥 𝑷𝑷(𝒙𝒙)

Q(𝒙𝒙)

Let 𝑦𝑦 = 𝑢𝑢 𝑥𝑥 𝑦𝑦1 𝑥𝑥

⇒ 𝑦𝑦 ′ = 𝑢𝑢𝑦𝑦1′ + 𝑢𝑢′ 𝑦𝑦1 & 𝑦𝑦 ′′ = 𝑢𝑢𝑦𝑦 ′′ + 2𝑢𝑢′ 𝑦𝑦1′ + 𝑢𝑢′′ 𝑦𝑦1 9/30/2014

Dr. Eli Saber

141

Cauchy-Euler Equations Case 2: Repeated Real Roots 𝑑𝑑 2 𝑦𝑦 𝑏𝑏 𝑑𝑑𝑑𝑑 𝑐𝑐 + + 𝑦𝑦 = 0 𝑑𝑑𝑥𝑥 2 𝑎𝑎𝑎𝑎 𝑑𝑑𝑑𝑑 𝑎𝑎𝑥𝑥 2 Replace 𝑦𝑦1 , 𝑦𝑦1′ , 𝑦𝑦1′ ′

⇒ 𝑢𝑢𝑦𝑦1′′ + 2𝑢𝑢′ 𝑦𝑦1′ + 𝑢𝑢′′ 𝑦𝑦1 + ⇒ 𝑢𝑢 = 𝑦𝑦1

𝑦𝑦1′′ 𝑢𝑢′′

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𝑦𝑦1 = 𝑥𝑥 𝑚𝑚1 𝑦𝑦 = 𝑢𝑢𝑦𝑦1 ′ ′ 𝑦𝑦 = 𝑢𝑢𝑦𝑦1 + 𝑢𝑢′ 𝑦𝑦1 𝑦𝑦 ′′ = 𝑢𝑢𝑦𝑦 ′′ + 2𝑢𝑢′ 𝑦𝑦1′ + 𝑢𝑢′′ 𝑦𝑦1

𝑏𝑏 𝑐𝑐 𝑢𝑢𝑦𝑦1′ + 𝑢𝑢′ 𝑦𝑦1 + 2 𝑢𝑢𝑦𝑦1 = 0 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎

𝑏𝑏 ′ 𝑐𝑐 𝑏𝑏 ′ ′′ + 𝑦𝑦 + 𝑦𝑦 + 𝑦𝑦1 𝑢𝑢 + 2𝑦𝑦1 + 𝑦𝑦 𝑎𝑎𝑎𝑎 1 𝑎𝑎𝑥𝑥 2 1 𝑎𝑎𝑎𝑎 1

+

2𝑦𝑦 ′

=0  since 𝒚𝒚𝟏𝟏 = 𝒙𝒙𝒎𝒎 is a solution

𝑢𝑢′ = 0

𝑏𝑏 + 𝑦𝑦1 𝑢𝑢′ = 0 𝑎𝑎𝑎𝑎

Dr. Eli Saber

142

Cauchy-Euler Equations Case 2: Repeated Real Roots 𝑦𝑦1

𝑢𝑢′′

+

2𝑦𝑦 ′

𝑏𝑏 + 𝑦𝑦1 𝑢𝑢′ = 0 𝑎𝑎𝑎𝑎

Let 𝑤𝑤 = 𝑢𝑢′ ⇒ 𝑦𝑦1 𝑤𝑤 ′ + 2𝑦𝑦1′ + ⇒ 𝑦𝑦1

𝑏𝑏 𝑎𝑎𝑎𝑎

𝑏𝑏 𝑑𝑑𝑑𝑑 = − 2 𝑦𝑦1′ + 𝑦𝑦 𝑤𝑤 𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎 1

𝑦𝑦1 𝑤𝑤 = 0

𝑦𝑦1 = 𝑥𝑥 𝑚𝑚1 𝑦𝑦 = 𝑢𝑢𝑦𝑦1 ′ ′ 𝑦𝑦 = 𝑢𝑢𝑦𝑦1 + 𝑢𝑢′ 𝑦𝑦1 𝑦𝑦 ′′ = 𝑢𝑢𝑦𝑦 ′′ + 2𝑢𝑢′ 𝑦𝑦1′ + 𝑢𝑢′′ 𝑦𝑦1

𝑑𝑑𝑑𝑑 1 𝑏𝑏 ′ ⇒ =− 2𝑦𝑦1 + 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑤𝑤 𝑦𝑦1 𝑎𝑎𝑎𝑎 1 𝑑𝑑𝑑𝑑 𝑦𝑦1′ 𝑏𝑏 ⇒ = −2 𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑑𝑑 𝑦𝑦1 𝑤𝑤 𝑎𝑎𝑎𝑎 9/30/2014

Dr. Eli Saber

143

Cauchy-Euler Equations Case 2: Repeated Real Roots 𝑑𝑑𝑑𝑑 𝑚𝑚1 𝑥𝑥 𝑚𝑚1−1 𝑏𝑏 ⇒ = −2 𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑑𝑑 𝑥𝑥 𝑚𝑚1 𝑤𝑤 𝑎𝑎𝑎𝑎

𝑑𝑑𝑑𝑑 𝑚𝑚1 𝑏𝑏 1 ⇒ � = � −2 𝑑𝑑𝑑𝑑 − � 𝑑𝑑𝑑𝑑 𝑥𝑥 𝑤𝑤 𝑎𝑎 𝑥𝑥 𝑏𝑏 ⇒ ln |𝑤𝑤| = −2 𝑚𝑚1 ln 𝑥𝑥 − ln 𝑥𝑥 + 𝑐𝑐 𝑎𝑎

𝑑𝑑𝑑𝑑 𝑦𝑦1′ 𝑏𝑏 = −2 𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑑𝑑 𝑦𝑦1 𝑤𝑤 𝑎𝑎𝑎𝑎

𝑦𝑦1 = 𝑥𝑥 𝑚𝑚1 ⇒ 𝑦𝑦1′ = 𝑚𝑚1 𝑥𝑥 𝑚𝑚1−1

𝑏𝑏 ⇒ ln |𝑤𝑤| + 2 𝑚𝑚1 ln 𝑥𝑥 + ln 𝑥𝑥 = 𝑐𝑐 𝑎𝑎 ⇒ ln 𝑤𝑤 + ln 𝑥𝑥 ⇒ ln

𝑏𝑏 2𝑚𝑚 𝑤𝑤𝑥𝑥 1 𝑥𝑥 𝑎𝑎

9/30/2014

2𝑚𝑚1

+ ln 𝑥𝑥

= 𝑐𝑐

𝑏𝑏 𝑎𝑎

= 𝑐𝑐

Dr. Eli Saber

144

Cauchy-Euler Equations Case 2: Repeated Real Roots ⇒ ln ⇒

𝑏𝑏 2𝑚𝑚 1 𝑤𝑤𝑥𝑥 𝑥𝑥 𝑎𝑎

𝑏𝑏 2𝑚𝑚 1 𝑎𝑎 𝑤𝑤𝑥𝑥 𝑥𝑥

But 𝑤𝑤 =

𝑢𝑢′

= 𝑐𝑐

= 𝑒𝑒 𝑐𝑐

⇒

𝑏𝑏

𝑢𝑢′ 𝑥𝑥 2𝑚𝑚1 𝑥𝑥 𝑎𝑎 𝑏𝑏

⇒ 𝑢𝑢′ = 𝑒𝑒 𝑐𝑐 𝑥𝑥 −2𝑚𝑚1 𝑥𝑥 − 𝑎𝑎 ⇒ 𝑢𝑢 =

𝑏𝑏 − 𝑎𝑎 𝑐𝑐 −2𝑚𝑚 1 � 𝑒𝑒 𝑥𝑥 𝑥𝑥 𝑑𝑑𝑑𝑑

Now, 𝑦𝑦2 = 𝑢𝑢𝑦𝑦1 = =

𝑥𝑥 𝑚𝑚1

= 𝑒𝑒 𝑐𝑐

� 𝑒𝑒 𝑐𝑐

9/30/2014

𝑥𝑥

𝑥𝑥 𝑚𝑚1

𝑏𝑏 𝑏𝑏 −1−𝑎𝑎 𝑎𝑎

𝒃𝒃−𝒂𝒂

∫ 𝑒𝑒 𝑐𝑐 𝑥𝑥 𝒂𝒂

𝑥𝑥

−

𝑏𝑏 𝑎𝑎

𝑏𝑏 − 𝑎𝑎 2𝑎𝑎 𝑏𝑏 − 𝑎𝑎 ⇒ 2𝑚𝑚1 = − 𝑎𝑎 𝑚𝑚1 = −

𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = 𝑥𝑥 𝑚𝑚1 � 𝑒𝑒 𝑐𝑐 𝑥𝑥 −1 𝑑𝑑𝑑𝑑

Dr. Eli Saber

145

Cauchy-Euler Equations Case 2: Repeated Real Roots 𝑦𝑦2 = 𝑥𝑥 𝑚𝑚1 � 𝑒𝑒 𝑐𝑐 𝑥𝑥 −1 𝑑𝑑𝑑𝑑 𝑦𝑦2 = 𝑒𝑒 𝑐𝑐 𝑥𝑥 𝑚𝑚1 ln 𝑥𝑥 = 𝑐𝑐2 𝑥𝑥 𝑚𝑚1 ln 𝑥𝑥 General solution:

9/30/2014

𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙𝒎𝒎𝟏𝟏 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝒎𝒎𝟏𝟏 𝒍𝒍𝒍𝒍 𝒙𝒙

Dr. Eli Saber

146

Cauchy-Euler Equations Case 2: Repeated Real Roots E.g.

4𝑥𝑥 2 𝑦𝑦 ′′ + 8𝑥𝑥𝑦𝑦 ′ + 𝑦𝑦 = 0

𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 𝑚𝑚1 + 𝑐𝑐2 𝑥𝑥 𝑚𝑚1 𝑙𝑙𝑙𝑙 𝑥𝑥

Let 𝑦𝑦 = 𝑥𝑥 𝑚𝑚 ⇒ 𝑦𝑦 ′ = 𝑚𝑚𝑥𝑥 𝑚𝑚−1 ⇒ 𝑦𝑦 ′′ = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 ⇒ 4𝑥𝑥 2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 + 8𝑥𝑥 𝑚𝑚𝑥𝑥 𝑚𝑚−1 + 𝑥𝑥 𝑚𝑚 = 0 ⇒ 4𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚 + 8𝑚𝑚 𝑥𝑥 𝑚𝑚 + 𝑥𝑥 𝑚𝑚 = 0

⇒ 4𝑚𝑚2 − 4𝑚𝑚 + 8𝑚𝑚 + 1 𝑥𝑥 𝑚𝑚 = 0 ⇒ 4𝑚𝑚2 + 4𝑚𝑚 + 1 = 0 ⇒ 2𝑚𝑚 + 1

9/30/2014

2

= 0 ⇒ 𝑚𝑚 = −

1 2

⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙

Repeated roots 𝟏𝟏 − 𝟐𝟐

+ 𝒄𝒄𝟐𝟐 𝒙𝒙

Dr. Eli Saber

𝟏𝟏 − 𝟐𝟐

𝐥𝐥𝐥𝐥 𝒙𝒙

147

Cauchy-Euler Equations Case 2: Repeated Real Roots Note: For higher order equations, if 𝑚𝑚1 is a root of multiplicity 𝐾𝐾

⇒ 𝑥𝑥 𝑚𝑚1 , 𝑥𝑥 𝑚𝑚1 ln 𝑥𝑥 , 𝑥𝑥 𝑚𝑚1 ln 𝑥𝑥 2 , … , 𝑥𝑥 𝑚𝑚1 ln 𝑥𝑥

are 𝐾𝐾 linearly independent solutions

⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙𝒎𝒎𝟏𝟏 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝒎𝒎𝟏𝟏 𝒍𝒍𝒍𝒍 𝒙𝒙 + 𝒄𝒄𝟑𝟑 𝒙𝒙𝒎𝒎𝟏𝟏 𝒍𝒍𝒍𝒍 𝒙𝒙

9/30/2014

Dr. Eli Saber

𝟐𝟐

𝑘𝑘−1

+ ⋯ + 𝒄𝒄𝒌𝒌 𝒙𝒙𝒎𝒎𝟏𝟏 𝒍𝒍𝒍𝒍 𝒙𝒙

𝒌𝒌−𝟏𝟏

148

Cauchy-Euler Equations Case 3: Conjugate Complex Roots Case 3: Conjugate Complex Roots If the roots are conjugate pairs i.e. 𝑚𝑚1 = 𝛼𝛼 + 𝑗𝑗𝑗𝑗 & 𝑚𝑚2 = 𝛼𝛼 − 𝑗𝑗𝑗𝑗 ⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙𝜶𝜶+𝒋𝒋𝒋𝒋 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝜶𝜶−𝒋𝒋𝒋𝒋

(𝛼𝛼, 𝛽𝛽 > 0)

We can rewrite that in terms of 𝑐𝑐𝑐𝑐𝑐𝑐 & 𝑠𝑠𝑠𝑠𝑠𝑠 as: 𝑥𝑥 𝑗𝑗𝑗𝑗 = 𝑒𝑒 ln 𝑥𝑥

𝑥𝑥 −𝑗𝑗𝑗𝑗 = 𝑒𝑒 ln 𝑥𝑥

9/30/2014

𝑗𝑗𝛽𝛽

−𝑗𝑗𝑗𝑗

= 𝑒𝑒 ln 𝑥𝑥𝑗𝑗𝑗𝑗 = cos 𝛽𝛽 ln 𝑥𝑥 + 𝑗𝑗 sin(𝛽𝛽 ln 𝑥𝑥)

= 𝑒𝑒 − ln 𝑥𝑥𝑗𝑗𝑗𝑗 = cos 𝛽𝛽 ln 𝑥𝑥 − 𝑗𝑗 sin(𝛽𝛽 ln 𝑥𝑥)

Dr. Eli Saber

149

Cauchy-Euler Equations Case 3: Conjugate Complex Roots We have, 𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 𝛼𝛼+𝑗𝑗𝑗𝑗 + 𝑐𝑐2 𝑥𝑥 𝛼𝛼−𝑗𝑗𝑗𝑗

⇒ 𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 𝛼𝛼 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽 ln 𝑥𝑥 + 𝑗𝑗 𝑠𝑠𝑠𝑠𝑠𝑠(𝛽𝛽 ln 𝑥𝑥) + 𝑐𝑐2 𝑥𝑥 𝛼𝛼 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽 ln 𝑥𝑥 − 𝑗𝑗 𝑠𝑠𝑠𝑠𝑠𝑠(𝛽𝛽 ln 𝑥𝑥)

= 𝑐𝑐1 𝑥𝑥 𝛼𝛼 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽 ln 𝑥𝑥 + 𝑗𝑗𝑐𝑐1 𝑥𝑥 𝛼𝛼 𝑠𝑠𝑠𝑠𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥 𝛼𝛼 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽 ln 𝑥𝑥 − 𝑗𝑗𝑐𝑐2 𝑥𝑥 𝛼𝛼 𝑠𝑠𝑠𝑠𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 = 𝑥𝑥 𝛼𝛼 𝑐𝑐1 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽 ln 𝑥𝑥 + 𝑐𝑐2 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽 ln 𝑥𝑥 + 𝑗𝑗𝑐𝑐1 𝑠𝑠𝑠𝑠𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 − 𝑗𝑗𝑐𝑐2 𝑠𝑠𝑠𝑠𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 = 𝑥𝑥 𝛼𝛼 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽 ln 𝑥𝑥 {𝑐𝑐1 + 𝑐𝑐2 } + 𝑗𝑗{𝑐𝑐1 − 𝑐𝑐2 } 𝑠𝑠𝑠𝑠𝑠𝑠 𝛽𝛽 ln 𝑥𝑥 𝒚𝒚 = 𝒙𝒙𝜶𝜶 ∝𝟏𝟏 𝒄𝒄𝒄𝒄𝒄𝒄 𝜷𝜷 𝒍𝒍𝒍𝒍 𝒙𝒙 +∝𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔 𝜷𝜷 𝒍𝒍𝒍𝒍 𝒙𝒙

9/30/2014

Dr. Eli Saber

150

Cauchy-Euler Equations Case 3: Conjugate Complex Roots E.g. 4𝑥𝑥 2 𝑦𝑦 ′′ + 17𝑦𝑦 = 0

with I.C. 𝑦𝑦 1 = −1; 𝑦𝑦 ′ 1 = −

Let 𝑦𝑦 = 𝑥𝑥 𝑚𝑚 ⇒ 𝑦𝑦 ′ = 𝑚𝑚𝑥𝑥 𝑚𝑚−1 ⇒ 𝑦𝑦 ′′ = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2

1 2

⇒ 4𝑥𝑥 2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 + 17𝑥𝑥 𝑚𝑚 = 0 ⇒ 4𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚 + 17𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑥𝑥 𝑚𝑚 4𝑚𝑚2 − 4𝑚𝑚 + 17 = 0

Auxiliary Eqn. : 𝟒𝟒𝒎𝒎𝟐𝟐 − 𝟒𝟒𝟒𝟒 + 𝟏𝟏𝟏𝟏 = 𝟎𝟎

9/30/2014

Dr. Eli Saber

151

Cauchy-Euler Equations Case 3: Conjugate Complex Roots Auxiliary Eqn. : 4𝑚𝑚2 − 4𝑚𝑚 + 17 = 0

− −4 ± 16 − 4(4)(17) 4 ± 16 − 272 4 ± 256 𝑗𝑗 2 𝑚𝑚 = = = 8 8 8 4 ± 4 ∗ 64 𝑗𝑗 2 4 ± 𝑗𝑗𝑗(8) 1 𝑚𝑚 = = = ± 2𝑗𝑗 8 8 2 ⇒ 𝒎𝒎𝟏𝟏 =

𝟏𝟏 𝟏𝟏 + 𝟐𝟐𝟐𝟐 & 𝒎𝒎𝟐𝟐 = − 𝟐𝟐𝟐𝟐 𝟐𝟐 𝟐𝟐

1 𝛼𝛼 = & 𝛽𝛽 = 2 2 ⇒ 𝒚𝒚 =

9/30/2014

𝟏𝟏 𝒄𝒄𝟏𝟏 𝒙𝒙𝟐𝟐+𝟐𝟐𝟐𝟐

+

𝟏𝟏 𝒄𝒄𝟐𝟐 𝒙𝒙𝟐𝟐−𝟐𝟐𝟐𝟐

𝒐𝒐𝒐𝒐 𝒚𝒚 =

𝟏𝟏 𝒙𝒙𝟐𝟐

∝𝟏𝟏 𝒄𝒄𝒄𝒄𝒄𝒄 𝟐𝟐 𝒍𝒍𝒍𝒍 𝒙𝒙 +∝𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔 𝟐𝟐 𝒍𝒍𝒍𝒍 𝒙𝒙

Dr. Eli Saber

152

Cauchy-Euler Equations Case 3: Conjugate Complex Roots Now, I.C. 𝑦𝑦 1 = −1; 𝑦𝑦 ′ 1 = − 𝑦𝑦 1 = −1 ⇒ −1 = 1

1 2

⇒ −1 = 1 ∝1 1 +∝2 0

𝑦𝑦 ′ 1 = − ′

𝑦𝑦 =∝1

9/30/2014

1 2

𝑦𝑦 =

1 2

1 𝑥𝑥 2

∝1 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑙𝑙𝑙𝑙 𝑥𝑥 +∝2 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑙𝑙𝑙𝑙 𝑥𝑥

∝1 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑙𝑙𝑙𝑙 1 +∝2 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑙𝑙𝑙𝑙 1 ⇒ ∝1 = −1

1 1 −1 2 𝑥𝑥 cos 2 ln 𝑥𝑥 + 𝑥𝑥 2 − sin 2 ln 𝑥𝑥 2 1 −1 ∝2 2 𝑥𝑥 2 sin

1 2

2 + 𝑥𝑥

2 ln 𝑥𝑥 + 𝑥𝑥 cos 2 ln 𝑥𝑥 Dr. Eli Saber

2 𝑥𝑥 153

Cauchy-Euler Equations Case 3: Conjugate Complex Roots

𝑦𝑦 ′ =∝1

1 1 −1 𝑥𝑥 2 cos 2 ln 𝑥𝑥 + 𝑥𝑥 2 − sin 2 ln 𝑥𝑥 2

1 ⇒ − = −1 2

⇒−

𝑦𝑦 =

1

2 1

∝2

1 −1 𝑥𝑥 2 sin 2

1 cos 0 + 2

+∝2

1 12

1

2 1

1 2

2 + 𝑥𝑥 1

1 𝑥𝑥 2

∝1 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑙𝑙𝑙𝑙 𝑥𝑥 +∝2 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑙𝑙𝑙𝑙 𝑥𝑥 ∝1 = −1

2 ln 𝑥𝑥 + 𝑥𝑥 2 cos 2 ln 𝑥𝑥

− sin 0

2 1

sin 0 + 1

1 2

+ cos 0

2 𝑥𝑥

2 1

1 1 1 1 = −1 +∝2 ⇒∝2 = − + = 0 ⇒∝2 = 0 2 2 2 2

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𝟏𝟏 𝟐𝟐

⇒ 𝒚𝒚 = −𝒙𝒙 𝐜𝐜𝐜𝐜𝐜𝐜(𝟐𝟐 𝐥𝐥𝐥𝐥 𝒙𝒙) Dr. Eli Saber

154

Cauchy-Euler Equations Case 3: Conjugate Complex Roots E.g. 𝑥𝑥 3

𝑑𝑑𝑑𝑑 𝑑𝑑 3 𝑦𝑦 𝑑𝑑 2 𝑦𝑦 2 + 5𝑥𝑥 + 7𝑥𝑥 + 8𝑦𝑦 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 3 𝑑𝑑𝑥𝑥 2

Assume 𝑦𝑦 = 𝑥𝑥 𝑚𝑚 ⇒ 𝑦𝑦 ′ = 𝑚𝑚𝑥𝑥 𝑚𝑚−1 ⇒ 𝑦𝑦 ′′ = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 ⇒ 𝑦𝑦 ′′′ = 𝑚𝑚 𝑚𝑚 − 1 𝑚𝑚 − 2 𝑥𝑥 𝑚𝑚−3

⇒ 𝑥𝑥 3 𝑚𝑚 𝑚𝑚 − 1 𝑚𝑚 − 2 𝑥𝑥 𝑚𝑚−3 + 5𝑥𝑥 2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 +7𝑥𝑥 𝑚𝑚𝑥𝑥 𝑚𝑚−1 + 8𝑥𝑥 𝑚𝑚 = 0

⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚 𝑚𝑚 − 1 𝑚𝑚 − 2 + 5𝑚𝑚 𝑚𝑚 − 1 + 7𝑚𝑚 + 8 = 0 ⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚 𝑚𝑚2 − 3𝑚𝑚 + 2 + 5𝑚𝑚2 − 5𝑚𝑚 + 7𝑚𝑚 + 8 = 0 ⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚3 − 3𝑚𝑚2 + 2𝑚𝑚 + 5𝑚𝑚2 + 2𝑚𝑚 + 8 = 0 ⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚3 + 2𝑚𝑚2 + 4𝑚𝑚 + 8 = 0 9/30/2014

Dr. Eli Saber

155

Cauchy-Euler Equations Case 3: Conjugate Complex Roots ⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚3 + 2𝑚𝑚2 + 4𝑚𝑚 + 8 = 0 ⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚 + 2 𝑚𝑚2 + 4

Auxiliary Equation

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

=0

𝑚𝑚 + 2 𝑚𝑚2 + 4 = 0

⇒ 𝑚𝑚 + 2 𝑚𝑚 + 2𝑗𝑗 𝑚𝑚 − 2𝑗𝑗 = 0 ⇒ 𝑚𝑚1 = −2, 𝑚𝑚2 = −2𝑗𝑗, 𝑚𝑚3 = 2𝑗𝑗

Solution:

9/30/2014

𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙−𝟐𝟐 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟑𝟑 𝒙𝒙−𝟐𝟐𝟐𝟐 or 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙−𝟐𝟐 + 𝒄𝒄𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜(𝟐𝟐 𝐥𝐥𝐥𝐥 𝒙𝒙) + 𝒄𝒄𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬(𝟐𝟐 𝐥𝐥𝐥𝐥 𝒙𝒙) Dr. Eli Saber

156

Cauchy-Euler Equations Case 3: Conjugate Complex Roots E.g.

•

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 3𝑦𝑦 = 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥

Non Homogeneous Eqn. solve associated Homogeneous Eqn. 𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 3𝑦𝑦 = 0

Assume 𝑦𝑦 = 𝑥𝑥 𝑚𝑚 ⇒ 𝑦𝑦 ′ = 𝑚𝑚𝑥𝑥 𝑚𝑚−1 ⇒ 𝑦𝑦 ′′ = 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 ⇒ 𝑥𝑥 2 𝑚𝑚 𝑚𝑚 − 1 𝑥𝑥 𝑚𝑚−2 − 3𝑥𝑥 𝑚𝑚𝑥𝑥 𝑚𝑚−1 + 3𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑚𝑚2 − 𝑚𝑚 𝑥𝑥 𝑚𝑚 − 3𝑚𝑚𝑥𝑥 𝑚𝑚 + 3𝑥𝑥 𝑚𝑚 = 0 ⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚2 − 𝑚𝑚 − 3𝑚𝑚 + 3 = 0 ⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚2 − 4𝑚𝑚 + 3 = 0 9/30/2014

Dr. Eli Saber

157

Cauchy-Euler Equations Case 3: Conjugate Complex Roots ⇒ 𝑥𝑥 𝑚𝑚 𝑚𝑚2 − 4𝑚𝑚 + 3 = 0

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 3𝑦𝑦 = 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥

Auxiliary Equation

Auxiliary Eqn. 𝑚𝑚2 − 4𝑚𝑚 + 3 = 0 ⇒ 𝑚𝑚 − 1 𝑚𝑚 − 3 = 0 ⇒ 𝑚𝑚1 = 1 & 𝑚𝑚2 = 3 •

⇒ 𝒚𝒚𝒄𝒄 = 𝒄𝒄𝟏𝟏 𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝟑𝟑

Utilize Variation of Parameters to solve for particular solution 𝑦𝑦𝑝𝑝

,where 𝑦𝑦1 = 𝑥𝑥 & 𝑦𝑦2 = 𝑥𝑥 3

9/30/2014

𝑦𝑦𝑝𝑝 = 𝑢𝑢1 𝑦𝑦1 + 𝑢𝑢2 𝑦𝑦2

Dr. Eli Saber

158

Cauchy-Euler Equations Case 3: Conjugate Complex Roots Note: To use Variation of Parameters  must transform the equation

Divide by 𝑥𝑥 2 ,

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 3𝑦𝑦 = 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥

𝑦𝑦 ′′

3𝑥𝑥 ′ 3 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥 − 2 𝑦𝑦 + 2 𝑦𝑦 = 𝑥𝑥 2 𝑥𝑥 𝑥𝑥

⇒ 𝑦𝑦 ′′ −

3 ′ 3 𝑦𝑦 + 2 𝑦𝑦 = 2𝑥𝑥 2 𝑒𝑒 𝑥𝑥 𝑥𝑥 𝑥𝑥

𝑷𝑷(𝒙𝒙)

9/30/2014

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 3𝑦𝑦 = 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥 3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥 3

𝑸𝑸(𝒙𝒙)

Dr. Eli Saber

𝒇𝒇(𝒙𝒙)

159

Cauchy-Euler Equations Case 3: Conjugate Complex Roots Form all Wronskians: 𝑦𝑦1 𝑊𝑊 = 𝑦𝑦 ′ 1

𝑦𝑦2 𝑥𝑥 ′ = 𝑦𝑦2 1

0 𝑊𝑊1 = 𝑓𝑓(𝑥𝑥) 𝑥𝑥 1

𝑊𝑊2 =

𝑥𝑥 3 = 3𝑥𝑥 3 − 𝑥𝑥 3 = 2𝑥𝑥 3 3𝑥𝑥 2

𝑦𝑦2 0 ′ = 𝑦𝑦2 2𝑥𝑥 2 𝑒𝑒 𝑥𝑥

0 = 2𝑥𝑥 3 𝑒𝑒 𝑥𝑥 2 𝑥𝑥 2𝑥𝑥 𝑒𝑒

𝑥𝑥 3 = −2𝑥𝑥 5 𝑒𝑒 𝑥𝑥 3𝑥𝑥 2

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 3𝑦𝑦 = 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥 3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥 3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥 2 𝑒𝑒 𝑥𝑥

5 𝑒𝑒 𝑥𝑥 𝑊𝑊 2𝑥𝑥 1 ⇒ 𝑢𝑢1′ = =− = −𝑥𝑥 2 𝑒𝑒 𝑥𝑥 3 𝑊𝑊 2𝑥𝑥

⇒

𝑢𝑢2′

𝑊𝑊2 2𝑥𝑥 3 𝑒𝑒 𝑥𝑥 = = = 𝑒𝑒 𝑥𝑥 3 𝑊𝑊 2𝑥𝑥

9/30/2014

Dr. Eli Saber

160

Cauchy-Euler Equations Case 3: Conjugate Complex Roots Integrate 𝑢𝑢1′ & 𝑢𝑢2′ to get 𝑢𝑢1 &𝑢𝑢2 :

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 3𝑦𝑦 = 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥 3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥 3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥 2 𝑒𝑒 𝑥𝑥 𝑢𝑢1′ = −𝑥𝑥 2 𝑒𝑒 𝑥𝑥 𝑢𝑢2′ = 𝑒𝑒 𝑥𝑥

𝑢𝑢2 = � 𝑢𝑢2′ 𝑑𝑑𝑑𝑑 = � 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝒆𝒆𝒙𝒙

𝑢𝑢1 = � 𝑢𝑢1′ 𝑑𝑑𝑑𝑑 = − � 𝑥𝑥 2 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑

Integration by parts

Let 𝛼𝛼 = 𝑥𝑥 2 ⇒ 𝑑𝑑𝛼𝛼 = 2𝑥𝑥 𝑑𝑑𝑑𝑑 ; 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 ⇒ 𝛽𝛽 = 𝑒𝑒 𝑥𝑥 ⇒ 𝑢𝑢1 = − 𝑥𝑥 2 𝑒𝑒 𝑥𝑥 − � 𝑒𝑒 𝑥𝑥 2𝑥𝑥𝑥𝑥𝑥𝑥 ⇒ 𝑢𝑢1 = −𝑥𝑥 2 𝑒𝑒 𝑥𝑥 + 2 � 𝑒𝑒 𝑥𝑥 𝑥𝑥𝑥𝑥𝑥𝑥 9/30/2014

Dr. Eli Saber

161

Cauchy-Euler Equations Case 3: Conjugate Complex Roots ⇒ 𝑢𝑢1 =

−𝑥𝑥 2 𝑒𝑒 𝑥𝑥

+

𝑥𝑥 2 𝑦𝑦 ′′ − 3𝑥𝑥𝑦𝑦 ′ + 3𝑦𝑦 = 2𝑥𝑥 4 𝑒𝑒 𝑥𝑥 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑥𝑥 + 𝑐𝑐2 𝑥𝑥 3 𝑦𝑦1 = 𝑥𝑥 &𝑦𝑦2 = 𝑥𝑥 3 𝑓𝑓 𝑥𝑥 = 2𝑥𝑥 2 𝑒𝑒 𝑥𝑥 𝑢𝑢1′ = −𝑥𝑥 2 𝑒𝑒 𝑥𝑥 𝑢𝑢2 = 𝑒𝑒 𝑥𝑥

2 � 𝑒𝑒 𝑥𝑥 𝑥𝑥𝑥𝑥𝑥𝑥

⇒ 𝑢𝑢1 = −𝑥𝑥 2 𝑒𝑒 𝑥𝑥 + 2 𝑥𝑥𝑒𝑒 𝑥𝑥 − 𝑒𝑒 𝑥𝑥

⇒ 𝒖𝒖𝟏𝟏 = −𝒙𝒙𝟐𝟐 𝒆𝒆𝒙𝒙 + 𝟐𝟐𝟐𝟐𝒆𝒆𝒙𝒙 − 𝟐𝟐𝒆𝒆𝒙𝒙

Now, 𝑦𝑦𝑝𝑝 = 𝑢𝑢1 𝑦𝑦1 + 𝑢𝑢2 𝑦𝑦2 = −𝑥𝑥 2 𝑒𝑒 𝑥𝑥 + 2𝑥𝑥𝑒𝑒 𝑥𝑥 − 2𝑒𝑒 𝑥𝑥 𝑥𝑥 + 𝑒𝑒 𝑥𝑥 𝑥𝑥 3

= −𝑥𝑥 3 𝑒𝑒 𝑥𝑥 + 2𝑥𝑥 2 𝑒𝑒 𝑥𝑥 − 2𝑒𝑒 𝑥𝑥 𝑥𝑥 + 𝑒𝑒 𝑥𝑥 𝑥𝑥 3 = 𝟐𝟐𝒙𝒙𝟐𝟐 𝒆𝒆𝒙𝒙 − 𝟐𝟐𝟐𝟐𝒆𝒆𝒙𝒙

⇒ 𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

9/30/2014

⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 𝒙𝒙 + 𝒄𝒄𝟐𝟐 𝒙𝒙𝟑𝟑 + 𝟐𝟐𝒙𝒙𝟐𝟐 𝒆𝒆𝒙𝒙 − 𝟐𝟐𝟐𝟐𝒆𝒆𝒙𝒙 Dr. Eli Saber

162

Cauchy-Euler Equations Summary •

Identified when 𝒙𝒙𝒏𝒏 matches the order of the differentiation

– Case 2: Repeated Real Roots – 𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 𝑚𝑚1 + 𝑐𝑐2 𝑥𝑥 𝑚𝑚1 𝑙𝑙𝑙𝑙 𝑥𝑥 + 𝑐𝑐3 𝑥𝑥 𝑚𝑚1 𝑙𝑙𝑙𝑙 𝑥𝑥 2 + ⋯ + 𝑐𝑐𝑘𝑘 𝑥𝑥 𝑚𝑚1 𝑙𝑙𝑙𝑙 𝑥𝑥 𝑘𝑘−1

𝒅𝒅𝒏𝒏 𝒚𝒚 𝒅𝒅𝒙𝒙𝒏𝒏

𝑑𝑑𝑛𝑛 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑎𝑎𝑛𝑛 𝑥𝑥 + … + 𝑎𝑎 𝑥𝑥 + 𝑎𝑎0 𝑦𝑦 = 𝑔𝑔(𝑥𝑥) 1 𝑑𝑑𝑥𝑥 𝑛𝑛 𝑑𝑑𝑑𝑑 𝑛𝑛

Step1: Obtain Complementary Solution(𝑦𝑦𝑐𝑐 ) • Consider 𝒈𝒈 𝒙𝒙 = 𝟎𝟎 • •

•

Try the form 𝒚𝒚 = 𝒙𝒙𝒎𝒎

Auxiliary Equation: 𝒂𝒂𝒎𝒎𝟐𝟐 + 𝒃𝒃 − 𝒂𝒂 𝒎𝒎 + 𝒄𝒄 = 𝟎𝟎

Obtain roots for the equation – Case 1: Distinct Real Roots – 𝑦𝑦 = 𝑐𝑐1 𝑥𝑥 𝑚𝑚1 + 𝑐𝑐2 𝑥𝑥 𝑚𝑚2 + ⋯ + 𝑐𝑐𝑛𝑛 𝑥𝑥 𝑚𝑚𝑛𝑛

9/30/2014

– Case 3: Conjugate Complex Roots – 𝑦𝑦 = 𝑥𝑥 𝛼𝛼 ∝1 𝑐𝑐𝑐𝑐𝑐𝑐 𝛽𝛽 𝑙𝑙𝑙𝑙 𝑥𝑥 +∝2 𝑠𝑠𝑠𝑠𝑠𝑠 𝛽𝛽 𝑙𝑙𝑙𝑙 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 Dr. Eli Saber

163

Section 3.8 Linear Models

9/30/2014

Dr. Eli Saber

164

Linear Models 3.8.4. : Series Circuit (LRC) R

Note: 𝑉𝑉𝑅𝑅 = 𝑅𝑅𝑅𝑅

𝑉𝑉𝐿𝐿 = 𝐿𝐿

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑑𝑑𝑉𝑉𝑐𝑐 𝑖𝑖𝑐𝑐 = 𝑐𝑐 ; 𝑑𝑑𝑑𝑑

E

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑞𝑞 𝑖𝑖 = ⇒ = 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 2

Kirchoff’s Voltage Law:

9/30/2014

L

C

𝑑𝑑𝑑𝑑 = 𝑖𝑖 𝑑𝑑𝑑𝑑 ⇒ 𝑞𝑞 = ∫ 𝑖𝑖 𝑑𝑑𝑑𝑑

𝐸𝐸 = 𝑅𝑅𝑅𝑅 + 𝐿𝐿

𝑖𝑖(𝑡𝑡)

𝑑𝑑𝑑𝑑 1 + � 𝑖𝑖 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐶𝐶

𝑞𝑞 charge

Dr. Eli Saber

165

Linear Models 3.8.4. : Series Circuit (LRC) R

𝑑𝑑𝑑𝑑 1 𝐸𝐸 = 𝑅𝑅𝑅𝑅 + 𝐿𝐿 + � 𝑖𝑖 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐶𝐶

E

𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑞𝑞 𝑞𝑞 ⇒ 𝐸𝐸 = 𝑅𝑅 + 𝐿𝐿 2 + 𝑑𝑑𝑑𝑑 𝐶𝐶 𝑑𝑑𝑡𝑡

𝑖𝑖(𝑡𝑡)

L

C

𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑞𝑞 𝟏𝟏 ⇒ 𝑹𝑹 + 𝑳𝑳 2 + 𝑞𝑞 = 𝐸𝐸 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 𝑪𝑪 • • • •

𝑖𝑖 =

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑉𝑉𝐿𝐿 = 𝐿𝐿 1

; 𝑉𝑉𝑅𝑅 = 𝑅𝑅𝑅𝑅

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑉𝑉𝐶𝐶 = ∫ 𝑖𝑖 𝑑𝑑𝑑𝑑 𝐶𝐶

𝐸𝐸(𝑡𝑡): forcing function

9/30/2014

Dr. Eli Saber

166

Linear Models 3.8.4. : Series Circuit (LRC) 𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑞𝑞 𝑞𝑞 𝑅𝑅 + 𝐿𝐿 2 + = 𝐸𝐸 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 𝐶𝐶

R E

Rearranging the equation, we get: 𝑑𝑑 2 𝑞𝑞 𝑑𝑑𝑑𝑑 1 𝐿𝐿 2 + 𝑅𝑅 + 𝑞𝑞 = 𝐸𝐸 𝑑𝑑𝑑𝑑 𝐶𝐶 𝑑𝑑𝑡𝑡 1 𝐶𝐶

Auxiliary Eqn: 𝐿𝐿𝑚𝑚2 + 𝑅𝑅𝑅𝑅 + = 0 ⇒ 𝑚𝑚 =

−𝑅𝑅 ± 𝑅𝑅2 − 4 𝐿𝐿 2𝐿𝐿

4𝐿𝐿 −𝑅𝑅 ± 𝑅𝑅2 − 𝐶𝐶 ⇒ 𝑚𝑚 = 2𝐿𝐿 9/30/2014

1 𝐶𝐶

𝑖𝑖(𝑡𝑡)

L

C

(Assume 𝐸𝐸 𝑡𝑡 = 0)

Dr. Eli Saber

167

Linear Models 3.8.4. : Series Circuit (LRC) 𝑚𝑚 = •

•

•

−𝑅𝑅 ±

If 𝑹𝑹𝟐𝟐 − If 𝑹𝑹𝟐𝟐 − If 𝑹𝑹𝟐𝟐 −

Now,

𝑅𝑅2

2𝐿𝐿

𝟒𝟒𝟒𝟒 𝑪𝑪 𝟒𝟒𝟒𝟒 𝑪𝑪 𝟒𝟒𝟒𝟒 𝑪𝑪

R

4𝐿𝐿 − 𝐶𝐶

E

> 𝟎𝟎  over damped

L

C

= 𝟎𝟎 critically damped < 𝟎𝟎 under damped 𝜶𝜶

𝜷𝜷

4𝐿𝐿 2 − 4𝐿𝐿 −𝑅𝑅 ± 𝑅𝑅2 − 𝑅𝑅 𝑅𝑅 𝐶𝐶 𝐶𝐶 𝑚𝑚 = =− ± 2𝐿𝐿 2𝐿𝐿 2𝐿𝐿 9/30/2014

𝑖𝑖(𝑡𝑡)

Dr. Eli Saber

168

Linear Models 3.8.4. : Series Circuit (LRC) E.g. 𝐿𝐿 = 0.25𝐻𝐻; 𝑅𝑅 = 10Ω; 𝐶𝐶 = 0.001𝐹𝐹

E=0V

𝐸𝐸 𝑡𝑡 = 0𝑉𝑉; 𝑞𝑞 0 = 𝑞𝑞0 ; 𝑖𝑖 0 = 0𝐴𝐴

R=10 Ω 𝑖𝑖(𝑡𝑡)

L= 0.25𝐻𝐻

C= 0.001𝐹𝐹

Solution:

𝑑𝑑 2 𝑞𝑞 𝑑𝑑𝑑𝑑 1 𝐿𝐿 2 + 𝑅𝑅 + 𝑞𝑞 = 0 ⇒ 0.25𝑞𝑞 ′′ + 10𝑞𝑞 ′ + 1000𝑞𝑞 = 0 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑 𝐶𝐶 ⇒ 𝑞𝑞 ′′ + 40𝑞𝑞 ′ + 4000𝑞𝑞 = 0

Aux. Eq.: 𝑚𝑚2 + 40𝑚𝑚 + 4000 = 0 𝑚𝑚 =

−40 ± 1600 − 4(1)(4000) −40 ± 1600 − 16000 = 2 2

9/30/2014

Dr. Eli Saber

169

Linear Models 3.8.4. : Series Circuit (LRC) −40 ± −14400 𝑚𝑚 = 2

−40 ± 14400𝑗𝑗 2 −40 ± 16(900)𝑗𝑗 2 𝑚𝑚 = = 2 2 =

𝐿𝐿 = 0.25𝐻𝐻; 𝑅𝑅 = 10Ω; 𝐶𝐶 = 0.001𝐹𝐹 𝐸𝐸 𝑡𝑡 = 0𝑉𝑉; 𝑞𝑞 0 = 𝑞𝑞0 ; 𝑖𝑖 0 = 0𝐴𝐴 Aux. Eqn. 𝑚𝑚2 + 40𝑚𝑚 + 4000 = 0

−40 ± 4 30 𝑗𝑗 = −𝟐𝟐𝟐𝟐 ± 𝟔𝟔𝟔𝟔𝟔𝟔 2

𝑚𝑚1 = −20 + 60𝑗𝑗 & 𝑚𝑚2 = −20 − 60𝑗𝑗 Hence: 𝛼𝛼 = −20 & 𝛽𝛽 = 60

⇒ 𝑞𝑞 𝑡𝑡 = 𝑒𝑒 −20𝑡𝑡 𝑐𝑐1 cos 60𝑡𝑡 + 𝑐𝑐2 sin 60𝑡𝑡 9/30/2014

Dr. Eli Saber

170

Linear Models 3.8.4. : Series Circuit (LRC) •

𝑞𝑞 0 = 𝑞𝑞0

𝐿𝐿 = 0.25𝐻𝐻; 𝑅𝑅 = 10Ω; 𝐶𝐶 = 0.001𝐹𝐹 𝐸𝐸 𝑡𝑡 = 0𝑉𝑉; 𝑞𝑞 0 = 𝑞𝑞0 ; 𝑖𝑖 0 = 0𝐴𝐴 Aux. Eqn. 𝑚𝑚2 + 40𝑚𝑚 + 4000 = 0 𝑞𝑞 𝑡𝑡 = 𝑒𝑒 −20𝑡𝑡 𝑐𝑐1 cos 60𝑡𝑡 + 𝑐𝑐2 sin 60𝑡𝑡

⇒ 𝑞𝑞0 = 𝑒𝑒 0 𝑐𝑐1 cos 0 + 𝑐𝑐2 sin 0 ⇒ 𝑞𝑞0 = 1 𝑐𝑐1 + 0 ⇒ 𝑐𝑐1 = 𝑞𝑞0

Hence, we now have: 𝒒𝒒 𝒕𝒕 = 𝒆𝒆−𝟐𝟐𝟐𝟐𝟐𝟐 𝒒𝒒𝟎𝟎 𝒄𝒄𝒄𝒄𝒄𝒄 𝟔𝟔𝟔𝟔𝟔𝟔 + 𝒄𝒄𝟐𝟐 𝒔𝒔𝒔𝒔𝒔𝒔 𝟔𝟔𝟔𝟔𝟔𝟔

𝑑𝑑𝑑𝑑 −20𝑒𝑒 −20𝑡𝑡 [𝑞𝑞0 cos 60𝑡𝑡 + 𝑐𝑐2 sin 60𝑡𝑡] ⇒ 𝑖𝑖 𝑡𝑡 = = 𝑑𝑑𝑑𝑑 +𝑒𝑒 −20𝑡𝑡 [−60𝑞𝑞0 sin 60𝑡𝑡 + 60𝑐𝑐2 cos 60𝑡𝑡] 9/30/2014

Dr. Eli Saber

171

Linear Models 3.8.4. : Series Circuit (LRC) −20𝑒𝑒 −20𝑡𝑡 [𝑞𝑞0 cos 60𝑡𝑡 + 𝑐𝑐2 sin 60𝑡𝑡] 𝑖𝑖 𝑡𝑡 = +𝑒𝑒 −20𝑡𝑡 [−60𝑞𝑞0 sin 60𝑡𝑡 + 60𝑐𝑐2 cos 60𝑡𝑡] But 𝑖𝑖 0 = 0

𝐿𝐿 = 0.25𝐻𝐻; 𝑅𝑅 = 10Ω; 𝐶𝐶 = 0.001𝐹𝐹 𝐸𝐸 𝑡𝑡 = 0𝑉𝑉; 𝑞𝑞 0 = 𝑞𝑞0 ; 𝑖𝑖 0 = 0𝐴𝐴 Aux. Eqn. 𝑚𝑚2 + 40𝑚𝑚 + 4000 = 0 𝑞𝑞 𝑡𝑡 = 𝑒𝑒 −20𝑡𝑡 𝑞𝑞0 𝑐𝑐𝑐𝑐𝑐𝑐 60𝑡𝑡 + 𝑐𝑐2 𝑠𝑠𝑠𝑠𝑠𝑠 60𝑡𝑡

⇒ 0 = −20 𝑞𝑞0 + 0 + 1 0 + 60 𝑐𝑐2 ⇒ 0 = −20 𝑞𝑞0 + 60𝑐𝑐2 ⇒ 60𝑐𝑐2 = 20𝑞𝑞0 20 1 ⇒ 𝑐𝑐2 = 𝑞𝑞 ⇒ 𝑐𝑐2 = 𝑞𝑞0 60 0 3 ⇒ 𝑞𝑞 𝑡𝑡 =

𝑒𝑒 −20𝑡𝑡

𝑞𝑞0 𝑞𝑞0 𝑐𝑐𝑐𝑐𝑐𝑐 60𝑡𝑡 + 𝑠𝑠𝑠𝑠𝑠𝑠 60𝑡𝑡 3

1 ⇒ 𝑞𝑞 𝑡𝑡 = 𝑞𝑞0 𝑒𝑒 −20𝑡𝑡 𝑐𝑐𝑐𝑐𝑐𝑐 60𝑡𝑡 + 𝑠𝑠𝑠𝑠𝑠𝑠 60𝑡𝑡 3 9/30/2014

Dr. Eli Saber

172

Linear Models 3.8.4. : Series Circuit (LRC) 1 𝑞𝑞 𝑡𝑡 = 𝑞𝑞0 𝑒𝑒 −20𝑡𝑡 𝑐𝑐𝑐𝑐𝑐𝑐 60𝑡𝑡 + 𝑠𝑠𝑠𝑠𝑠𝑠 60𝑡𝑡 3

We know, sin 𝐴𝐴 + 𝐵𝐵 = sin 𝐴𝐴 cos 𝐵𝐵 + cos 𝐴𝐴 sin 𝐵𝐵 We can transform 𝑞𝑞(𝑡𝑡) into an alternate form: 𝑞𝑞 𝑡𝑡 = 𝑞𝑞0

𝑒𝑒 −20𝑡𝑡

1 (1) 𝑐𝑐𝑐𝑐𝑐𝑐 60𝑡𝑡 + 𝑠𝑠𝑠𝑠𝑠𝑠 60𝑡𝑡 3

⇒ 𝑞𝑞 𝑡𝑡 = 𝑞𝑞0 𝑒𝑒 −20𝑡𝑡

9/30/2014

10 3

1 𝑐𝑐𝑐𝑐𝑐𝑐 60𝑡𝑡 + 3 𝑠𝑠𝑠𝑠𝑠𝑠 60𝑡𝑡 10 10 3 3

1

𝐬𝐬𝐬𝐬𝐬𝐬 𝝓𝝓

1

2

1 + 3

2

=

𝟏𝟏𝟏𝟏 𝟑𝟑

𝝓𝝓

1

1 3

1 sin 𝜙𝜙 = ; cos 𝜙𝜙 = 3 10 10 3 3 1

𝐜𝐜𝐜𝐜𝐜𝐜 𝝓𝝓

Dr. Eli Saber

173

Linear Models 3.8.4. : Series Circuit (LRC) ⇒ 𝑞𝑞 𝑡𝑡 = 𝑞𝑞0 Note: sin 𝜙𝜙 = ⇒ 𝑞𝑞 𝑡𝑡 = 𝑞𝑞0

9/30/2014

10 −20𝑡𝑡 𝑒𝑒 sin[60𝑡𝑡 + 𝜙𝜙] 3 3 10

⇒ 𝜙𝜙 = sin−1

3 10

= 1.249 rad

10 −20𝑡𝑡 𝑒𝑒 sin[60𝑡𝑡 + 1.249] 3

Dr. Eli Saber

1

2

1 + 3

2

=

𝟏𝟏𝟏𝟏 𝟑𝟑

𝝓𝝓

1

1 3

1 sin 𝜙𝜙 = ; cos 𝜙𝜙 = 3 10 10 3 3 1

174

Linear Models 3.8.4. : Series Circuit (LRC) Note: • 𝑞𝑞𝑐𝑐 (𝑡𝑡): solution to the homogeneous equation is called the transient solution •

𝑞𝑞𝑝𝑝 𝑡𝑡 : solution to the non-homogeneous equation (i.e. 𝐸𝐸(𝑡𝑡) ≠ 0) is called the steady-state solution

9/30/2014

Dr. Eli Saber

175

Linear Models 3.8.4. : Series Circuit (LRC) E.g. 𝐿𝐿 = 1𝐻𝐻; 𝑅𝑅 = 2Ω; 𝐶𝐶 = 0.25𝐹𝐹; 𝐸𝐸 𝑡𝑡 = 50 cos 𝑡𝑡 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣

E=50 cos (t) V

Find the steady-state charge and the steady-state current in the LRC Circuit (Advanced Eng. Mathematics –

5th

Edition – Ex. 3.8 Prob. 49)

R=2 Ω 𝑖𝑖(𝑡𝑡)

L= 1𝐻𝐻

C= 0.25𝐹𝐹

Solution: 𝑑𝑑𝑑𝑑 𝑞𝑞 𝐸𝐸 𝑡𝑡 = 𝐿𝐿 + 𝑅𝑅𝑅𝑅 + 𝑑𝑑𝑑𝑑 𝐶𝐶

𝑑𝑑 2 𝑞𝑞 𝑑𝑑𝑑𝑑 𝑞𝑞 ⇒ 𝐿𝐿 2 + 𝑅𝑅 + = 𝐸𝐸 𝑡𝑡 𝑑𝑑𝑑𝑑 𝐶𝐶 𝑑𝑑𝑡𝑡

𝑑𝑑 2 𝑞𝑞 𝑑𝑑𝑑𝑑 𝑞𝑞 ⇒1 2 +2 + = 50 cos 𝑡𝑡 𝑑𝑑𝑑𝑑 0.25 𝑑𝑑𝑡𝑡 9/30/2014

Dr. Eli Saber

176

Linear Models 3.8.4. : Series Circuit (LRC) 𝑑𝑑 2 𝑞𝑞 𝑑𝑑𝑑𝑑 ⇒1 2 +2 + 4 𝑞𝑞 = 50 cos 𝑡𝑡 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑 Homogeneous Eqn.

𝑑𝑑2 𝑞𝑞 𝑑𝑑𝑡𝑡 2

⇒ 𝑚𝑚2 + 2𝑚𝑚 + 4 = 0

+2

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

+ 4 𝑞𝑞 = 0

−2 ± 4 − 4(1)(4) −2 ± 12 −2 ± 4 3 𝑗𝑗 2 ⇒ 𝑚𝑚 = = = 2 2 2 ⇒ 𝑚𝑚 = −1 ± 𝑗𝑗 3 ⇒ 𝛼𝛼 = −1 & 𝛽𝛽 = 3 𝑞𝑞𝑐𝑐 𝑡𝑡 = 𝑐𝑐1 𝑒𝑒

9/30/2014

𝑞𝑞𝑐𝑐 𝑡𝑡 = 𝑒𝑒 −𝑡𝑡

−1+𝑗𝑗 3 𝑡𝑡

+ 𝑐𝑐2 𝑒𝑒

−1−𝑗𝑗 3 𝑡𝑡

or ∝1 cos 3𝑡𝑡 +∝2 sin 3𝑡𝑡 Dr. Eli Saber

177

Linear Models 3.8.4. : Series Circuit (LRC) From table 3.4.1., we can write: 𝑞𝑞𝑝𝑝 𝑡𝑡 = 𝐴𝐴 cos 𝑡𝑡 + 𝐵𝐵 sin 𝑡𝑡

𝑞𝑞 ′ 𝑡𝑡 = −𝐴𝐴 sin 𝑡𝑡 + 𝐵𝐵 cos 𝑡𝑡

𝑑𝑑 2 𝑞𝑞 𝑑𝑑𝑑𝑑 1 2 +2 + 4 𝑞𝑞 = 50 cos 𝑡𝑡 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡

𝑞𝑞 ′′ 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 = 9/30/2014

3 𝐵𝐵 2

Dr. Eli Saber

178

Linear Models 3.8.4. : Series Circuit (LRC) 𝑑𝑑 2 𝑞𝑞 𝑑𝑑𝑑𝑑 1 2 +2 + 4 𝑞𝑞 = 50 cos 𝑡𝑡 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 3𝐴𝐴 + 2𝐵𝐵 = 50 3 9 13 100 3 3 𝐵𝐵 + 2𝐵𝐵 = 50 ⇒ 𝐵𝐵 + 2𝐵𝐵 = 50 ⇒ 𝐵𝐵 = 50 ⇒ 𝐵𝐵 = 𝐴𝐴 = 𝐵𝐵 2 2 2 13 2

3𝐴𝐴 + 2𝐵𝐵 = 50

𝐴𝐴 =

3 3 100 300 150 𝐵𝐵 = ∗ = ⇒ 𝐴𝐴 = 2 2 13 26 13

𝑞𝑞𝑝𝑝 𝑡𝑡 =

150 100 cos 𝑡𝑡 + sin 𝑡𝑡 13 13

We already have: Hence,

𝑞𝑞𝑐𝑐 𝑡𝑡 = 𝑒𝑒 −𝑡𝑡 ∝1 cos 3𝑡𝑡 +∝2 sin 3𝑡𝑡

𝑞𝑞 𝑡𝑡 = 𝑒𝑒 −𝑡𝑡 𝑐𝑐1 cos 3𝑡𝑡 + 𝑐𝑐2 sin 3𝑡𝑡 +

Transient Solution

9/30/2014

Dr. Eli Saber

150 100 cos 𝑡𝑡 + sin 𝑡𝑡 13 13

Steady-State Solution 179

Linear Models 3.8.4. : Series Circuit (LRC) 𝑞𝑞𝑠𝑠𝑠𝑠 𝑡𝑡 = 𝑞𝑞𝑝𝑝 𝑡𝑡 ⇒ 𝑞𝑞𝑠𝑠𝑠𝑠 𝑖𝑖 𝑡𝑡 =

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

150 100 𝑡𝑡 = cos 𝑡𝑡 + sin 𝑡𝑡 13 13

𝑞𝑞 𝑡𝑡 = 𝑒𝑒 −𝑡𝑡 𝑐𝑐1 cos 3𝑡𝑡 + 𝑐𝑐2 sin 3𝑡𝑡 +

= −𝑒𝑒 −𝑡𝑡 𝑐𝑐1 cos 3𝑡𝑡 + 𝑐𝑐2 sin 3𝑡𝑡 + 𝑒𝑒 −𝑡𝑡 −𝑐𝑐1 3 sin 3𝑡𝑡 + 𝑐𝑐2 3 cos 3𝑡𝑡 −

150 100 cos 𝑡𝑡 + sin 𝑡𝑡 13 13

150 100 sin 𝑡𝑡 + cos 𝑡𝑡 13 13

150 ⇒ 𝑖𝑖 𝑡𝑡 = −𝑐𝑐1 𝑒𝑒 −𝑡𝑡 cos 3𝑡𝑡 − 𝑐𝑐2 𝑒𝑒 −𝑡𝑡 sin 3𝑡𝑡 − 𝑐𝑐1 𝑒𝑒 −𝑡𝑡 3 sin 3𝑡𝑡 + 𝑐𝑐2 𝑒𝑒 −𝑡𝑡 3 cos 3𝑡𝑡 − sin 𝑡𝑡 13 100 + cos 𝑡𝑡 13 ⇒ 𝑖𝑖 𝑡𝑡 = 𝑒𝑒 −𝑡𝑡 −𝑐𝑐1 + 3𝑐𝑐2 cos 3𝑡𝑡 + 𝑒𝑒 −𝑡𝑡 −𝑐𝑐2 − 3𝑐𝑐1 sin 3𝑡𝑡 − ⇒ 𝑖𝑖𝑠𝑠𝑠𝑠 𝑡𝑡 = −

9/30/2014

150 sin 𝑡𝑡 13

+

100 cos 𝑡𝑡 13

Dr. Eli Saber

150 100 sin 𝑡𝑡 + cos 𝑡𝑡 13 13

180

Linear Models Summary R

𝑑𝑑𝑑𝑑 1 𝐿𝐿 + 𝑅𝑅𝑅𝑅 + � 𝑖𝑖 𝑑𝑑𝑑𝑑 = 𝐸𝐸(𝑡𝑡) 𝑑𝑑𝑑𝑑 𝐶𝐶 𝑑𝑑2 𝑞𝑞 𝑑𝑑𝑑𝑑 𝑞𝑞 ⇒ 𝐿𝐿 2 + 𝑅𝑅 + = 𝐸𝐸 𝑡𝑡 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑 𝐶𝐶

E

𝑖𝑖(𝑡𝑡)

L

C 1 𝐶𝐶

Auxiliary Eqn: 𝐿𝐿𝑚𝑚2 + 𝑅𝑅𝑅𝑅 + = 0

𝜶𝜶

𝜷𝜷

4𝐿𝐿 2 − 4𝐿𝐿 𝑅𝑅 −𝑅𝑅 ± 𝑅𝑅 2 − 𝑅𝑅 𝐶𝐶 𝐶𝐶 𝑚𝑚 = =− ± 2𝐿𝐿 2𝐿𝐿 2𝐿𝐿

⇒ obtain 𝑦𝑦𝑐𝑐

Use already known methods to obtain 𝑦𝑦𝑝𝑝

𝒚𝒚 = 𝒚𝒚𝒄𝒄 + 𝒚𝒚𝒑𝒑

9/30/2014

Dr. Eli Saber

181

Section 3.12 Solving Systems of Linear Equations

9/30/2014

Dr. Eli Saber

182

Solving Systems of Linear Equations Coupled Spring/Mass Systems Newton’s 2nd Law: 𝑑𝑑 2 𝑥𝑥1 𝑚𝑚1 = −𝑘𝑘1 𝑥𝑥1 + 𝑘𝑘2 𝑥𝑥2 − 𝑥𝑥1 𝑑𝑑𝑡𝑡 2 𝑑𝑑 2 𝑥𝑥2 𝑚𝑚2 = −𝑘𝑘2 𝑥𝑥2 − 𝑥𝑥1 𝑑𝑑𝑡𝑡 2 Also can be written as: 𝑚𝑚1 𝑥𝑥1′′ = −𝑘𝑘1 𝑥𝑥1 + 𝑘𝑘2 𝑥𝑥2 − 𝑥𝑥1 𝑚𝑚2 𝑥𝑥2′′

= −𝑘𝑘2 𝑥𝑥2 − 𝑥𝑥1

9/30/2014

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

A coupled system of Differential Equations

Dr. Eli Saber

183

Solving Systems of Linear Equations Systematic Elimination Given

𝑎𝑎𝑛𝑛 𝑦𝑦

𝑛𝑛

+ 𝑎𝑎𝑛𝑛−1 𝑦𝑦

𝑛𝑛−1

,where 𝑎𝑎𝑖𝑖 , 𝑖𝑖 = 0,1,2,3, … , 𝑛𝑛 are constants

+ ⋯ + 𝑎𝑎1 𝑦𝑦 ′ + 𝑎𝑎0 𝑦𝑦 = 𝑔𝑔 𝑥𝑥

Rewrite as: 𝑎𝑎𝑛𝑛 𝐷𝐷𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝐷𝐷𝑛𝑛−1 + ⋯ + 𝑎𝑎1 𝐷𝐷 + 𝑎𝑎0 𝑦𝑦 = 𝑔𝑔 𝑥𝑥

Then group like terms for solving.

9/30/2014

Dr. Eli Saber

184

Solving Systems of Linear Equations 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

⇒ 𝐷𝐷2 𝑥𝑥 + 2𝐷𝐷𝐷𝐷 + 𝐷𝐷2 𝑦𝑦 − 𝑥𝑥 − 3𝑦𝑦 = sin 𝑡𝑡 ⇒ 𝐷𝐷𝐷𝐷 + 𝐷𝐷𝐷𝐷 + 4𝑥𝑥 − 2𝑦𝑦 = 𝑒𝑒 −𝑡𝑡

⇒ 𝑫𝑫𝟐𝟐 + 𝟐𝟐𝟐𝟐 − 𝟏𝟏 𝑥𝑥 + 𝑫𝑫𝟐𝟐 − 𝟑𝟑 𝑦𝑦 = sin 𝑡𝑡 ⇒ 𝑫𝑫 + 𝟒𝟒 𝑥𝑥 + 𝑫𝑫 − 𝟐𝟐 𝑦𝑦 = 𝑒𝑒 −𝑡𝑡

9/30/2014

Dr. Eli Saber

185

Solving Systems of Linear Equations Solution of System A solution of a system of D.E. is a set of sufficiently differentiable functions 𝑥𝑥 = ∅1 𝑡𝑡 𝑦𝑦 = ∅2 𝑡𝑡 𝑧𝑧 = ∅3 𝑡𝑡 ⋮ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜

that satisfies each equation in the system on some common interval 𝐼𝐼

9/30/2014

Dr. Eli Saber

186

Solving Systems of Linear Equations E.g. Linear 1st order equations:

Solution: 𝑑𝑑𝑑𝑑 = 3𝑦𝑦 ⇒ 𝐷𝐷𝐷𝐷 − 3𝑦𝑦 = 0 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = 3𝑦𝑦 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 2𝑥𝑥 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = 2𝑥𝑥 ⇒ 𝐷𝐷𝐷𝐷 − 2𝑥𝑥 = 0 𝑑𝑑𝑑𝑑

𝐷𝐷𝐷𝐷 − 3𝑦𝑦 = 0 → 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝐷𝐷 → 𝐷𝐷2 𝑥𝑥 − 3𝐷𝐷𝐷𝐷 = 0 𝐷𝐷𝐷𝐷 − 2𝑥𝑥 = 0 → 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏 3 → +3𝐷𝐷𝐷𝐷 − 6𝑥𝑥 = 0 9/30/2014

𝑫𝑫𝟐𝟐 𝒙𝒙 − 𝟔𝟔𝟔𝟔 = 𝟎𝟎 Dr. Eli Saber

187

Solving Systems of Linear Equations ⇒ 𝐷𝐷2 𝑥𝑥 − 6𝑥𝑥 = 0

Auxiliary Equation: 𝑚𝑚2 − 6 = 0 ⇒ 𝑚𝑚2 = 6 ⇒ 𝑚𝑚 = ± 6 Now, to obtain 𝑦𝑦(𝑡𝑡):

𝒙𝒙 𝒕𝒕 = 𝒄𝒄𝟏𝟏 𝒆𝒆−

𝟔𝟔𝒕𝒕

+ 𝒄𝒄𝟐𝟐 𝒆𝒆

𝟔𝟔𝒕𝒕

𝐷𝐷𝐷𝐷 − 3𝑦𝑦 = 0 → 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏 2 → 2𝐷𝐷𝐷𝐷 − 6𝑦𝑦 = 0

𝐷𝐷𝐷𝐷 − 2𝑥𝑥 = 0 → 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝐷𝐷 → 𝐷𝐷2 𝑦𝑦 − 2𝐷𝐷𝐷𝐷 = 0

𝑫𝑫𝟐𝟐 𝒚𝒚 − 𝟔𝟔𝒚𝒚 = 𝟎𝟎

Auxiliary Equation: 𝑚𝑚2 − 6 = 0 ⇒ 𝑚𝑚2 = 6 ⇒ 𝑚𝑚 = ± 6 9/30/2014

𝒚𝒚 𝒕𝒕 = 𝒄𝒄𝟑𝟑 𝒆𝒆−

𝟔𝟔𝒕𝒕

+ 𝒄𝒄𝟒𝟒 𝒆𝒆

Dr. Eli Saber

𝟔𝟔𝒕𝒕 188

Solving Systems of Linear Equations 𝑥𝑥𝑥 𝑡𝑡 = − 6𝑐𝑐1 𝑒𝑒 − We know:

6𝑡𝑡

𝑑𝑑𝑑𝑑 = 3𝑦𝑦 𝑑𝑑𝑑𝑑

⇒ − 6𝑐𝑐1 𝑒𝑒 −

6𝑡𝑡

+ 6𝑐𝑐2 𝑒𝑒

⇒ − 6𝑐𝑐1 − 3𝑐𝑐3 𝑒𝑒 −

6𝑡𝑡

+ 6𝑐𝑐2 𝑒𝑒

6𝑡𝑡

6𝑡𝑡

+

= 3𝑐𝑐3 𝑒𝑒 −

9/30/2014

𝒙𝒙 𝒕𝒕 = 𝒄𝒄𝟏𝟏 𝒆𝒆−

6𝑡𝑡

+ 3𝑐𝑐4 𝑒𝑒

6𝑐𝑐2 − 3𝑐𝑐4 𝑒𝑒

⇒ − 6𝑐𝑐1 − 3𝑐𝑐3 = 0 ⇒ 𝑐𝑐3 = − ⇒ 6𝑐𝑐2 − 3𝑐𝑐4 = 0 ⇒ 𝑐𝑐4 =

𝑥𝑥 𝑡𝑡 = 𝑐𝑐1 𝑒𝑒 − 𝑦𝑦 𝑡𝑡 = 𝑐𝑐3 𝑒𝑒 −

6 𝑐𝑐 3 1

6 c 3 2 𝟔𝟔𝒕𝒕

+ 𝒄𝒄𝟐𝟐 𝒆𝒆

𝟔𝟔𝒕𝒕

6𝑡𝑡

6𝑡𝑡

+ 𝑐𝑐2 𝑒𝑒 6𝑡𝑡 + 𝑐𝑐 𝑒𝑒 4

6𝑡𝑡

6𝑡𝑡

6𝑡𝑡

=0

∀𝑡𝑡

& 𝒚𝒚 𝒕𝒕 = − Dr. Eli Saber

𝟔𝟔 𝒄𝒄𝟏𝟏 𝒆𝒆− 𝟑𝟑

𝟔𝟔𝒕𝒕

+

𝟔𝟔 𝒄𝒄 𝒆𝒆 𝟑𝟑 𝟐𝟐

𝟔𝟔𝒕𝒕

189

Solving Systems of Linear Equations E.g.

Solution:

𝑥𝑥 ′ − 4𝑥𝑥 + 𝑦𝑦 ′′ = 𝑡𝑡 2 𝑥𝑥 ′ + 𝑥𝑥 + 𝑦𝑦 ′ = 0

𝐷𝐷𝐷𝐷 − 4𝑥𝑥 + 𝐷𝐷2 𝑦𝑦 = 𝑡𝑡 2 ⇒ 𝐷𝐷 − 4 𝑥𝑥 + 𝐷𝐷2 𝑦𝑦 = 𝑡𝑡 2 𝐷𝐷𝐷𝐷 + 𝑥𝑥 + 𝐷𝐷𝐷𝐷 = 0 ⇒ 𝐷𝐷 + 1 𝑥𝑥 + 𝐷𝐷𝐷𝐷 = 0 Solving for 𝑦𝑦 first:

(𝟏𝟏)

(𝟐𝟐)

1 ∗ 𝐷𝐷 + 1 ⇒ 𝐷𝐷 − 4 𝐷𝐷 + 1 𝑥𝑥 + 𝐷𝐷2 𝐷𝐷 + 1 𝑦𝑦 = 𝐷𝐷 + 1 𝑡𝑡 2

2 ∗ 𝐷𝐷 − 4 ⇒ 𝐷𝐷 + 1 𝐷𝐷 − 4 𝑥𝑥 + 𝐷𝐷 𝐷𝐷 − 4 𝑦𝑦 = 0 (−) (−) (−) 𝑫𝑫𝟐𝟐 𝑫𝑫 + 𝟏𝟏 𝒚𝒚 − 𝑫𝑫 𝑫𝑫 − 𝟒𝟒 𝒚𝒚 = 𝑫𝑫 + 𝟏𝟏 𝒕𝒕𝟐𝟐

9/30/2014

Dr. Eli Saber

190

Solving Systems of Linear Equations ⇒ 𝐷𝐷2 𝐷𝐷 + 1 𝑦𝑦 − 𝐷𝐷 𝐷𝐷 − 4 𝑦𝑦 = 𝐷𝐷 + 1 𝑡𝑡 2 ⇒ 𝐷𝐷3 + 𝐷𝐷2 − 𝐷𝐷2 + 4𝐷𝐷 𝑦𝑦 = 𝐷𝐷𝑡𝑡 2 + 𝑡𝑡 2 ⇒ 𝐷𝐷3 + 4𝐷𝐷 𝑦𝑦 = 2𝑡𝑡 + 𝑡𝑡 2 𝑑𝑑 3 𝑦𝑦 𝑑𝑑𝑑𝑑 → 3 +4 = 2𝑡𝑡 + 𝑡𝑡 2 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑

Here, 𝐷𝐷𝑡𝑡 2 =

𝑑𝑑 𝑑𝑑𝑑𝑑

𝑡𝑡 2 = 2𝑡𝑡

⇒ 𝐷𝐷3 + 4𝐷𝐷 𝑦𝑦 = 2𝑡𝑡 + 𝑡𝑡 2

Aux. Equation: 𝑚𝑚3 + 4𝑚𝑚 = 0 ⇒ 𝑚𝑚 𝑚𝑚2 + 4 = 0 ⇒ 𝑚𝑚1 = 0; 𝑚𝑚2 = 2𝑗𝑗; 𝑚𝑚3 = −2𝑗𝑗 𝑦𝑦𝑐𝑐 = 𝑐𝑐1 𝑒𝑒 0𝑡𝑡 + 𝑐𝑐2 cos 2𝑡𝑡 + 𝑐𝑐3 sin 2𝑡𝑡

9/30/2014

𝑦𝑦𝑐𝑐 = 𝑐𝑐1 + 𝑐𝑐2 𝑐𝑐𝑐𝑐𝑐𝑐 2𝑡𝑡 + 𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑡𝑡 Dr. Eli Saber

191

Solving Systems of Linear Equations Particular Solution 𝒚𝒚: use undetermined coefficient ⇒ 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑡𝑡 3 + 𝐵𝐵𝑡𝑡 2 + 𝐶𝐶𝐶𝐶

⇒ 𝑦𝑦𝑝𝑝′ = 3𝐴𝐴𝑡𝑡 2 + 2𝐵𝐵𝐵𝐵 + 𝐶𝐶; 𝑦𝑦𝑝𝑝′′ = 6𝐴𝐴𝐴𝐴 + 2𝐵𝐵; 𝑦𝑦𝑝𝑝′′′ = 6𝐴𝐴

𝑑𝑑3 𝑦𝑦 𝑑𝑑𝑑𝑑 + 4 = 𝑡𝑡 2 + 2𝑡𝑡 3 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑

𝑦𝑦𝑐𝑐 = 𝑐𝑐1 + 𝑐𝑐2 𝑐𝑐𝑐𝑐𝑐𝑐 2𝑡𝑡 + 𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑡𝑡

⇒ 6𝐴𝐴 + 4 3𝐴𝐴𝑡𝑡 2 + 2𝐵𝐵𝐵𝐵 + 𝐶𝐶 = 𝑡𝑡 2 + 2𝑡𝑡 ⇒ 6𝐴𝐴 + 12𝐴𝐴𝑡𝑡 2 + 8𝐵𝐵𝐵𝐵 + 4𝐶𝐶 = 𝑡𝑡 2 + 2𝑡𝑡 ⇒

12𝐴𝐴𝑡𝑡 2

+ 8𝐵𝐵𝐵𝐵 + 6𝐴𝐴 + 4𝐶𝐶 =

⇒ 12𝐴𝐴 = 1 → 𝐴𝐴 = 9/30/2014

1 12

𝑡𝑡 2

Note: 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑡𝑡 3 + 𝐵𝐵𝑡𝑡 2 + 𝐶𝐶𝐶𝐶

+ 2𝑡𝑡

Here, +𝐷𝐷 is not considered since 𝑦𝑦𝑐𝑐 already has a constant term Dr. Eli Saber

192

Solving Systems of Linear Equations 1 ⇒ 8𝐵𝐵 = 2 → 𝐵𝐵 = 4

1 ⇒ 6𝐴𝐴 + 4𝐶𝐶 = 0 ⇒ 4𝐶𝐶 = −6𝐴𝐴 = −6 12 1 ⇒ 𝐶𝐶 = − 8 Hence,

𝑦𝑦 = 𝑦𝑦𝑐𝑐 + 𝑦𝑦𝑝𝑝

9/30/2014

𝑑𝑑3 𝑦𝑦 𝑑𝑑𝑑𝑑 + 4 = 𝑡𝑡 2 + 2𝑡𝑡 3 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑

𝑦𝑦𝑐𝑐 = 𝑐𝑐1 + 𝑐𝑐2 𝑐𝑐𝑐𝑐𝑐𝑐 2𝑡𝑡 + 𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑡𝑡 𝑦𝑦𝑝𝑝 = 𝐴𝐴𝑡𝑡 3 + 𝐵𝐵𝑡𝑡 2 + 𝐶𝐶𝐶𝐶

1 3 1 2 1 𝑦𝑦𝑝𝑝 = 𝑡𝑡 + 𝑡𝑡 − 𝑡𝑡 12 4 8 𝟏𝟏 𝟑𝟑 𝟏𝟏 𝟐𝟐 𝟏𝟏 ⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 + 𝒄𝒄𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟐𝟐𝟐𝟐 + 𝒕𝒕 + 𝒕𝒕 − 𝒕𝒕 𝟏𝟏𝟏𝟏 𝟒𝟒 𝟖𝟖 Dr. Eli Saber

𝐴𝐴 =

1 12

193

Solving Systems of Linear Equations We have: (1) ≡ 𝐷𝐷𝐷𝐷 − 4𝑥𝑥 + 𝐷𝐷2 𝑦𝑦 = 𝑡𝑡 2 ⇒ 𝐷𝐷 − 4 𝑥𝑥 + 𝐷𝐷2 𝑦𝑦 = 𝑡𝑡 2 2 ≡ 𝐷𝐷𝐷𝐷 + 𝑥𝑥 + 𝐷𝐷𝐷𝐷 = 0 ⇒ 𝐷𝐷 + 1 𝑥𝑥 + 𝐷𝐷𝐷𝐷 = 0

Solving for 𝑥𝑥 now: 1

⇒ 𝐷𝐷 − 4 𝑥𝑥 + 𝐷𝐷2 𝑦𝑦 = 𝑡𝑡 2

2 ∗ 𝐷𝐷 ⇒ 𝐷𝐷 𝐷𝐷 + 1 𝑥𝑥 + 𝐷𝐷2 𝑦𝑦 = 0 (−) (−) (−) 𝑫𝑫 − 𝟒𝟒 − 𝑫𝑫 𝑫𝑫 + 𝟏𝟏 𝒙𝒙 = 𝒕𝒕𝟐𝟐 9/30/2014

Dr. Eli Saber

194

Solving Systems of Linear Equations ⇒

𝐷𝐷 − 4 − 𝐷𝐷 𝐷𝐷 + 1 𝑥𝑥 = 𝑡𝑡 2

⇒ 𝐷𝐷 − 4 − 𝐷𝐷2 − 𝐷𝐷 𝑥𝑥 = 𝑡𝑡 2 ⇒ − 4 + 𝐷𝐷2

= 𝑡𝑡 2

⇒ 𝐷𝐷2 + 4 𝑥𝑥 = −𝑡𝑡 2

Aux. Equation: 𝑚𝑚2 + 4 = 0 ⇒ 𝑚𝑚1 = 2𝑗𝑗; 𝑚𝑚2 = −2𝑗𝑗

𝑥𝑥𝑐𝑐 = 𝑐𝑐4 cos 2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡

9/30/2014

Dr. Eli Saber

195

Solving Systems of Linear Equations Particular Solution 𝒙𝒙: use undetermined coefficient 2

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑥𝑥𝑝𝑝 = 𝐴𝐴𝑡𝑡 + 𝐵𝐵𝐵𝐵 + 𝐶𝐶 (Table 3.4.1)

𝐷𝐷2 + 4 𝑥𝑥 = −𝑡𝑡 2

𝑥𝑥𝑐𝑐 = 𝑐𝑐4 cos 2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡

⇒ 𝑥𝑥𝑝𝑝′ = 2𝐴𝐴𝐴𝐴 + 𝐵𝐵; 𝑥𝑥𝑝𝑝′′ = 2𝐴𝐴 𝐷𝐷2 𝑥𝑥 + 4𝑥𝑥 = −𝑡𝑡 2

𝑑𝑑 2 𝑥𝑥𝑝𝑝 → + 4𝑥𝑥𝑝𝑝 = −𝑡𝑡 2 2 𝑑𝑑𝑡𝑡

⇒ 2𝐴𝐴 + 4 𝐴𝐴𝑡𝑡 2 + 𝐵𝐵𝐵𝐵 + 𝐶𝐶 = −𝑡𝑡 2

⇒ 2𝐴𝐴 + 4𝐴𝐴𝑡𝑡 2 + 4𝐵𝐵𝐵𝐵 + 4𝐶𝐶 = −𝑡𝑡 2 4𝐴𝐴𝑡𝑡 2 + 4𝐵𝐵𝐵𝐵 + 2𝐴𝐴 + 4𝐶𝐶 = −𝑡𝑡 2 9/30/2014

Dr. Eli Saber

196

Solving Systems of Linear Equations 𝐷𝐷2 + 4 𝑥𝑥 = −𝑡𝑡 2

4𝐴𝐴𝑡𝑡 2 + 4𝐵𝐵𝐵𝐵 + 2𝐴𝐴 + 4𝐶𝐶 = −𝑡𝑡 2

𝑥𝑥𝑐𝑐 = 𝑐𝑐4 cos 2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡

1 4𝐴𝐴 = −1 ⇒ 𝐴𝐴 = − 4 4𝐵𝐵 = 0 ⇒ 𝐵𝐵 = 0

1 1 1 1 1 2𝐴𝐴 + 4𝐶𝐶 = 0 ⇒ 𝐶𝐶 = − 𝐴𝐴 = − − = ⇒ 𝐶𝐶 = 2 2 4 8 8

𝑥𝑥 = 𝑥𝑥𝑐𝑐 + 𝑥𝑥𝑝𝑝

9/30/2014

1 1 𝑥𝑥𝑝𝑝 = − 𝑡𝑡 2 + 4 8

𝟏𝟏 𝟐𝟐 𝟏𝟏 ⇒ 𝒙𝒙 = 𝒄𝒄𝟒𝟒 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟓𝟓 𝐬𝐬𝐬𝐬𝐬𝐬 𝟐𝟐𝟐𝟐 − 𝒕𝒕 + 𝟒𝟒 𝟖𝟖 Dr. Eli Saber

197

Solving Systems of Linear Equations 1 2 1 𝑥𝑥 = 𝑐𝑐4 cos 2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡 − 𝑡𝑡 + 4 8 𝑦𝑦 = 𝑐𝑐1 + 𝑐𝑐2 cos 2𝑡𝑡 + 𝑐𝑐3 sin 2𝑡𝑡 +

1 3 1 2 1 𝑡𝑡 + 𝑡𝑡 − 𝑡𝑡 12 4 8

Re-substituting 𝑥𝑥, 𝑦𝑦 in 𝒙𝒙′ + 𝒙𝒙 + 𝒚𝒚′ = 𝟎𝟎

1 1 1 ⇒ −2𝑐𝑐4 cos 2𝑡𝑡 + 2𝑐𝑐5 cos 2𝑡𝑡 − 𝑡𝑡 + 𝑐𝑐4 cos 2𝑡𝑡 + 𝑐𝑐5 sin 2𝑡𝑡 − 𝑡𝑡 2 + 2 4 8 1 1 1 + −2𝑐𝑐2 cos 2𝑡𝑡 + 2𝑐𝑐3 cos 2𝑡𝑡 + 𝑡𝑡 3 + 𝑡𝑡 − =0 4 2 8 ⇒ sin 2𝑡𝑡 −2 𝑐𝑐4 + 𝑐𝑐5 − 2𝑐𝑐2 + cos 2𝑡𝑡 2𝑐𝑐5 + 𝑐𝑐4 + 2𝑐𝑐3 = 0 ⇒ −2 𝑐𝑐4 + 𝑐𝑐5 − 2𝑐𝑐2 = 0 & 2𝑐𝑐5 + 𝑐𝑐4 + 2𝑐𝑐3 = 0

9/30/2014

Dr. Eli Saber

198

Solving Systems of Linear Equations ⇒ 𝑐𝑐5 − 2𝑐𝑐4 = 2𝑐𝑐2

⇒ 2𝑐𝑐5 + 𝑐𝑐4 = −2𝑐𝑐3

1 ⇒ 𝑐𝑐4 = − 4𝑐𝑐2 + 2𝑐𝑐3 5 ⇒ 𝑐𝑐5 =

1 2𝑐𝑐2 − 4𝑐𝑐3 5 ⇒ 𝒙𝒙 = −

𝟏𝟏 𝟓𝟓

1 1 𝑥𝑥 = 𝑐𝑐4 𝑐𝑐𝑐𝑐𝑐𝑐 2𝑡𝑡 + 𝑐𝑐5 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑡𝑡 − 𝑡𝑡 2 + 4 8 1 3 1 2 1 𝑦𝑦 = 𝑐𝑐1 + 𝑐𝑐2 𝑐𝑐𝑐𝑐𝑐𝑐 2𝑡𝑡 + 𝑐𝑐3 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑡𝑡 + 𝑡𝑡 + 𝑡𝑡 − 𝑡𝑡 12 4 8

𝟒𝟒𝒄𝒄𝟐𝟐 + 𝟐𝟐𝒄𝒄𝟑𝟑 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 +

𝟏𝟏 𝟓𝟓

⇒ 𝒚𝒚 = 𝒄𝒄𝟏𝟏 + 𝒄𝒄𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜 𝟐𝟐𝟐𝟐 + 𝒄𝒄𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟐𝟐𝟐𝟐 + 9/30/2014

𝟏𝟏 𝟒𝟒

𝟐𝟐𝒄𝒄𝟐𝟐 − 𝟒𝟒𝒄𝒄𝟑𝟑 𝐬𝐬𝐬𝐬𝐬𝐬 𝟐𝟐𝟐𝟐 − 𝒕𝒕𝟐𝟐 +

Dr. Eli Saber

𝟏𝟏 𝟑𝟑 𝒕𝒕 𝟏𝟏𝟏𝟏

𝟏𝟏 𝟒𝟒

𝟏𝟏 𝟖𝟖

+ 𝒕𝒕𝟐𝟐 − 𝒕𝒕

𝟏𝟏 𝟖𝟖

199

End of Chapter 3

9/30/2014

Dr. Eli Saber

200