chapter 3-higher order differential equations
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Higher order differential EquationsTRANSCRIPT
EEEE707: Engineering Analysis
Dr. Eli Saber Department of Electrical and Microelectronic Engineering
Chester F. Carlson Center for Imaging Science Rochester Institute of Technology, Rochester, NY 14623 USA
Chapter 3 Higher Order Differential
Equations
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Section 3.1 Theory of Linear Equations
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Theory of Linear Equations
Objective: Investigate Differential Equations of Order 2++
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Theory of Linear Equations
IVP: Initial Value Problem BVP: Boundary Value Problem โข IVP: Solve:
๐๐๐๐ ๐ฅ๐ฅ ๐๐๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐ + ๐๐๐๐โ1 ๐ฅ๐ฅ
๐๐๐๐โ1๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐โ1 + โฆ + ๐๐1 ๐ฅ๐ฅ
๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ + ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)
Subject to:
๐ฆ๐ฆ ๐ฅ๐ฅ0 = ๐ฆ๐ฆ0, ๐ฆ๐ฆโฒ ๐ฅ๐ฅ0 = ๐ฆ๐ฆ, โฆ . . ,๐ฆ๐ฆ ๐๐โ1 (๐ฅ๐ฅ0) = ๐ฆ๐ฆ๐๐โ1 i.e. seek a function defined on interval ๐ผ๐ผ containing ๐ฅ๐ฅ0 that satisfies the D.E. and the ๐๐ initial conditions
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Initial Value and Boundary Value Problems
Theory of Linear Equations
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Initial Value and Boundary Value Problems
Theorem: Existence of a Unique Solution (for 1st order D.E.) Let ๐ ๐ be a Rectangular region in the x-y plane defined by ๐๐ โค ๐ฅ๐ฅ โค ๐๐; ๐๐ โค ๐ฆ๐ฆ โค ๐๐ that contains the point ๐ฅ๐ฅ0,๐ฆ๐ฆ0 . If ๐๐ ๐ฅ๐ฅ, ๐ฆ๐ฆ & ๐๐๐ฆ๐ฆ/๐๐๐ฅ๐ฅ are continuous on ๐ ๐ , then there exists some Interval ๐ผ๐ผ0: ๐ฅ๐ฅ0 โ โ, ๐ฅ๐ฅ0 + โ ; โ > 0 contained in [๐๐, ๐๐] and a unique function ๐ฆ๐ฆ ๐ฅ๐ฅ defined on ๐ผ๐ผ0 that is a solution of the Initial Value Problem.
Theory of Linear Equations
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Initial Value and Boundary Value Problems
Theorem: Existence of a Unique Solution (for nth order D.E.)
๐๐๐๐ ๐ฅ๐ฅ ๐๐๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐ + ๐๐๐๐โ1 ๐ฅ๐ฅ
๐๐๐๐โ1๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐โ1 + โฆ + ๐๐1 ๐ฅ๐ฅ
๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)
Let ๐๐๐๐ ๐ฅ๐ฅ , ๐๐๐๐โ1 ๐ฅ๐ฅ , โฆ , ๐๐1 ๐ฅ๐ฅ ,๐๐0 ๐ฅ๐ฅ & ๐๐ ๐ฅ๐ฅ be continuous on an interval ๐ผ๐ผ, and let ๐๐๐๐ ๐ฅ๐ฅ โ 0 โ ๐ฅ๐ฅ ๐๐๐ผ๐ผ If ๐ฅ๐ฅ = ๐ฅ๐ฅ0 is any point in ๐ผ๐ผ, a solution ๐ฆ๐ฆ(๐ฅ๐ฅ) of the IVP exists on the interval and is unique.
Theory of Linear Equations
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Initial Value and Boundary Value Problems
E.g. 3๐ฆ๐ฆโฒโฒโฒ + 5๐ฆ๐ฆโฒโฒ โ ๐ฆ๐ฆโฒ + 7๐ฆ๐ฆ = 0 ๐ฆ๐ฆ 1 = 0; ๐ฆ๐ฆโฒ 1 = 0; ๐ฆ๐ฆโฒโฒ 1 = 0 Solution: ๐๐ = ๐๐ Since D.E. is linear with constant coefficients, the unique solution theorem is fulfilled. Hence, ๐๐ = ๐๐ is the only solution on any interval containing x=1
Theory of Linear Equations
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Initial Value and Boundary Value Problems
E.g. ๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆ = 12๐ฅ๐ฅ ๐ฆ๐ฆ 0 = 4;๐ฆ๐ฆโฒ 0 = 1 Solution: ๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐โ๐๐๐๐ โ ๐๐๐๐ 1. D.E. is linear with constant coefficients 2. The coefficients as well as ๐๐(๐ฅ๐ฅ) are continuous 3. ๐๐2(๐ฅ๐ฅ) = 1 โ 0 on any interval ๐ผ๐ผ containing ๐ฅ๐ฅ = 0
The unique solution theorem is fulfilled. Hence, ๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐โ๐๐๐๐ โ ๐๐๐๐ is the unique solution on interval ๐ฐ๐ฐ
Theory of Linear Equations
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Initial Value and Boundary Value Problems
Check: ๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆ = 12๐ฅ๐ฅ; Solution: ๐ฆ๐ฆ = 3๐๐2๐ฅ๐ฅ + ๐๐โ2๐ฅ๐ฅ โ 3๐ฅ๐ฅ Now, ๐ฆ๐ฆโฒ = 6๐๐2๐ฅ๐ฅ โ 2๐๐โ2๐ฅ๐ฅ โ 3 And, ๐ฆ๐ฆโฒโฒ = 12๐๐2๐ฅ๐ฅ + 4๐๐โ2๐ฅ๐ฅ ๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆ = 12๐๐2๐ฅ๐ฅ + 4๐๐โ2๐ฅ๐ฅ โ 4 3๐๐2๐ฅ๐ฅ + ๐๐โ2๐ฅ๐ฅ โ 3๐ฅ๐ฅ = 12๐๐2๐ฅ๐ฅ + 4๐๐โ2๐ฅ๐ฅ โ 12๐๐2๐ฅ๐ฅ โ 4๐๐โ2๐ฅ๐ฅ + 12๐ฅ๐ฅ
Theory of Linear Equations
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Initial Value and Boundary Value Problems
Check: ๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆ = 12๐ฅ๐ฅ;๐ฆ๐ฆ = 3๐๐2๐ฅ๐ฅ + ๐๐โ2๐ฅ๐ฅ โ 3๐ฅ๐ฅ Now, ๐ฆ๐ฆโฒ = 6๐๐2๐ฅ๐ฅ โ 2๐๐โ2๐ฅ๐ฅ โ 3 And, ๐ฆ๐ฆโฒโฒ = 12๐๐2๐ฅ๐ฅ + 4๐๐โ2๐ฅ๐ฅ ๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆ = 12๐๐2๐ฅ๐ฅ + 4๐๐โ2๐ฅ๐ฅ โ 4 3๐๐2๐ฅ๐ฅ + ๐๐โ2๐ฅ๐ฅ โ 3๐ฅ๐ฅ = 12๐๐2๐ฅ๐ฅ + 4๐๐โ2๐ฅ๐ฅ โ 12๐๐2๐ฅ๐ฅ โ 4๐๐โ2๐ฅ๐ฅ + 12๐ฅ๐ฅ = 12๐ฅ๐ฅ Verified.
Theory of Linear Equations
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Initial Value and Boundary Value Problems
E.g. ๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 2๐ฅ๐ฅ๐ฆ๐ฆโฒ + 2๐ฆ๐ฆ = 6 ๐ฆ๐ฆ 0 = 3; ๐ฆ๐ฆโฒ 0 = 1 Solution: ๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ2 + ๐ฅ๐ฅ + 3 in interval ๐ผ๐ผ โก (โโ,โ)
Theory of Linear Equations
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Initial Value and Boundary Value Problems
Check: ๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ2 + ๐ฅ๐ฅ + 3 โ ๐ฆ๐ฆโฒ = 2๐๐๐ฅ๐ฅ + 1; ๐ฆ๐ฆโฒโฒ = 2๐๐ ๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 2๐ฅ๐ฅ๐ฆ๐ฆโฒ + 2๐ฆ๐ฆ = ๐ฅ๐ฅ2 2๐๐ โ 2๐ฅ๐ฅ 2๐๐๐ฅ๐ฅ + 1 + 2 ๐๐๐ฅ๐ฅ2 + ๐ฅ๐ฅ + 3 = 2๐๐๐ฅ๐ฅ2 โ 4๐๐๐ฅ๐ฅ2 โ 2๐ฅ๐ฅ + 2๐๐๐ฅ๐ฅ2 + 2๐ฅ๐ฅ + 6 = 2๐๐๐ฅ๐ฅ2 โ 4๐๐๐ฅ๐ฅ2 โ 2๐ฅ๐ฅ + 2๐๐๐ฅ๐ฅ2 + 2๐ฅ๐ฅ + 6 ๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 2๐ฅ๐ฅ๐ฆ๐ฆโฒ + 2๐ฆ๐ฆ = 6
Theory of Linear Equations
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Initial Value and Boundary Value Problems
IVP Check: ๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ2 + ๐ฅ๐ฅ + 3 โข ๐ฆ๐ฆ 0 = 3 โ 3 = ๐๐ 0 2 + 0 + 3 โ 3 = 3
โข ๐ฆ๐ฆโฒ 0 = 1
๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ2 + ๐ฅ๐ฅ + 3 โ ๐ฆ๐ฆโฒ = 2๐๐๐ฅ๐ฅ + 1 1 = 2๐๐ 0 + 1 โ 1 = 1
โข Note:
For ๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ2 + ๐ฅ๐ฅ + 3 , the initial conditions of ๐ฆ๐ฆ 0 = 3 & ๐ฆ๐ฆโฒ 0 = 1 did not provide a unique value for ๐๐
Hence: ๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ2 + ๐ฅ๐ฅ + 3 is a solution for the D.E. ๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 2๐ฅ๐ฅ๐ฆ๐ฆโฒ + 2๐ฆ๐ฆ = 6 โ๐๐ i.e. there is no unique solution But what w.r.t. unique solution theorem?
Theory of Linear Equations
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Initial Value and Boundary Value Problems
Let us apply the theorem towards this example.
๐ฅ๐ฅ2 ๐ฆ๐ฆโฒโฒ โ 2๐ฅ๐ฅ ๐ฆ๐ฆโฒ + 2 ๐ฆ๐ฆ = 6 Note: ๐๐2 ๐ฅ๐ฅ = ๐ฅ๐ฅ2 = 0 ๐๐๐๐๐๐ ๐ฅ๐ฅ = 0 and ๐ฅ๐ฅ โ ๐ผ๐ผ = (โโ,โ) ๐๐๐๐ ๐๐ โ ๐๐โ ๐๐ = ๐๐๐๐ โ ๐ฐ๐ฐ this condition is NOT satisfied
๐๐2 ๐ฅ๐ฅ = ๐ฅ๐ฅ2
๐๐1 ๐ฅ๐ฅ = โ2๐ฅ๐ฅ
๐๐0 ๐ฅ๐ฅ = 2
๐๐ ๐ฅ๐ฅ = 6
Theory of Linear Equations
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Initial Value and Boundary Value Problems
โข BVP (Boundary Value Problem):
๐ท๐ท.๐ธ๐ธ. โถ ๐๐2 ๐ฅ๐ฅ๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
+ ๐๐1 ๐ฅ๐ฅ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ
With ๐ฆ๐ฆ ๐๐ = ๐ฆ๐ฆ0 & ๐ฆ๐ฆ ๐๐ = ๐ฆ๐ฆ1 Other boundary value conditions could be: ๐ฆ๐ฆ๐ฆ ๐๐ = ๐ฆ๐ฆ0 & ๐ฆ๐ฆ ๐๐ = ๐ฆ๐ฆ1 ๐ฆ๐ฆ ๐๐ = ๐ฆ๐ฆ0 & ๐ฆ๐ฆ๐ฆ ๐๐ = ๐ฆ๐ฆ1 ๐ฆ๐ฆ๐ฆ ๐๐ = ๐ฆ๐ฆ0 & ๐ฆ๐ฆ๐ฆ ๐๐ = ๐ฆ๐ฆ1 General Boundary Conditions: ๐จ๐จ๐๐๐๐ ๐๐ + ๐ฉ๐ฉ๐๐๐๐โฒ ๐๐ = ๐ช๐ช๐๐ ๐จ๐จ๐๐๐๐ ๐๐ + ๐ฉ๐ฉ๐๐๐๐โฒ ๐๐ = ๐ช๐ช๐๐
Boundary conditions
Theory of Linear Equations
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Initial Value and Boundary Value Problems
Note: Even when the conditions for Unique Solution theorem are met, a BVP may have: 1) Many solutions 2) Unique Solution 3) No solution
Theory of Linear Equations
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Initial Value and Boundary Value Problems
E.g. ๐๐2๐ฅ๐ฅ๐๐๐ก๐ก2
+ 16๐ฅ๐ฅ = 0
Solution: ๐ฅ๐ฅ = ๐๐1 cos4๐ก๐ก + ๐๐2 sin 4๐ก๐ก Check: ๐๐๐ฅ๐ฅ๐๐๐ก๐ก
= โ4๐๐1 sin4๐ก๐ก + 4๐๐2 cos4๐ก๐ก
๐๐2๐ฅ๐ฅ๐๐๐ก๐ก2
= โ16๐๐1 cos4๐ก๐ก โ 16๐๐2 sin 4๐ก๐ก
๐๐2๐ฅ๐ฅ๐๐๐ก๐ก2
+ 16๐ฅ๐ฅ = โ16๐๐1 cos4๐ก๐ก โ 16๐๐2 sin 4๐ก๐ก + 16 ๐๐1 cos4๐ก๐ก + ๐๐2 sin4๐ก๐ก = 0
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 19
Initial Value and Boundary Value Problems
Now, let us consider these different sets of BV Conditions: ๐ฅ๐ฅ = ๐๐1 cos4๐ก๐ก + ๐๐2 sin 4๐ก๐ก
1) ๐ฅ๐ฅ 0 = 0; ๐ฅ๐ฅ๐๐2
= 0
โข ๐ฅ๐ฅ 0 = 0 โ 0 = ๐๐1 1 + ๐๐2 0 โ ๐๐๐๐ = ๐๐
โข ๐ฅ๐ฅ ๐๐2
= 0 โ 0 = ๐๐1 cos(2๐๐) + ๐๐2 sin(2๐๐) โ 0 = ๐๐1 1 + ๐๐2 0
โ But ๐๐1 = 0
โ That means, ๐๐2 0 = 0
โ Implies ๐๐2 can be anything
โข Infinite solutions since ๐๐2 can be anything
Theory of Linear Equations
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Initial Value and Boundary Value Problems
2) ๐ฅ๐ฅ 0 = 0; ๐ฅ๐ฅ๐๐8
= 0
โข ๐ฅ๐ฅ 0 = 0 โ 0 = ๐๐1 1 + ๐๐2 0 โ ๐๐๐๐ = ๐๐
โข ๐ฅ๐ฅ ๐๐8
= 0 โ 0 = ๐๐1 cos ๐๐2
+ ๐๐2 sin ๐๐2โ 0 = ๐๐1 0 + ๐๐2 1
โ But ๐๐1 = 0
โ That means, ๐๐2 1 = 0
โ Implies ๐๐๐๐ = ๐๐
โ ๐๐ = ๐๐ is the solution of this new boundary problem
โข Unique solution โก ๐๐ = ๐๐
๐ฅ๐ฅ = ๐๐1 cos4๐ก๐ก + ๐๐2 sin4๐ก๐ก
Theory of Linear Equations
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Initial Value and Boundary Value Problems
3) ๐ฅ๐ฅ 0 = 0;๐ฅ๐ฅ๐๐2
= 1
โข ๐ฅ๐ฅ 0 = 0 โ 0 = ๐๐1 1 + ๐๐2 0 โ ๐๐๐๐ = ๐๐
โข ๐ฅ๐ฅ ๐๐2
= 1 โ 1 = ๐๐1 cos 4๐๐ โ ๐๐2
+ ๐๐2 sin 4๐๐ โ ๐๐2โ 1 = ๐๐1 cos(2๐๐) + ๐๐2 sin(2๐๐)
โ 1 = 0 1 + ๐๐2 0
โ That means, ๐๐2 0 = 1
โ Implies ๐๐๐๐ = ๐๐๐๐
= ๐ต๐ต.๐ซ๐ซ.
โ Not possible to find ๐๐2
โข No solution for BVP
๐ฅ๐ฅ = ๐๐1 cos4๐ก๐ก + ๐๐2 sin4๐ก๐ก
Theory of Linear Equations
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Differential Operators โDโ
E.g. ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ = ๐ท๐ท๐ฆ๐ฆ
๐๐๐ฆ๐ฆ2
๐๐๐ฅ๐ฅ2 =๐๐๐๐๐ฅ๐ฅ
๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ = ๐ท๐ท ๐ท๐ท๐ฆ๐ฆ = ๐ท๐ท2๐ฆ๐ฆ
i.e. ๐๐๐๐๐ฅ๐ฅ
cos4๐ฅ๐ฅ = โ4 sin4๐ฅ๐ฅ โ ๐ท๐ท cos4๐ฅ๐ฅ = โ4 sin 4๐ฅ๐ฅ
โ ๐ผ๐ผ๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐๐๐: ๐ ๐ ๐๐๐๐๐ ๐ ๐๐๐๐ = ๐ซ๐ซ๐๐๐๐
Theory of Linear Equations
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Differential Equations
Note: โข ๐ท๐ท ๐๐ ๐๐ ๐ฅ๐ฅ = ๐๐ ๐ท๐ท๐๐(๐ฅ๐ฅ)
โข ๐ท๐ท ๐๐ ๐ฅ๐ฅ + ๐๐ ๐ฅ๐ฅ = ๐ท๐ท๐๐ ๐ฅ๐ฅ + ๐ท๐ท๐๐(๐ฅ๐ฅ)
โข ๐ท๐ท ๐ผ๐ผ ๐๐ ๐ฅ๐ฅ + ๐ฝ๐ฝ ๐๐ ๐ฅ๐ฅ = ๐ผ๐ผ ๐ท๐ท ๐๐ ๐ฅ๐ฅ + ๐ฝ๐ฝ ๐ท๐ท ๐๐ ๐ฅ๐ฅ
โ ๐ผ๐ผ,๐ฝ๐ฝ are constants
Linear
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 24
Differential Equations
Let ๐ฆ๐ฆโฒโฒ + 5๐ฆ๐ฆโฒ + 6๐ฆ๐ฆ = 5๐ฅ๐ฅ โ 3
This can be written was ๐๐2๐ฆ๐ฆ
๐๐๐ฅ๐ฅ2+ 5 ๐๐๐ฆ๐ฆ
๐๐๐ฅ๐ฅ+ 6๐ฆ๐ฆ = 5๐ฅ๐ฅ โ 3
Which can also be written as: ๐ท๐ท2๐ฆ๐ฆ + 5๐ท๐ท๐ฆ๐ฆ + 6๐ฆ๐ฆ = 5๐ฅ๐ฅ โ 3
Similarly, ๐๐๐๐ ๐ฅ๐ฅ ๐๐๐๐๐ฆ๐ฆ
๐๐๐ฅ๐ฅ๐๐ + ๐๐๐๐โ1 ๐ฅ๐ฅ ๐๐
๐๐โ1๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐โ1
+ โฆ + ๐๐1 ๐ฅ๐ฅ ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = 0
can be written as ๐ฟ๐ฟ ๐ฆ๐ฆ = 0
And, ๐๐๐๐ ๐ฅ๐ฅ ๐๐๐๐๐ฆ๐ฆ
๐๐๐ฅ๐ฅ๐๐ + ๐๐๐๐โ1 ๐ฅ๐ฅ ๐๐
๐๐โ1๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐โ1
+ โฆ + ๐๐1 ๐ฅ๐ฅ ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)
can be written as ๐ฟ๐ฟ ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 25
Differential Equations Definition: ๐๐๐ก๐ก๐กorder differential operator is:
๐ฟ๐ฟ = ๐๐๐๐ ๐ฅ๐ฅ ๐ท๐ท๐๐ + ๐๐๐๐โ1 ๐ฅ๐ฅ ๐ท๐ท๐๐โ1 + โฏ+ ๐๐1 ๐ฅ๐ฅ ๐ท๐ท + ๐๐0(๐ฅ๐ฅ)
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 26
Superposition Principle
Theorem: Superposition Principle โ Homogeneous Equations Let ๐ฆ๐ฆ1, ๐ฆ๐ฆ2, โฆ ,๐ฆ๐ฆ๐๐ be solutions of the Homogeneous ๐๐๐ก๐ก๐ก order differential equation on an interval ๐ผ๐ผ. Then the linear combination
๐ฆ๐ฆ = ๐๐1๐ฆ๐ฆ1 ๐ฅ๐ฅ + ๐๐2๐ฆ๐ฆ2 ๐ฅ๐ฅ + โฏ+ ๐๐๐๐๐ฆ๐ฆ๐๐ ๐ฅ๐ฅ ,where ๐๐1, ๐๐2, โฆ , ๐๐๐๐ as are arbitrary constants, is also a solution on ๐ผ๐ผ Corollaries: โข A constant multiple ๐ฆ๐ฆ = ๐๐1๐ฆ๐ฆ1(๐ฅ๐ฅ) of the solution ๐ฆ๐ฆ1(๐ฅ๐ฅ) of a homogeneous
linear differential equation is also a solution
โข A homogeneous linear differential equation always possesses the trivial solution ๐ฆ๐ฆ = 0
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 27
Superposition Principle
E.g. ๐ฅ๐ฅ3 ๐๐3๐ฆ๐ฆ
๐๐๐ฅ๐ฅ3โ 2๐ฅ๐ฅ ๐๐๐ฆ๐ฆ
๐๐๐ฅ๐ฅ+ 4๐ฆ๐ฆ = 0
And ๐ฆ๐ฆ1 = ๐ฅ๐ฅ2 & ๐ฆ๐ฆ2 = ๐ฅ๐ฅ2 ln ๐ฅ๐ฅ are both solutions Check: First solution: ๐ฆ๐ฆ1 = ๐ฅ๐ฅ2
๐ฆ๐ฆ = ๐ฅ๐ฅ2 โ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ = 2๐ฅ๐ฅ โ
๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 = 2 &
๐๐3๐ฆ๐ฆ๐๐๐ฅ๐ฅ3 = 0
Implies, ๐ฅ๐ฅ3 0 โ 2๐ฅ๐ฅ 2๐ฅ๐ฅ + 4 ๐ฅ๐ฅ2 = โ4๐ฅ๐ฅ2 + 4๐ฅ๐ฅ2 = 0
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 28
Superposition Principle
E.g. ๐ฅ๐ฅ3 ๐๐3๐ฆ๐ฆ
๐๐๐ฅ๐ฅ3โ 2๐ฅ๐ฅ ๐๐๐ฆ๐ฆ
๐๐๐ฅ๐ฅ+ 4๐ฆ๐ฆ = 0
And ๐ฆ๐ฆ1 = ๐ฅ๐ฅ2 & ๐ฆ๐ฆ2 = ๐ฅ๐ฅ2 ln ๐ฅ๐ฅ are both solutions Check: Second solution: ๐ฆ๐ฆ1 = ๐ฅ๐ฅ2 ln ๐ฅ๐ฅ
๐ฆ๐ฆ = ๐ฅ๐ฅ2 ln ๐ฅ๐ฅ โ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
= 2๐ฅ๐ฅ ln ๐ฅ๐ฅ + ๐ฅ๐ฅ21๐ฅ๐ฅ
= 2๐ฅ๐ฅ ln ๐ฅ๐ฅ + ๐ฅ๐ฅ
๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
= 2 ln ๐ฅ๐ฅ +๐ฅ๐ฅ๐ฅ๐ฅ
+ 1 = 2 ln ๐ฅ๐ฅ + 3 โ๐๐3๐ฆ๐ฆ๐๐๐ฅ๐ฅ3
=2๐ฅ๐ฅ
Implies, ๐ฅ๐ฅ3 2๐ฅ๐ฅโ 2๐ฅ๐ฅ 2๐ฅ๐ฅ ln ๐ฅ๐ฅ + ๐ฅ๐ฅ + 4 ๐ฅ๐ฅ2 ln ๐ฅ๐ฅ = 2๐ฅ๐ฅ2 โ 4๐ฅ๐ฅ2 ln ๐ฅ๐ฅ โ 2๐ฅ๐ฅ2 + 4๐ฅ๐ฅ2 ln ๐ฅ๐ฅ = 0
By superposition ๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐ ๐ฅ๐ฅ๐ฅ๐ฅ ๐๐ is also a solution
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 29
Linear (Dependence & Independence)
Definition: โข A set of functions ๐๐1 ๐ฅ๐ฅ ,๐๐2 ๐ฅ๐ฅ , โฆ ,๐๐๐๐(๐ฅ๐ฅ) is said to be
linearly dependent on an Interval ๐ฐ๐ฐ if there exists constants ๐๐๐๐, ๐๐๐๐, โฆ , ๐๐๐๐ not all zero such that:
๐๐๐๐๐๐๐๐ ๐๐ + ๐๐๐๐๐๐๐๐ ๐๐ + โฏ+ ๐๐๐๐๐๐๐๐ ๐๐ = ๐๐ โ๐๐ โ ๐ฐ๐ฐ
โข A set of functions is linearly independent on an interval if the only constants for which
๐๐๐๐๐๐๐๐ ๐๐ + ๐๐๐๐๐๐๐๐ ๐๐ + โฏ+ ๐๐๐๐๐๐๐๐ ๐๐ = ๐๐ โ๐๐ โ ๐ฐ๐ฐ are ๐๐๐๐ = ๐๐๐๐ = ๐๐๐๐ = โฏ = ๐๐๐๐ = ๐๐
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 30
Linear (Dependence & Independence)
Definition: โข A set of functions ๐๐1 ๐ฅ๐ฅ ,๐๐2 ๐ฅ๐ฅ , โฆ ,๐๐๐๐(๐ฅ๐ฅ) is said to be
linearly dependent on an Interval ๐ฐ๐ฐ if there exists constants ๐๐๐๐, ๐๐๐๐, โฆ , ๐๐๐๐ not all zero such that:
๐๐๐๐๐๐๐๐ ๐๐ + ๐๐๐๐๐๐๐๐ ๐๐ + โฏ+ ๐๐๐๐๐๐๐๐ ๐๐ = ๐๐ โ๐๐ โ ๐ฐ๐ฐ
โข A set of functions is linearly independent on an interval if the only constants for which
๐๐๐๐๐๐๐๐ ๐๐ + ๐๐๐๐๐๐๐๐ ๐๐ + โฏ+ ๐๐๐๐๐๐๐๐ ๐๐ = ๐๐ โ๐๐ โ ๐ฐ๐ฐ are ๐๐๐๐ = ๐๐๐๐ = ๐๐๐๐ = โฏ = ๐๐๐๐ = ๐๐
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 31
Wronskian
Definition: Suppose each of the functions ๐๐1 ๐ฅ๐ฅ ,๐๐2 ๐ฅ๐ฅ , โฆ ,๐๐๐๐(๐ฅ๐ฅ) possesses at least ๐๐ โ 1 derivatives Then
๐๐ ๐๐1,๐๐2, โฆ ,๐๐๐๐ =
๐๐1 ๐๐2 โฆ โฏ ๐๐๐๐๐๐1โฒ ๐๐2โฒ โฆ โฆ ๐๐๐๐โฒ
โฎ๐๐1
(๐๐โ1) ๐๐2(๐๐โ1) โฆ ๐๐๐๐
(๐๐โ1)
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 32
Wronskian
Criterion for Linearly Independent solutions Let ๐ฆ๐ฆ1, ๐ฆ๐ฆ2, โฆ ,๐ฆ๐ฆ๐๐ be n-solutions of the homogeneous linear nth order differential equation on an interval ๐ผ๐ผ. Then the set of solutions is linearly independent on ๐ผ๐ผ if and only if
๐พ๐พ ๐๐๐๐,๐๐๐๐, โฆ ,๐๐๐๐ โ ๐๐ โ๐๐ โ ๐ฐ๐ฐ
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 33
Wronskian
E.g. ๐ฆ๐ฆ1 = ๐๐3๐ฅ๐ฅ and ๐ฆ๐ฆ2 = ๐๐โ3๐ฅ๐ฅ are both the solutions of the homogeneous linear equation ๐ฆ๐ฆโฒโฒ โ 9๐ฆ๐ฆ = 0; ๐ผ๐ผ = (โโ,โ) Check: ๐๐ ๐๐3๐ฅ๐ฅ, ๐๐โ3๐ฅ๐ฅ = ๐๐3๐ฅ๐ฅ ๐๐โ3๐ฅ๐ฅ
3๐๐3๐ฅ๐ฅ โ3๐๐โ3๐ฅ๐ฅ
= ๐๐3๐ฅ๐ฅ โ3๐๐โ3๐ฅ๐ฅ โ ๐๐โ3๐ฅ๐ฅ 3๐๐3๐ฅ๐ฅ = โ3 โ 3 = โ6 โ 0 Thus, ๐๐ = ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐โ๐๐๐๐ is the general solution
๐พ๐พ ๐๐๐๐,๐๐๐๐, โฆ ,๐๐๐๐ โ ๐๐ โ๐๐ โ ๐ฐ๐ฐ
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 34
Wronskian
E.g. ๐ฆ๐ฆโฒโฒโฒ โ 6๐ฆ๐ฆโฒโฒ + 11๐ฆ๐ฆโฒ โ 6๐ฆ๐ฆ = 0 The functions ๐ฆ๐ฆ1 = ๐๐๐ฅ๐ฅ;๐ฆ๐ฆ2 = ๐๐2๐ฅ๐ฅ & ๐ฆ๐ฆ3 = ๐๐3๐ฅ๐ฅ satisfy the D.E. above Check:
๐๐ ๐๐๐ฅ๐ฅ, ๐๐2๐ฅ๐ฅ, ๐๐3๐ฅ๐ฅ =๐๐๐ฅ๐ฅ ๐๐2๐ฅ๐ฅ ๐๐3๐ฅ๐ฅ๐๐๐ฅ๐ฅ 2๐๐2๐ฅ๐ฅ 3๐๐3๐ฅ๐ฅ๐๐๐ฅ๐ฅ 4๐๐2๐ฅ๐ฅ 9๐๐3๐ฅ๐ฅ
= ๐๐๐ฅ๐ฅ 2๐๐2๐ฅ๐ฅ 3๐๐3๐ฅ๐ฅ
4๐๐2๐ฅ๐ฅ 9๐๐3๐ฅ๐ฅโ ๐๐2๐ฅ๐ฅ ๐๐๐ฅ๐ฅ 3๐๐3๐ฅ๐ฅ
๐๐๐ฅ๐ฅ 9๐๐3๐ฅ๐ฅ+ ๐๐3๐ฅ๐ฅ ๐๐๐ฅ๐ฅ 2๐๐2๐ฅ๐ฅ
๐๐๐ฅ๐ฅ 4๐๐2๐ฅ๐ฅ= ๐๐๐๐๐ก๐ก๐๐๐๐ ๐ ๐ ๐๐๐๐๐ ๐ ๐ ๐ ๐๐๐๐
= 2๐๐6๐ฅ๐ฅ โ 0 Hence, ๐๐๐๐, ๐๐๐๐๐๐,๐๐๐๐๐๐ form a fundamental set & ๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ is the general solution
๐พ๐พ ๐๐๐๐,๐๐๐๐, โฆ ,๐๐๐๐ โ ๐๐ โ๐๐ โ ๐ฐ๐ฐ
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 35
Non Homogeneous Equations
๐๐๐๐ ๐ฅ๐ฅ ๐๐๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐
+ ๐๐๐๐โ1 ๐ฅ๐ฅ ๐๐๐๐โ1๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐โ1
+ โฆ + ๐๐1 ๐ฅ๐ฅ ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)
,where ๐๐(๐ฅ๐ฅ) โ 0 โข If ๐ฆ๐ฆ๐๐ (free of arbitrary parameter) satisfies the equation above, ๐ฆ๐ฆ๐๐ is called
particular solution E.g. ๐ฆ๐ฆโฒโฒ + 9๐ฆ๐ฆ = 27 Let ๐ฆ๐ฆ๐๐ = 3 โ ๐ฆ๐ฆโฒโฒ + 9๐ฆ๐ฆ = 0 + 9 3 = ๐๐๐๐ โข If ๐ฆ๐ฆ1,๐ฆ๐ฆ2, โฆ , ๐ฆ๐ฆ๐๐ are solutions of Homogeneous equations and ๐ฆ๐ฆ๐๐ is any particular
solution,
๐ฆ๐ฆ = ๐๐1๐ฆ๐ฆ1 ๐ฅ๐ฅ + ๐๐2๐ฆ๐ฆ2 ๐ฅ๐ฅ + โฏ+ ๐๐๐๐๐ฆ๐ฆ๐๐ ๐ฅ๐ฅ + ๐ฆ๐ฆ๐๐ a
General solution Complementary S๐๐๐๐๐๐๐ก๐ก๐ ๐ ๐๐๐๐ ๐๐๐๐
Particular Solution ๐๐๐๐
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 36
Non Homogeneous Equations
E.g. ๐ฆ๐ฆโฒโฒโฒ โ 6๐ฆ๐ฆโฒโฒ + 11๐ฆ๐ฆโฒ โ 6๐ฆ๐ฆ = 3๐ฅ๐ฅ non-homogeneous equation
Let ๐๐๐๐ = โ๐๐๐๐๐๐๐๐โ ๐๐
๐๐๐๐. Is it a solution?
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 37
Non Homogeneous Equations
E.g.
๐ฆ๐ฆ๐๐โฒ = โ12
; ๐ฆ๐ฆ๐๐โฒโฒ = 0; ๐ฆ๐ฆ๐๐โฒโฒโฒ = 0
โ ๐ฆ๐ฆโฒโฒโฒ โ 6๐ฆ๐ฆโฒโฒ + 11๐ฆ๐ฆโฒ โ 6๐ฆ๐ฆ = 0 โ 6 0 + 11 โ12
โ 6 โ1112
โ12๐ฅ๐ฅ
= โ112 +
112 + 3๐ฅ๐ฅ = ๐๐๐๐
Verified.
๐ฆ๐ฆโฒโฒโฒ โ 6๐ฆ๐ฆโฒโฒ + 11๐ฆ๐ฆโฒ โ 6๐ฆ๐ฆ = 3๐ฅ๐ฅ ; ๐ฆ๐ฆ๐๐= โ1112โ 1
2๐ฅ๐ฅ
Theory of Linear Equations
9/30/2014 Dr. Eli Saber 38
Non Homogeneous Equations
Homogeneous Equation: ๐ฆ๐ฆโฒโฒโฒ โ 6๐ฆ๐ฆโฒโฒ + 11๐ฆ๐ฆโฒ โ 6๐ฆ๐ฆ = 0 Let ๐ฆ๐ฆ๐๐ = ๐๐1๐๐๐ฅ๐ฅ + ๐๐2๐๐2๐ฅ๐ฅ + ๐๐3๐๐3๐ฅ๐ฅ be a complimentary solution Hence, the general solution is given by:
๐ฆ๐ฆ = ๐๐1๐๐๐ฅ๐ฅ + ๐๐2๐๐2๐ฅ๐ฅ + ๐๐3๐๐3๐ฅ๐ฅ + (โ1112 โ
12 ๐ฅ๐ฅ)
๐ฆ๐ฆโฒโฒโฒ โ 6๐ฆ๐ฆโฒโฒ + 11๐ฆ๐ฆโฒ โ 6๐ฆ๐ฆ = 3๐ฅ๐ฅ ; ๐ฆ๐ฆ๐๐= โ1112โ 1
2๐ฅ๐ฅ
๐๐๐๐ ๐๐๐๐
Section 3.2 Reduction of Order
9/30/2014 Dr. Eli Saber 39
Reduction of Order
9/30/2014 Dr. Eli Saber 40
Introduction
2nd ๐๐๐๐๐๐๐๐๐๐ ๐ป๐ป๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ ๐ท๐ท.๐ธ๐ธ.: ๐๐2 ๐ฅ๐ฅ ๐ฆ๐ฆโฒโฒ + ๐๐1 ๐ฅ๐ฅ ๐ฆ๐ฆโฒ + ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = 0 Solution: ๐ฆ๐ฆ = ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2 Where ๐ฆ๐ฆ1&๐ฆ๐ฆ2 are linearly independent (L.I.) solutions on ๐ผ๐ผ Objective: Assume that we know ๐ฆ๐ฆ1(๐ฅ๐ฅ) solution seek a 2nd solution ๐ฆ๐ฆ2(๐ฅ๐ฅ) such that ๐ฆ๐ฆ1 ๐ฅ๐ฅ & ๐ฆ๐ฆ2(๐ฅ๐ฅ) are independent on ๐ผ๐ผ
Reduction of Order
9/30/2014 Dr. Eli Saber 41
Introduction
Approach:
โข Recall if ๐ฆ๐ฆ1 ๐ฅ๐ฅ & ๐ฆ๐ฆ2(๐ฅ๐ฅ) are L.I. => ๐ฆ๐ฆ2๐ฆ๐ฆ1
is non-constant
๐ฆ๐ฆ2๐ฆ๐ฆ1
= ๐๐ ๐ฅ๐ฅ ๐๐๐๐ ๐ฆ๐ฆ2 ๐ฅ๐ฅ = ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ1(๐ฅ๐ฅ)
Seek to find ๐๐(๐ฅ๐ฅ) in order to find
๐๐๐๐ ๐๐ = ๐๐ ๐๐ ๐๐๐๐(๐๐)
Reduction of Order
9/30/2014 Dr. Eli Saber 42
E.g. Given ๐๐2๐ฆ๐ฆ
๐๐๐ฅ๐ฅ2โ ๐ฆ๐ฆ = 0; ๐ผ๐ผ = (โโ,โ) and assume that ๐ฆ๐ฆ1 = ๐๐๐ฅ๐ฅ is a solution. Find
a second solution ๐ฆ๐ฆ2 Check:
๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ โ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ = ๐๐๐ฅ๐ฅ โ
๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 = ๐๐๐ฅ๐ฅ
And substituting back in the equation, ๐๐2๐ฆ๐ฆ
๐๐๐ฅ๐ฅ2โ ๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ โ ๐๐๐ฅ๐ฅ = ๐๐
Reduction of Order
9/30/2014 Dr. Eli Saber 43
Let ๐ฆ๐ฆ ๐ฅ๐ฅ = ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐ฅ๐ฅ = ๐๐ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
โ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
= ๐๐ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + ๐๐โฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
โ๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 = ๐๐ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + ๐๐โฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + ๐๐โฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + ๐๐โฒโฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
โ๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 = ๐๐ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + 2๐๐โฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + ๐๐โฒโฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
Hence, ๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 โ ๐ฆ๐ฆ = 0 โ ๐๐ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + 2๐๐โฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + ๐๐โฒโฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ โ ๐๐ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ = 0
Reduction of Order
9/30/2014 Dr. Eli Saber 44
๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
โ ๐ฆ๐ฆ = 0 โ 2๐๐โฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + ๐๐โฒโฒ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ = 0
โ ๐๐๐ฅ๐ฅ ๐๐โฒโฒ + 2๐๐โฒ = 0 But ๐๐๐ฅ๐ฅ โ 0. โ ๐๐โฒโฒ + 2๐๐โฒ = 0 Let ๐ค๐ค = ๐๐๐ฆ change of variable โ ๐ค๐คโฒ + 2๐ค๐ค = 0 (Linear First Order D.E.) โ๐๐๐ค๐ค๐๐๐ฅ๐ฅ
+ 2๐ค๐ค = 0
โ๐๐๐ค๐ค๐๐๐ฅ๐ฅ
= โ2๐ค๐ค
Reduction of Order
9/30/2014 Dr. Eli Saber 45
โ๐๐๐ค๐ค๐๐๐ฅ๐ฅ
= โ2๐ค๐ค
โ๐๐๐ค๐ค๐ค๐ค
= โ2 ๐๐๐ฅ๐ฅ
โ ๏ฟฝ๐๐๐ค๐ค๐ค๐ค = ๏ฟฝโ2๐๐๐ฅ๐ฅ
โ ln ๐ค๐ค = โ2๐ฅ๐ฅ + ๐๐ โ ๐ค๐ค = ๐๐โ2๐ฅ๐ฅ+๐๐ = ๐๐โ2๐ฅ๐ฅ ๐๐๐๐ = ๐๐โ2๐ฅ๐ฅ ๐๐1 โ ๐ค๐ค = ๐๐1๐๐โ2๐ฅ๐ฅ
Reduction of Order
9/30/2014 Dr. Eli Saber 46
Introduction
๐ค๐ค = ๐๐1๐๐โ2๐ฅ๐ฅ
But ๐ค๐ค = ๐๐โฒ โ ๐๐โฒ = ๐๐1๐๐โ2๐ฅ๐ฅ โ๐๐๐๐๐๐๐ฅ๐ฅ
= ๐๐1๐๐โ2๐ฅ๐ฅ
Hence, โซ๐๐๐๐ = โซ ๐๐1๐๐โ2๐ฅ๐ฅ๐๐๐ฅ๐ฅ
โ ๐๐ = โ 12 ๐๐1๐๐
โ2๐ฅ๐ฅ + ๐๐2
Hence, ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ = โ๐๐12๐๐โ2๐ฅ๐ฅ + ๐๐2 ๐๐๐ฅ๐ฅ
โ ๐ฆ๐ฆ = โ๐๐12 ๐๐โ๐ฅ๐ฅ + ๐๐2๐๐๐ฅ๐ฅ
Reduction of Order
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๐ฆ๐ฆ = โ๐๐12 ๐๐โ๐ฅ๐ฅ + ๐๐2๐๐๐ฅ๐ฅ
Let, ๐๐1 = โ2 & ๐๐2 = 0 โ ๐ฆ๐ฆ2 ๐ฅ๐ฅ = ๐๐โ๐ฅ๐ฅ Let us check for independence in the two solutions ๐๐ ๐๐๐ฅ๐ฅ, ๐๐โ๐ฅ๐ฅ = ๐๐๐ฅ๐ฅ ๐๐โ๐ฅ๐ฅ
๐๐๐ฅ๐ฅ โ๐๐โ๐ฅ๐ฅ = โ๐๐๐ฅ๐ฅ๐๐โ๐ฅ๐ฅ โ ๐๐๐ฅ๐ฅ๐๐โ๐ฅ๐ฅ = โ1 โ 1 = โ2 โ 0
๐๐๐ฅ๐ฅ & ๐๐โ๐ฅ๐ฅ are independent
General solution: ๐ฆ๐ฆ = ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐โ๐ฅ๐ฅ
Wronskian
Reduction of Order
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Check: ๐ฆ๐ฆ = ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐โ๐ฅ๐ฅ
โ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
= ๐ผ๐ผ1๐๐๐ฅ๐ฅ โ ๐ผ๐ผ2๐๐โ๐ฅ๐ฅ
โ๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 = ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐โ๐ฅ๐ฅ
Hence, ๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 โ ๐ฆ๐ฆ = ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐โ๐ฅ๐ฅ โ ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐โ๐ฅ๐ฅ = ๐๐
Reduction of Order
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General case: ๐๐2 ๐ฅ๐ฅ ๐ฆ๐ฆโฒโฒ + ๐๐1 ๐ฅ๐ฅ ๐ฆ๐ฆ๐ฆ + ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = 0
๐๐๐ ๐ ๐ ๐ ๐ ๐ ๐๐๐๐ ๐๐๐ฆ๐ฆ ๐๐2 ๐ฅ๐ฅ โ ๐ฆ๐ฆโฒโฒ +๐๐1 ๐ฅ๐ฅ๐๐2 ๐ฅ๐ฅ
๐ฆ๐ฆโฒ +๐๐0 ๐ฅ๐ฅ๐๐2 ๐ฅ๐ฅ
๐ฆ๐ฆ = 0
โ ๐ฆ๐ฆโฒโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ = 0 ๐๐ ๐ฅ๐ฅ & ๐๐(๐ฅ๐ฅ) are continuous on ๐ผ๐ผ
Assume ๐ฆ๐ฆ1 ๐ฅ๐ฅ is a known solution on ๐ผ๐ผ and ๐ฆ๐ฆ1 ๐ฅ๐ฅ โ 0โ๐ฅ๐ฅ โ ๐ผ๐ผ
P(x) Q(x)
Reduction of Order
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Introduction
Let ๐ฆ๐ฆ ๐ฅ๐ฅ = ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐ฅ๐ฅ โ ๐ฆ๐ฆโฒ ๐ฅ๐ฅ = ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ1โฒ ๐ฅ๐ฅ + ๐๐โฒ ๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐ฅ๐ฅ โ ๐๐โฒ = ๐๐๐๐๐๐โฒ + ๐๐โฒ๐๐๐๐ โ ๐ฆ๐ฆโฒโฒ = ๐๐๐ฆ๐ฆ1โฒโฒ + ๐๐โฒ๐ฆ๐ฆ1โฒ + ๐๐โฒ๐ฆ๐ฆ1โฒ + ๐๐โฒโฒ๐ฆ๐ฆ1 โ ๐๐โฒโฒ = ๐๐๐๐๐๐โฒโฒ + ๐๐๐๐โฒ๐๐๐๐โฒ + ๐๐โฒโฒ๐๐๐๐ Now, ๐ฆ๐ฆโฒโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ = 0 Replacing, ๐๐๐ฆ๐ฆ1โฒโฒ + 2๐๐โฒ๐ฆ๐ฆ1โฒ + ๐๐โฒโฒ๐ฆ๐ฆ1 + ๐๐ ๐ฅ๐ฅ ๐๐๐ฆ๐ฆ1โฒ + ๐๐โฒ๐ฆ๐ฆ1 + ๐๐ ๐ฅ๐ฅ ๐๐๐ฆ๐ฆ1 = 0
Reduction of Order
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Rearranging terms, โ ๐๐ ๐ฆ๐ฆ1โฒโฒ + ๐๐๐ฆ๐ฆ1โฒ + ๐๐๐ฆ๐ฆ1 + ๐ฆ๐ฆ1๐๐โฒโฒ + 2๐ฆ๐ฆ1โฒ + ๐๐๐ฆ๐ฆ1 ๐๐โฒ = 0 โ ๐ฆ๐ฆ1๐๐โฒโฒ + 2๐ฆ๐ฆ1โฒ + ๐๐๐ฆ๐ฆ1 ๐๐โฒ = 0 Let ๐ค๐ค = ๐๐๐ฆ change of variables ๐ค๐คโฒ = ๐๐๐ฆ๐ฆ ๐ฆ๐ฆ1๐ค๐คโฒ + 2๐ฆ๐ฆ1โฒ + ๐๐๐ฆ๐ฆ1 ๐ค๐ค = 0 linear and separable
โ ๐ฆ๐ฆ1๐ค๐คโฒ = โ 2๐ฆ๐ฆ1โฒ + ๐๐๐ฆ๐ฆ1 ๐ค๐ค ๐๐๐๐ ๐ฆ๐ฆ1๐๐๐ค๐ค๐๐๐ฅ๐ฅ
= โ 2๐๐๐ฆ๐ฆ1๐๐๐ฅ๐ฅ
+ ๐๐ ๐ฆ๐ฆ1 ๐ค๐ค
โ๐๐๐ค๐ค๐ค๐ค
= โ1๐ฆ๐ฆ1
2๐๐๐ฆ๐ฆ1๐๐๐ฅ๐ฅ
+ ๐๐ ๐ฆ๐ฆ1 ๐๐๐ฅ๐ฅ
=0 since ๐ฆ๐ฆ1 is a solution
๐๐๐ฆ๐ฆ1โฒโฒ + 2๐๐โฒ๐ฆ๐ฆ1โฒ + ๐๐โฒโฒ๐ฆ๐ฆ1 + ๐๐ ๐ฅ๐ฅ ๐๐๐ฆ๐ฆ1โฒ + ๐๐โฒ๐ฆ๐ฆ1 + ๐๐ ๐ฅ๐ฅ ๐๐๐ฆ๐ฆ1 = 0
Reduction of Order
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๏ฟฝ๐๐๐ค๐ค๐ค๐ค
= ๏ฟฝโ1๐ฆ๐ฆ1
2๐๐๐ฆ๐ฆ1๐๐๐ฅ๐ฅ
+ ๐๐ ๐ฆ๐ฆ1 ๐๐๐ฅ๐ฅ
โ ๏ฟฝ๐๐๐ค๐ค๐ค๐ค
= ๏ฟฝโ2๐๐๐ฆ๐ฆ1๐ฆ๐ฆ1
โ๏ฟฝ๐๐๐๐๐ฅ๐ฅ
โ ln ๐ค๐ค = โ2 ln ๐ฆ๐ฆ1 โ ๏ฟฝ๐๐๐๐๐ฅ๐ฅ + ๐๐
โ ln ๐ค๐ค + 2 ln ๐ฆ๐ฆ1 = โ๏ฟฝ๐๐๐๐๐ฅ๐ฅ + ๐๐ โ ln ๐ค๐ค + ln ๐ฆ๐ฆ12 = โ๏ฟฝ๐๐๐๐๐ฅ๐ฅ + ๐๐
โ ln ๐ค๐ค๐ฆ๐ฆ12 = โ๏ฟฝ๐๐๐๐๐ฅ๐ฅ + ๐๐
โ ๐ค๐ค๐ฆ๐ฆ12 = ๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ +๐๐ = ๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ ๐๐๐๐ = ๐๐1๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ
Reduction of Order
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๐ค๐ค๐ฆ๐ฆ12 = ๐๐1๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ โ ๐ค๐ค = ๐๐1๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ /๐ฆ๐ฆ12 But,
๐ค๐ค = ๐๐โฒ โ ๐ค๐ค = ๐๐โฒ =๐๐๐๐๐๐๐ฅ๐ฅ =
๐๐1๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ
๐ฆ๐ฆ12
โ ๐๐๐๐ =๐๐1๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ
๐ฆ๐ฆ12 ๐๐๐ฅ๐ฅ โ ๏ฟฝ๐๐๐๐ = ๏ฟฝ
๐๐1๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ
๐ฆ๐ฆ12 ๐๐๐ฅ๐ฅ
โ ๐๐ = ๐๐1 ๏ฟฝ๐๐โ โซ ๐๐๐๐๐ฅ๐ฅ
๐ฆ๐ฆ12๐๐๐ฅ๐ฅ + ๐๐2
Let ๐๐1 = 1 & ๐๐2 = 0 & note: ๐ฆ๐ฆ2 ๐ฅ๐ฅ = ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ1(๐ฅ๐ฅ)
โ ๐๐๐๐ ๐๐ = ๐๐๐๐ ๐๐ ๏ฟฝ๐๐โ โซ ๐ท๐ท๐ ๐ ๐๐
๐๐๐๐๐๐(๐๐)๐ ๐ ๐๐ ; ๐ท๐ท ๐๐ =
๐๐๐๐ ๐๐๐๐๐๐ ๐๐
Reduction of Order
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E.g. ๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 0; ๐ผ๐ผ โก 0,โ Let ๐ฆ๐ฆ1 ๐ฅ๐ฅ = ๐ฅ๐ฅ2 be a solution. Find a 2nd solution ๐ฆ๐ฆ2(๐ฅ๐ฅ) and the general solution ๐ฆ๐ฆ(๐ฅ๐ฅ) Solution: ๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 0
โ ๐ฆ๐ฆโฒโฒ โ3๐ฅ๐ฅ๐ฅ๐ฅ2 ๐ฆ๐ฆโฒ +
4๐ฅ๐ฅ2 ๐ฆ๐ฆ = 0
โ ๐ฆ๐ฆโฒโฒ + โ 3๐ฅ๐ฅ ๐ฆ๐ฆโฒ +
4๐ฅ๐ฅ2 ๐ฆ๐ฆ = 0
P(x) Q(x)
Reduction of Order
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According to our derivation, ๐ฆ๐ฆ2 ๐ฅ๐ฅ = ๐ฆ๐ฆ1 ๐ฅ๐ฅ โซ ๐๐โ โซ ๐๐๐๐๐๐
๐ฆ๐ฆ12(๐ฅ๐ฅ)๐๐๐ฅ๐ฅ
= ๐ฅ๐ฅ2 ๏ฟฝ๐๐โ โซ โ3๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
๐ฅ๐ฅ2 2 ๐๐๐ฅ๐ฅ
= ๐ฅ๐ฅ2 ๏ฟฝ๐๐โซ
3๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
๐ฅ๐ฅ4๐๐๐ฅ๐ฅ
= ๐ฅ๐ฅ2 ๏ฟฝ๐๐3 ln ๐ฅ๐ฅ
๐ฅ๐ฅ4๐๐๐ฅ๐ฅ = ๐ฅ๐ฅ2 ๏ฟฝ
๐๐ln ๐ฅ๐ฅ3
๐ฅ๐ฅ4๐๐๐ฅ๐ฅ = ๐ฅ๐ฅ2 ๏ฟฝ
๐ฅ๐ฅ3
๐ฅ๐ฅ4๐๐๐ฅ๐ฅ
= ๐ฅ๐ฅ2 ๏ฟฝ1๐ฅ๐ฅ๐๐๐ฅ๐ฅ = ๐ฅ๐ฅ2 ln ๐ฅ๐ฅ
โ ๐๐๐๐ ๐๐ = ๐๐๐๐ ๐ฅ๐ฅ๐ฅ๐ฅ ๐๐ General solution: ๐๐ ๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐๐๐ ๐๐๐๐๐ฅ๐ฅ๐ฅ๐ฅ ๐๐
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 0
Reduction of Order
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Check: ๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ2 + ๐๐2 ๐ฅ๐ฅ2 ln ๐ฅ๐ฅ
โ ๐ฆ๐ฆโฒ = 2 ๐๐1๐ฅ๐ฅ + ๐๐2 2๐ฅ๐ฅ ln ๐ฅ๐ฅ +๐ฅ๐ฅ2
๐ฅ๐ฅ
โ ๐ฆ๐ฆโฒ = 2๐๐1๐ฅ๐ฅ + 2๐๐2๐ฅ๐ฅ ln ๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ
๐ฆ๐ฆโฒโฒ = 2๐๐1 + 2๐๐2 ln ๐ฅ๐ฅ +๐ฅ๐ฅ๐ฅ๐ฅ + ๐๐2
โ ๐ฆ๐ฆโฒโฒ = 2๐๐1 + 2๐๐2 ln ๐ฅ๐ฅ + 2๐๐2 + ๐๐2 = 2๐๐1 + 3๐๐2 + 2๐๐2 ln ๐ฅ๐ฅ
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 0
Reduction of Order
9/30/2014 Dr. Eli Saber 57
We know, ๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ2 + ๐๐2 ๐ฅ๐ฅ2 ln ๐ฅ๐ฅ ๐ฆ๐ฆโฒ = 2๐๐1๐ฅ๐ฅ + 2๐๐2๐ฅ๐ฅ ln ๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ & ๐ฆ๐ฆโฒโฒ = 2๐๐1 + 2๐๐2 ln ๐ฅ๐ฅ + 2๐๐2 + ๐๐2 = 2๐๐1 + 3๐๐2 + 2๐๐2 ln ๐ฅ๐ฅ Replace in D.E.: ๐ฅ๐ฅ2 2๐๐1 + 3๐๐2 + 2๐๐2 ln ๐ฅ๐ฅ โ 3x 2๐๐1๐ฅ๐ฅ + 2๐๐2๐ฅ๐ฅ ln ๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ + 4 ๐๐1๐ฅ๐ฅ2 + ๐๐2๐ฅ๐ฅ2 ln ๐ฅ๐ฅ = 2๐๐1๐ฅ๐ฅ2 + 3๐๐2๐ฅ๐ฅ2 + 2๐๐2 ln ๐ฅ๐ฅ ๐ฅ๐ฅ2 โ 6๐๐1๐ฅ๐ฅ2 โ 6๐๐2๐ฅ๐ฅ2 ln ๐ฅ๐ฅ โ 3๐๐2๐ฅ๐ฅ2 + 4๐๐1๐ฅ๐ฅ2
+ 4๐๐2๐ฅ๐ฅ2 ln ๐ฅ๐ฅ = ๐๐
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 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
๐๐๐๐๐๐ ๐๐ + ๐๐๐๐โ๐๐๐๐(๐๐โ๐๐) + โฏ+ ๐๐๐๐๐๐โฒ + ๐๐๐๐๐๐ = ๐๐ โข ๐๐๐๐; ๐ ๐ = 0,1, โฆ ,๐๐ are real constant coefficients and ๐๐๐๐ โ 0 Objective: To find a solution to the above homogeneous solution
๐๐๐ก๐ก๐ก order Linear Constant Coefficients Differential Equation
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 60
Auxiliary Equation
Consider the special Case ( 2nd order LCCDE) given as: ๐๐๐ฆ๐ฆโฒโฒ + ๐๐๐ฆ๐ฆโฒ + ๐๐๐ฆ๐ฆ = 0 Try a solution of the form ๐ฆ๐ฆ = ๐๐๐๐๐ฅ๐ฅ โ ๐ฆ๐ฆโฒ = ๐๐๐๐๐๐๐ฅ๐ฅ โ ๐ฆ๐ฆโฒโฒ = ๐๐2๐๐๐๐๐ฅ๐ฅ Substituting back in the given D.E., ๐๐ ๐๐2๐๐๐๐๐ฅ๐ฅ + ๐๐ ๐๐๐๐๐๐๐ฅ๐ฅ + ๐๐ ๐๐๐๐๐ฅ๐ฅ = 0 โ ๐๐๐๐2 + ๐๐๐๐ + ๐๐ ๐๐๐๐๐ฅ๐ฅ = 0 Now, ๐๐๐๐๐๐ โ ๐๐ โ๐๐๐๐๐๐๐๐๐๐ โ ๐๐๐๐๐๐ + ๐๐๐๐ + ๐๐ = ๐๐
Auxiliary Eqn. of the LCCDE
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 61
Introduction
๐๐๐๐๐๐ + ๐๐๐๐ + ๐๐ = ๐๐ Auxiliary Eqn. of the LCCDE The only way that ๐ฆ๐ฆ = ๐๐๐๐๐ฅ๐ฅ can satisfy the D.E. is if ๐๐๐๐2 + ๐๐๐๐ + ๐๐ = 0 Hence, choose ๐๐ as the root of the equation to solve the problem
โ ๐๐1,2 =โ๐๐ ยฑ ๐๐2 โ 4๐๐๐๐
2๐๐
The ๐๐2 โ 4๐๐๐๐ leads to 3 cases:
1) ๐๐2 โ 4๐๐๐๐ > 0
2) ๐๐2 โ 4๐๐๐๐ = 0
3) ๐๐2 โ 4๐๐๐๐ < 0
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 62
Introduction
Case 1: ๐๐๐๐โ๐๐๐๐๐๐ > ๐๐ Here, ๐๐1& ๐๐2 are real and distinct 2 solutions: ๐ฆ๐ฆ1 = ๐๐๐๐1๐ฅ๐ฅ & ๐ฆ๐ฆ2 = ๐๐๐๐2๐ฅ๐ฅ
๐ฆ๐ฆ1&๐ฆ๐ฆ2 are linearly independent
๐ฆ๐ฆ = ๐๐1๐๐๐๐1๐ฅ๐ฅ + ๐๐2๐๐๐๐2๐ฅ๐ฅ is the general solution
๐๐1,2 =โ๐๐ ยฑ ๐๐2 โ 4๐๐๐๐
2๐๐
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 63
Introduction
Case 2: ๐๐๐๐โ๐๐๐๐๐๐ = ๐๐
๐๐1 = ๐๐2 = โ๐๐2๐๐
โ ๐ฆ๐ฆ1 = ๐๐๐๐1๐ฅ๐ฅ & ๐ฆ๐ฆ2 = ๐ฅ๐ฅ๐๐๐๐1๐ฅ๐ฅ
Digression:
๐๐๐ฆ๐ฆโฒโฒ + ๐๐๐ฆ๐ฆโฒ + ๐๐๐ฆ๐ฆ = 0 โ ๐ฆ๐ฆโฒโฒ +๐๐๐๐ ๐ฆ๐ฆโฒ +
๐๐๐๐ ๐ฆ๐ฆ = 0
โ ๐ฆ๐ฆ2 ๐ฅ๐ฅ = ๐ฆ๐ฆ1 ๐ฅ๐ฅ ๏ฟฝ๐๐โ โซ ๐๐ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
๐ฆ๐ฆ1 ๐ฅ๐ฅ 2 ๐๐๐ฅ๐ฅ = ๐๐๐๐1๐ฅ๐ฅ ๏ฟฝ๐๐โ โซ๐๐๐๐ ๐๐๐ฅ๐ฅ
๐๐2๐๐1๐ฅ๐ฅ๐๐๐ฅ๐ฅ = ๐๐๐๐1๐ฅ๐ฅ ๏ฟฝ
๐๐โซ 2๐๐1 ๐๐๐ฅ๐ฅ
๐๐2๐๐1๐ฅ๐ฅ๐๐๐ฅ๐ฅ
๐๐1,2 =โ๐๐ ยฑ ๐๐2 โ 4๐๐๐๐
2๐๐
P(x) Q(x)
(Note: ๐๐1 = โ ๐๐2๐๐โ โ ๐๐
๐๐= 2๐๐1)
Homogeneous Linear Eq. with Constant Coefficients
9/30/2014 Dr. Eli Saber 64
Introduction
๐ฆ๐ฆ2 ๐ฅ๐ฅ = ๐๐๐๐1๐ฅ๐ฅ ๏ฟฝ๐๐โซ 2๐๐1 ๐๐๐ฅ๐ฅ
๐๐2๐๐1๐ฅ๐ฅ๐๐๐ฅ๐ฅ = ๐๐๐๐1๐ฅ๐ฅ ๏ฟฝ
๐๐2๐๐1๐ฅ๐ฅ
๐๐2๐๐1๐ฅ๐ฅ๐๐๐ฅ๐ฅ
= ๐๐๐๐1๐ฅ๐ฅ ๏ฟฝ๐๐๐ฅ๐ฅ = ๐ฅ๐ฅ๐๐๐๐1๐ฅ๐ฅ
๐ฆ๐ฆ2 ๐ฅ๐ฅ = ๐ฅ๐ฅ๐๐๐๐1๐ฅ๐ฅ General solution: ๐ฆ๐ฆ = ๐๐1๐๐๐๐1๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ๐๐๐๐1๐ฅ๐ฅ
๐๐1,2 =โ๐๐ ยฑ ๐๐2 โ 4๐๐๐๐
2๐๐
Homogeneous Linear Eq. with Constant Coefficients
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Introduction
Case 3: ๐๐๐๐โ๐๐๐๐๐๐ < ๐๐ ๐๐1 & ๐๐2 are complex conjugate numbers
๐๐1 = ๐ผ๐ผ + ๐๐๐ฝ๐ฝ & ๐๐2 = ๐ผ๐ผ โ ๐๐๐ฝ๐ฝ
โข ๐ผ๐ผ,๐ฝ๐ฝ > 0 and are real โข ๐๐2 = โ1
General solution: ๐ฆ๐ฆ = ๐๐1๐๐๐๐1๐ฅ๐ฅ + ๐๐2๐๐๐๐2๐ฅ๐ฅ
๐ฆ๐ฆ = ๐๐1๐๐ ๐ผ๐ผ+๐๐๐๐ ๐ฅ๐ฅ + ๐๐2๐๐ ๐ผ๐ผโ๐๐๐๐ ๐ฅ๐ฅ
๐๐1,2 =โ๐๐ ยฑ ๐๐2 โ 4๐๐๐๐
2๐๐
Since ๐ฆ๐ฆ = ๐๐1๐๐ ๐ผ๐ผ+๐๐๐๐ ๐ฅ๐ฅ + ๐๐2๐๐ ๐ผ๐ผโ๐๐๐๐ ๐ฅ๐ฅ is a solution โ๐๐1 &โ๐๐2
Homogeneous Linear Eq. with Constant Coefficients
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Introduction
๐ฆ๐ฆ1 = ๐๐1๐๐ ๐ผ๐ผ+๐๐๐๐ ๐ฅ๐ฅ + ๐๐2๐๐ ๐ผ๐ผโ๐๐๐๐ ๐ฅ๐ฅ = ๐๐ ๐ผ๐ผ+๐๐๐๐ ๐ฅ๐ฅ + ๐๐ ๐ผ๐ผโ๐๐๐๐ ๐ฅ๐ฅ = ๐๐๐ผ๐ผ๐ฅ๐ฅ ๐๐๐๐๐๐๐ฅ๐ฅ + ๐๐โ๐๐๐๐๐ฅ๐ฅ = ๐๐๐ผ๐ผ๐ฅ๐ฅ2 cos๐ฝ๐ฝ๐ฅ๐ฅ ๐ฆ๐ฆ1 = 2๐๐๐ผ๐ผ๐ฅ๐ฅ cos๐ฝ๐ฝ๐ฅ๐ฅ
๐ฆ๐ฆ2 = ๐๐1๐๐ ๐ผ๐ผ+๐๐๐๐ ๐ฅ๐ฅ + ๐๐2๐๐ ๐ผ๐ผโ๐๐๐๐ ๐ฅ๐ฅ = ๐๐ ๐ผ๐ผ+๐๐๐๐ ๐ฅ๐ฅ โ ๐๐ ๐ผ๐ผโ๐๐๐๐ ๐ฅ๐ฅ = ๐๐๐ผ๐ผ๐ฅ๐ฅ ๐๐๐๐๐๐๐ฅ๐ฅ โ ๐๐โ๐๐๐๐๐ฅ๐ฅ = ๐๐๐ผ๐ผ๐ฅ๐ฅ2๐๐ sin๐ฝ๐ฝ๐ฅ๐ฅ ๐ฆ๐ฆ2 = 2๐๐๐๐๐ผ๐ผ๐ฅ๐ฅ sin๐ฝ๐ฝ๐ฅ๐ฅ
Choose ๐๐1 = ๐๐2 = 1 Choose ๐๐1 = 1 & ๐๐2 = โ1
General solution: ๐๐ = ๐๐๐๐๐๐๐ถ๐ถ๐๐ ๐๐๐๐๐๐๐ท๐ท๐๐ + ๐๐๐๐๐๐๐ถ๐ถ๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐ท๐ท๐๐
Homogeneous Linear Eq. with Constant Coefficients
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Alternate Derivation: ๐ฆ๐ฆ = ๐๐1๐๐ ๐ผ๐ผ+๐๐๐๐ ๐ฅ๐ฅ + ๐๐2๐๐ ๐ผ๐ผโ๐๐๐๐ ๐ฅ๐ฅ = ๐๐1๐๐๐ผ๐ผ๐ฅ๐ฅ๐๐๐๐๐๐๐ฅ๐ฅ + ๐๐2๐๐๐ผ๐ผ๐ฅ๐ฅ๐๐โ๐๐๐๐๐ฅ๐ฅ = ๐๐1๐๐๐ผ๐ผ๐ฅ๐ฅ cos๐ฝ๐ฝ๐ฅ๐ฅ + ๐๐ sin๐ฝ๐ฝ๐ฅ๐ฅ + ๐๐2๐๐๐ผ๐ผ๐ฅ๐ฅ cos๐ฝ๐ฝ๐ฅ๐ฅ โ ๐๐ sin๐ฝ๐ฝ๐ฅ๐ฅ = ๐๐1๐๐๐ผ๐ผ๐ฅ๐ฅ cos๐ฝ๐ฝ๐ฅ๐ฅ + ๐๐๐๐1๐๐๐ผ๐ผ๐ฅ๐ฅ sin๐ฝ๐ฝ๐ฅ๐ฅ + ๐๐2๐๐๐ผ๐ผ๐ฅ๐ฅ cos๐ฝ๐ฝ๐ฅ๐ฅ โ ๐๐๐๐2๐๐๐ผ๐ผ๐ฅ๐ฅ sin๐ฝ๐ฝ๐ฅ๐ฅ = ๐๐๐ผ๐ผ๐ฅ๐ฅ ๐๐1 + ๐๐2 cos๐ฝ๐ฝ๐ฅ๐ฅ + ๐๐๐ผ๐ผ๐ฅ๐ฅ ๐๐๐๐1 โ ๐๐๐๐2 sin๐ฝ๐ฝ๐ฅ๐ฅ Hence, ๐๐ = โ๐๐ ๐๐๐ถ๐ถ๐๐ ๐๐๐๐๐๐๐ท๐ท๐๐+โ๐๐ ๐๐๐ถ๐ถ๐๐ ๐๐๐๐๐๐๐ท๐ท๐๐
โ1 โ2
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Example: a) 2๐ฆ๐ฆโฒโฒ โ 5๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 0 Now, 2๐๐2 โ 5๐๐ โ 3 = 0 โ 2๐๐ + 1 ๐๐ โ 3 = 0
โ ๐๐1 = โ 12 ;๐๐2 = 3
General solution: ๐ฆ๐ฆ = ๐๐1๐๐โ 12๐ฅ๐ฅ + ๐๐2๐๐3๐ฅ๐ฅ
Homogeneous Linear Eq. with Constant Coefficients
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b) ๐ฆ๐ฆโฒโฒ โ 10๐ฆ๐ฆโฒ + 25๐ฆ๐ฆ = 0 ๐๐2 โ 10๐๐ + 25 = 0 โ ๐๐ โ 5 2 = 0 โ ๐๐1 = ๐๐2 = 5 General solution: ๐ฆ๐ฆ = ๐๐1๐๐5๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ๐๐5๐ฅ๐ฅ
Homogeneous Linear Eq. with Constant Coefficients
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c) ๐ฆ๐ฆโฒโฒ + 4๐ฆ๐ฆโฒ + 7๐ฆ๐ฆ = 0 โ ๐๐2 + 4๐๐ + 7 = 0
โ ๐๐ =โ4 ยฑ 4 2 โ 4(1)(7)
2(1)=โ4 ยฑ 16โ 28
2
โ ๐๐ = โ4 ยฑ โ12
2=โ4 ยฑ 12 โ1
2=โ4 ยฑ ๐๐ 12
2
โ ๐๐ =โ4 ยฑ ๐๐ 2 3
2= โ2 ยฑ ๐๐ 3
General solution: ๐ฆ๐ฆ = ๐๐1๐๐ โ2+๐๐ 3 ๐ฅ๐ฅ + ๐๐2๐๐ โ2โ๐๐ 3 ๐ฅ๐ฅ or ๐ฆ๐ฆ = ๐๐โ2๐ฅ๐ฅ ๐๐1 cos 3๐ฅ๐ฅ + ๐๐2 sin 3๐ฅ๐ฅ
Homogeneous Linear Eq. with Constant Coefficients
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๐ฆ๐ฆโฒโฒ + ๐พ๐พ2๐ฆ๐ฆ = 0 & ๐ฆ๐ฆโฒโฒ โ ๐พ๐พ2๐ฆ๐ฆ = 0 ๐พ๐พ:real Where do we see these equations??
D.E. Free of Undamped Motion: ๐๐2๐ฅ๐ฅ๐๐๐ก๐ก2
+ ๐๐2๐ฅ๐ฅ = 0 With the solution: ๐๐ = ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐๐๐
Two important Equations
HOW?
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed. Ref. D. Zill & W. Wright, Advanced
Engineering Mathematics. 5th Ed.
Homogeneous Linear Eq. with Constant Coefficients
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๐ฆ๐ฆโฒโฒ + ๐พ๐พ2๐ฆ๐ฆ = 0 & ๐ฆ๐ฆโฒโฒ โ ๐พ๐พ2๐ฆ๐ฆ = 0 ๐พ๐พ:real ๐๐2 + ๐พ๐พ2 = 0 โ ๐๐2 = โ๐พ๐พ2 = ๐พ๐พ2๐๐2 โ ๐๐ = ยฑ๐พ๐พ๐๐ Which results in: ๐ฆ๐ฆ = ๐๐1๐๐๐พ๐พ๐๐๐ฅ๐ฅ + ๐๐2๐๐โ๐พ๐พ๐๐๐ฅ๐ฅ or ๐ฆ๐ฆ = ๐๐1 cos๐พ๐พ๐ฅ๐ฅ + ๐๐2 sin๐พ๐พ๐ฅ๐ฅ
Two important Equations
Homogeneous Linear Eq. with Constant Coefficients
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๐ฆ๐ฆโฒโฒ + ๐พ๐พ2๐ฆ๐ฆ = 0 & ๐ฆ๐ฆโฒโฒ โ ๐พ๐พ2๐ฆ๐ฆ = 0 ๐พ๐พ:real ๐๐2 โ ๐พ๐พ2 = 0 โ ๐๐2 = ๐พ๐พ2 โ ๐๐ = ยฑ๐พ๐พ Which results in: ๐ฆ๐ฆ = ๐๐1๐๐๐พ๐พ๐ฅ๐ฅ + ๐๐2๐๐โ๐พ๐พ๐ฅ๐ฅ
Two important Equations
Homogeneous Linear Eq. with Constant Coefficients
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Note: ๐ฆ๐ฆโฒโฒ โ ๐พ๐พ2๐ฆ๐ฆ = 0 ๐๐ = ๐๐๐๐๐๐๐ฒ๐ฒ๐๐ + ๐๐๐๐๐๐โ๐ฒ๐ฒ๐๐
โข If ๐๐1 = ๐๐2 = 12โ ๐ฆ๐ฆ = 1
2 ๐๐๐พ๐พ๐ฅ๐ฅ + 1
2 ๐๐โ๐พ๐พ๐ฅ๐ฅ = cosh๐พ๐พ๐ฅ๐ฅ
โข If ๐๐1 = 12
& ๐๐2 = โ12โ ๐ฆ๐ฆ = 1
2 ๐๐๐พ๐พ๐ฅ๐ฅ โ 1
2 ๐๐โ๐พ๐พ๐ฅ๐ฅ = sinh๐พ๐พ๐ฅ๐ฅ
โข Since cosh๐พ๐พ๐ฅ๐ฅ & sinh๐พ๐พ๐ฅ๐ฅ are linearly independent
โ Alternate solution of ๐ฆ๐ฆโฒโฒ โ ๐พ๐พ2๐ฆ๐ฆ = 0 is ๐๐ = ๐๐๐๐ ๐๐๐๐๐๐๐๐๐ฒ๐ฒ๐๐ + ๐๐๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐๐ฒ๐ฒ๐๐
Two important Equations
Homogeneous Linear Eq. with Constant Coefficients
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๐๐๐๐ ๐ฅ๐ฅ ๐๐๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐
+ ๐๐๐๐โ1 ๐ฅ๐ฅ ๐๐๐๐โ1๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐โ1
+ โฆ + ๐๐1 ๐ฅ๐ฅ ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = 0
,where ๐๐๐๐ , ๐ ๐ = 0,1, โฆ ,๐๐ are real constants Auxiliary Equation: ๐๐๐๐๐๐๐๐ + ๐๐๐๐โ1๐๐๐๐โ1 + โฏ+ ๐๐2๐๐2 + ๐๐1๐๐ + ๐๐0 ๐๐0 = 0 Case 1: If all roots are distinct โ general solution is given by:
๐ฆ๐ฆ = ๐๐1๐๐๐๐1๐ฅ๐ฅ + ๐๐2๐๐๐๐2๐ฅ๐ฅ + โฏ+ ๐๐๐๐๐๐๐๐๐๐๐ฅ๐ฅ (similar to a 2nd order D.E.)
Higher Order Equations
Homogeneous Linear Eq. with Constant Coefficients
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Case 2: For multiple roots, if ๐๐1 is a root with multiplicity ๐พ๐พ i.e. ๐พ๐พ roots equal to ๐๐1
Then the general solution will have terms: ๐๐๐๐1๐ฅ๐ฅ, ๐ฅ๐ฅ๐๐๐๐1๐ฅ๐ฅ, ๐ฅ๐ฅ2๐๐๐๐1๐ฅ๐ฅ,โฆ, ๐ฅ๐ฅ๐๐โ1๐๐๐๐1๐ฅ๐ฅ Case 3: Complex roots appear in conjugate pairs when the coefficients of the D.E. are real
Higher Order Equations
Homogeneous Linear Eq. with Constant Coefficients
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E.g. ๐ฆ๐ฆโฒโฒโฒ + 3๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆ = 0 Auxiliary equation: ๐๐3 + 3๐๐2 โ 4 = 0 By inspection, ๐๐1 = 1 is a root since 1 3 + 3 1 2 โ 4 = 1 + 3 โ 4 = 4 โ 4 = ๐๐ Dividing the Auxiliary equation ๐๐3 + 3๐๐2 โ 4 = 0 by ๐๐ โ 1 , we get ๐๐2 + 4๐๐ + 4 โ ๐๐โ 1 ๐๐2 + 4๐๐ + 4 = ๐๐3 + 3๐๐2 โ 4 โ ๐๐โ 1 ๐๐2 + 4๐๐ + 4 = 0 โ ๐๐โ 1 ๐๐ + 2 2 = 0 Roots: ๐๐1 = 1,๐๐2 = ๐๐3 = โ2 General solution: ๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐โ๐๐๐๐ + ๐๐๐๐๐๐๐๐โ๐๐๐๐
Higher Order Equations
Section 3.4 Undetermined Coefficients
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Undetermined Coefficients
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By: 1. Finding a complementary solution ๐ฆ๐ฆ๐๐ for the homogeneous equation. 2. Finding a particular solution ๐ฆ๐ฆ๐๐.
โ ๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ ๐ ๐๐๐๐๐๐๐ก๐ก๐ ๐ ๐๐๐๐: ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
Solve a non-homogeneous Linear Differential Equation: ๐๐๐๐๐ฆ๐ฆ ๐๐ + ๐๐๐๐โ1๐ฆ๐ฆ ๐๐โ1 + โฏ+ ๐๐1๐ฆ๐ฆ1 + ๐๐0๐ฆ๐ฆ0 = ๐๐ ๐ฅ๐ฅ
Undetermined Coefficients
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Method of undetermined coefficient ๐๐๐๐ โEducated guess about the form of ๐๐๐๐โ Method is limited to non-homogeneous linear D.E. such that: 1. The coefficient ๐๐๐๐ , ๐ ๐ = 0,1,2, โฆ ,๐๐ are constant. 2. ๐๐ ๐ฅ๐ฅ is a constant, polynomial function, exponential function, sin or cos or
finite sums and products of these functions. E.g.: ๐๐ ๐ฅ๐ฅ = 10;๐๐ ๐ฅ๐ฅ = ๐ฅ๐ฅ2 โ 5๐ฅ๐ฅ, โฆ โฆ
Undetermined Coefficients
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Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.
Undetermined Coefficients
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E.g. 1: ๐ฆ๐ฆโฒโฒ + 4๐ฆ๐ฆโฒ โ 2๐ฆ๐ฆ = 2๐ฅ๐ฅ2 โ 3๐ฅ๐ฅ + 6 Step 1: Solve the associated Homogeneous equation. ๐ฆ๐ฆโฒโฒ + 4๐ฆ๐ฆโฒ โ 2๐ฆ๐ฆ = 0
๐๐2 + 4๐๐ โ 2 = 0 โ ๐๐ =โ4 ยฑ 16 โ 4 1 โ2
2
โ ๐๐ =โ4 ยฑ 24
2 =โ4 ยฑ 2 6
2
โ ๐๐ = โ2 ยฑ 6 โ ๐๐1 = โ2 โ 6 ๐๐๐๐๐๐ ๐๐2 = โ2 + 6
๐๐ ๐ฅ๐ฅ
Undetermined Coefficients
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๐ฆ๐ฆ๐๐ = ๐๐1๐๐๐๐1๐ฅ๐ฅ + ๐๐2๐๐๐๐2๐ฅ๐ฅ
โ ๐ฆ๐ฆ๐๐ = ๐๐1๐๐ โ2โ 6 ๐ฅ๐ฅ + ๐๐2๐๐ โ2+ 6 ๐ฅ๐ฅ Step 2: Note ๐๐(๐ฅ๐ฅ) is a quadratic โ assume a particular solution of quadratic form. (See Table 3.4.1) โ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ2 + ๐ต๐ต๐ฅ๐ฅ + ๐ถ๐ถ โ ๐ฆ๐ฆ๐๐โฒ = 2๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐โฒโฒ = 2๐ด๐ด Substitute into D.E. ๐ฆ๐ฆ๐๐โฒโฒ + 4๐ฆ๐ฆ๐๐โฒ โ 2๐ฆ๐ฆ๐๐ = 3๐ฅ๐ฅ2 โ 3๐ฅ๐ฅ + 6 โ 2๐ด๐ด + 4 2๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต โ 2 ๐ด๐ด๐ฅ๐ฅ2 + ๐ต๐ต๐ฅ๐ฅ + ๐ถ๐ถ = 2๐ฅ๐ฅ2 โ 3๐ฅ๐ฅ + 6
๐๐1 = โ2 โ 6 ๐๐๐๐๐๐ ๐๐2 = โ2 + 6
Undetermined Coefficients
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โ 2๐ด๐ด + 8๐ด๐ด๐ฅ๐ฅ + 4๐ต๐ต โ 2๐ด๐ด๐ฅ๐ฅ2 โ 2๐ต๐ต๐ฅ๐ฅ โ 2๐ถ๐ถ = 2๐ฅ๐ฅ2 โ 3๐ฅ๐ฅ + 6 โ โ๐๐๐จ๐จ๐ฅ๐ฅ2 + ๐๐๐จ๐จ โ ๐๐๐ฉ๐ฉ ๐ฅ๐ฅ + ๐๐๐จ๐จ + ๐๐๐ฉ๐ฉ โ ๐๐๐ช๐ช = 2๐ฅ๐ฅ2 โ 3๐ฅ๐ฅ + 6 โข โ2๐ด๐ด = 2 โ ๐ด๐ด = โ1
โข 8๐ด๐ด โ 2๐ต๐ต = โ3 โ 2๐ต๐ต = 8๐ด๐ด + 3 = 8 โ1 + 3 = โ5
โ 2๐ต๐ต = โ5 โ ๐ต๐ต =โ52
โข 2๐ด๐ด + 4๐ต๐ต โ 2๐ถ๐ถ = 6 โ 2๐ถ๐ถ = 2๐ด๐ด + 4๐ต๐ต โ 6 = 2 โ1 + 4 โ52
โ 6
2๐ถ๐ถ = โ2 โ 10 โ 6 โ 2๐ถ๐ถ = โ18 โ ๐ถ๐ถ = โ9
Undetermined Coefficients
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โ ๐ฆ๐ฆ๐๐ = โ๐ฅ๐ฅ2 โ52๐ฅ๐ฅ โ 9
โ ๐ฆ๐ฆ๐๐ = ๐๐1๐๐ โ2โ 6 ๐ฅ๐ฅ + ๐๐2๐๐ โ2+ 6 ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ ๐ ๐๐๐๐๐๐๐ก๐ก๐ ๐ ๐๐๐๐: ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โ ๐๐ = ๐๐๐๐๐๐โ ๐๐+ ๐๐ ๐๐ + ๐๐๐๐๐๐ โ๐๐+ ๐๐ ๐๐ โ ๐๐๐๐ โ๐๐๐๐๐๐ โ ๐๐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 86
E.g. 2: ๐ฆ๐ฆโฒโฒ โ ๐ฆ๐ฆโฒ + ๐ฆ๐ฆ = 2 sin 3๐ฅ๐ฅ Step 1: Find ๐ฆ๐ฆ๐๐ ๐๐๐๐๐๐ ๐ฆ๐ฆโฒโฒ โ ๐ฆ๐ฆโฒ + ๐ฆ๐ฆ = 0
๐๐2 โ๐๐ + 1 = 0 โ ๐๐ =1 ยฑ 1 โ 4 1 1
2
โ ๐๐ =1 ยฑ 3๐๐2
2 โ ๐๐ =12 ยฑ ๐๐
32
โ ๐๐1 =12 + ๐๐
32 ๐๐๐๐๐๐ ๐๐2 =
12 โ ๐๐
32
โ ๐ฆ๐ฆ๐๐ = ๐๐1๐๐12+๐๐
32 ๐ฅ๐ฅ + ๐๐2๐๐
12 โ ๐๐ 3
2 ๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 87
Step 2: ๐น๐น๐ ๐ ๐๐๐๐ ๐ฆ๐ฆ๐๐. ๐ด๐ด๐ ๐ ๐ ๐ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐ = ๐ด๐ด cos3๐ฅ๐ฅ + ๐ต๐ต sin 3๐ฅ๐ฅ (see Table 3.4.1) โ ๐ฆ๐ฆ๐๐โฒ = โ3๐ด๐ด sin3๐ฅ๐ฅ + 3๐ต๐ต cos3๐ฅ๐ฅ ๐ฆ๐ฆ๐๐โฒโฒ = โ9๐ด๐ด cos3๐ฅ๐ฅ โ 9๐ต๐ต sin 3๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐โฒโฒ โ ๐ฆ๐ฆ๐๐โฒ + ๐ฆ๐ฆ = 2 sin 3๐ฅ๐ฅ โ โ9๐ด๐ด cos3๐ฅ๐ฅ โ 9๐ต๐ต sin 3๐ฅ๐ฅ โ โ3๐ด๐ด sin 3๐ฅ๐ฅ + 3๐ต๐ต cos3๐ฅ๐ฅ + ๐ด๐ด cos3๐ฅ๐ฅ + ๐ต๐ต sin 3๐ฅ๐ฅ
= 2 sin 3๐ฅ๐ฅ โ โ9๐ด๐ด cos3๐ฅ๐ฅ โ 9๐ต๐ต sin3๐ฅ๐ฅ + 3๐ด๐ด sin 3๐ฅ๐ฅ โ 3๐ต๐ต cos3๐ฅ๐ฅ + ๐ด๐ด cos3๐ฅ๐ฅ + ๐ต๐ต sin3๐ฅ๐ฅ = 2 sin ๐ฅ๐ฅ โ โ9๐ด๐ด cos3๐ฅ๐ฅ โ 3๐ต๐ต cos3๐ฅ๐ฅ + ๐ด๐ด cos3๐ฅ๐ฅ โ 9๐ต๐ต sin 3๐ฅ๐ฅ + 3๐ด๐ด sin 3๐ฅ๐ฅ + ๐ต๐ต sin3๐ฅ๐ฅ = 2 sin ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐๐๐ โ๐๐๐จ๐จ โ ๐๐๐ฉ๐ฉ + ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐๐๐ โ๐๐๐ฉ๐ฉ + ๐๐๐จ๐จ = 2 sin ๐ฅ๐ฅ โ 3๐ด๐ด โ 8๐ต๐ต = 2 ๐๐๐๐๐๐ โ 8๐ด๐ด โ 3๐ต๐ต = 0
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 88
โ ๐ด๐ด =6
73 ๐๐๐๐๐๐ ๐ต๐ต = โ
1673
โ ๐ฆ๐ฆ๐๐ =6
73cos3๐ฅ๐ฅ โ
1673
sin 3๐ฅ๐ฅ
โ ๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ ๐ ๐๐๐๐๐๐๐ก๐ก๐ ๐ ๐๐๐๐: ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โ ๐๐ = ๐๐๐๐๐๐๐๐๐๐+๐๐
๐๐๐๐ ๐๐ + ๐๐๐๐๐๐
๐๐๐๐โ๐๐
๐๐๐๐ ๐๐ +
๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ โ
๐๐๐๐๐๐๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐๐๐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 89
E.g. 3: Using superposition ๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
Given:
๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
Step 1: ๐ฆ๐ฆ๐๐ = ๐๐1๐๐โ๐ฅ๐ฅ + ๐๐2๐๐3๐ฅ๐ฅ Step 2: Find ๐ฆ๐ฆ๐๐
polynomial exponential
๐๐๐๐(๐๐) ๐๐๐๐(๐๐)
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 90
๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 0 ๐๐2 โ 2๐๐ โ 3 = 0 โ (๐๐ โ 3)(๐๐ + 1) = 0 โ ๐๐1 = 3,๐๐2 = โ1 Complimentary Solution: ๐ฆ๐ฆ๐๐ = ๐๐1๐๐โ๐ฅ๐ฅ + ๐๐2๐๐3๐ฅ๐ฅ
๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 91
โ ๐๐๐ ๐ ๐ ๐ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐ = ๐ฆ๐ฆ๐๐1 + ๐ฆ๐ฆ๐๐2 ๐ ๐ ๐๐๐ ๐ ๐๐๐๐๐ ๐ ๐๐๐ ๐ ๐ ๐ ๐ก๐ก๐ ๐ ๐๐๐๐ โ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต + ๐ถ๐ถ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ + ๐ท๐ท๐๐2๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐โฒ = ๐ด๐ด + ๐ถ๐ถ ๐๐2๐ฅ๐ฅ + 2๐ฅ๐ฅ๐๐2๐ฅ๐ฅ + 2๐ท๐ท๐๐2๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐โฒโฒ = 2๐ถ๐ถ๐๐2๐ฅ๐ฅ + 2๐ถ๐ถ ๐๐2๐ฅ๐ฅ + 2๐ฅ๐ฅ๐๐2๐ฅ๐ฅ + 4๐ท๐ท๐๐2๐ฅ๐ฅ โ ๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ Substitute ๐ฆ๐ฆโฒโฒ ๐๐๐๐๐๐ ๐ฆ๐ฆโฒ๐๐๐๐๐๐ ๐ฆ๐ฆ: โ 2๐ถ๐ถ๐๐2๐ฅ๐ฅ + 2๐ถ๐ถ๐๐2๐ฅ๐ฅ + 4๐ถ๐ถ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ + 4๐ท๐ท๐๐2๐ฅ๐ฅ โ 2๐ด๐ด โ 2๐ถ๐ถ๐๐2๐ฅ๐ฅ โ 4๐ถ๐ถ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ โ 4๐ท๐ท๐๐2๐ฅ๐ฅ โ 3๐ด๐ด๐ฅ๐ฅ
โ 3๐ต๐ต โ 3๐ถ๐ถ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ โ 3๐ท๐ท๐๐2๐ฅ๐ฅ = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
๐๐๐๐๐๐ :for ๐๐๐๐(๐๐)
๐๐๐๐๐๐ :for ๐๐๐๐(๐๐)
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 92
โ โ3๐ด๐ด๐ฅ๐ฅ โ 2๐ด๐ด โ 3๐ต๐ต โ 3๐ถ๐ถ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ 2๐ถ๐ถ โ 3๐ท๐ท = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
โข โ3๐ด๐ด = 4 โ ๐ด๐ด = โ43
โข โ2๐ด๐ด โ 3๐ต๐ต = โ5 โ โ3๐ต๐ต = โ5 + 2๐ด๐ด = โ5 + 2 โ43
โ โ3๐ต๐ต = โ5 โ83 = โ
153 โ
83 = โ
233 โ ๐ต๐ต =
239
โข โ3๐ถ๐ถ = 6 โ ๐ถ๐ถ = โ2
โข 2๐ถ๐ถ โ 3๐ท๐ท = 0 โ 3๐ท๐ท = 2๐ถ๐ถ = 2 โ2 = โ4 โ D = โ43
๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 93
๐ฆ๐ฆ๐๐ = โ43๐ฅ๐ฅ +
239โ 2๐ฅ๐ฅ๐๐2๐ฅ๐ฅ โ
43๐๐2๐ฅ๐ฅ
๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ ๐ ๐๐๐๐๐๐๐ก๐ก๐ ๐ ๐๐๐๐: ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โ ๐๐ = ๐๐๐๐๐๐โ๐๐ + ๐๐๐๐๐๐๐๐๐๐ โ๐๐๐๐๐๐ +
๐๐๐๐๐๐ โ ๐๐๐๐๐๐๐๐๐๐ โ
๐๐๐๐๐๐
๐๐๐๐
๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ โ 3๐ฆ๐ฆ = 4๐ฅ๐ฅ โ 5 + 6๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 94
E.g. 4: ๐ฆ๐ฆโฒโฒ โ 5๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 8๐๐๐ฅ๐ฅ Step 1: Find ๐ฆ๐ฆ๐๐ โ ๐ฆ๐ฆ๐๐ = ๐ถ๐ถ1๐๐๐ฅ๐ฅ + ๐ถ๐ถ2๐๐4๐ฅ๐ฅ Step 2: Find ๐ฆ๐ฆ๐๐ โ ๐๐๐ ๐ ๐ ๐ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐๐๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐โฒ = ๐ด๐ด๐๐๐ฅ๐ฅ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐โฒโฒ = ๐ด๐ด๐๐๐ฅ๐ฅ Re-substituting back โ ๐ด๐ด๐๐๐ฅ๐ฅ โ 5๐ด๐ด๐๐๐ฅ๐ฅ + 4๐ด๐ด๐๐๐ฅ๐ฅ = 8๐๐๐ฅ๐ฅ โ 0๐ด๐ด๐๐๐ฅ๐ฅ = 8๐๐๐ฅ๐ฅ โ ๐๐ = ๐๐๐๐๐๐ โ ๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ Note: ๐ฆ๐ฆ๐๐ = ๐๐1๐๐๐ฅ๐ฅ + ๐๐2๐๐4๐ฅ๐ฅ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐๐๐ฅ๐ฅ โข ๐๐๐ฅ๐ฅ is already present in ๐ฆ๐ฆ๐๐ โ ๐๐๐ฅ๐ฅ is a solution of the homogeneous equation. โ ๐ด๐ด๐๐๐ฅ๐ฅ when substituted into the D.E. produces zero โ(see case II in section 3.3)
Not Independent
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 95
๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ ๐ฆ๐ฆ๐๐โฒ = ๐ด๐ด๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐โฒโฒ = ๐ด๐ด๐๐๐ฅ๐ฅ + ๐ด๐ด๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ = 2๐ด๐ด๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ 2๐ด๐ด๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ 5 ๐ด๐ด๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ + 4๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ = 8๐๐๐ฅ๐ฅ โ 2๐ด๐ด๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ 5๐ด๐ด๐๐๐ฅ๐ฅ โ 5๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ + 4๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ = 8๐๐๐ฅ๐ฅ
โ โ3๐ด๐ด๐๐๐ฅ๐ฅ = 8๐๐๐ฅ๐ฅ โ โ3๐ด๐ด = 8 โ ๐ด๐ด = โ83
โ ๐ฆ๐ฆ๐๐ = โ83๐ฅ๐ฅ๐๐๐ฅ๐ฅ
Now, ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โ ๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ โ๐๐๐๐๐๐๐๐๐๐
๐ฆ๐ฆ๐๐ = ๐๐1๐๐๐ฅ๐ฅ + ๐๐2๐๐4๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 96
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 97
E.g. 5.1 ๐ฆ๐ฆโฒโฒ โ 8๐ฆ๐ฆโฒ + 25๐ฆ๐ฆ = 5๐ฅ๐ฅ3๐๐โ๐ฅ๐ฅ โ 7๐๐โ๐ฅ๐ฅ โ ๐ฆ๐ฆโฒโฒ โ 8๐ฆ๐ฆโฒ + 25๐ฆ๐ฆ = 5๐ฅ๐ฅ3 โ 7 ๐๐โ๐ฅ๐ฅ Homogeneous solution: ๐ฆ๐ฆ๐๐ = ๐๐4๐ฅ๐ฅ ๐๐1 cos3๐ฅ๐ฅ + ๐๐2 sin 3๐ฅ๐ฅ Assume ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ3 + ๐ต๐ต๐ฅ๐ฅ2 + ๐ถ๐ถ๐ฅ๐ฅ + ๐ธ๐ธ ๐๐โ๐ฅ๐ฅ Note no duplication of terms between ๐ฆ๐ฆ๐๐๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐
Case I: No function in the assumed particular solution ๐ฆ๐ฆ๐๐ is a solution of the associated Homogeneous Differential Equation.
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 98
E.g. 5.2 ๐ฆ๐ฆโฒโฒ + 4๐ฆ๐ฆ = ๐ฅ๐ฅ cos๐ฅ๐ฅ
๐ฆ๐ฆ๐๐ = ๐๐1 cos2๐ฅ๐ฅ + ๐๐2 sin2๐ฅ๐ฅ ๐ด๐ด๐ ๐ ๐ ๐ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต cos๐ฅ๐ฅ + ๐ถ๐ถ๐ฅ๐ฅ + ๐ธ๐ธ sin ๐ฅ๐ฅ No duplication of terms between ๐ฆ๐ฆ๐๐ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 99
E.g. 6: ๐ฆ๐ฆโฒโฒ โ 9๐ฆ๐ฆโฒ + 14๐ฆ๐ฆ = 3๐ฅ๐ฅ2 โ 5 sin 2๐ฅ๐ฅ + 7๐ฅ๐ฅ๐๐6๐ฅ๐ฅ Given ๐ฆ๐ฆ๐๐ = ๐๐1๐๐2๐ฅ๐ฅ + ๐๐2๐๐7๐ฅ๐ฅ (computer earlier) Since ๐๐ ๐ฅ๐ฅ has various terms, form ๐ฆ๐ฆ๐๐ by superposition 3๐ฅ๐ฅ2 โ ๐ฆ๐ฆ๐๐1 = ๐ด๐ด๐ฅ๐ฅ2 + ๐ต๐ต๐ฅ๐ฅ + ๐ถ๐ถ โ5 sin 2๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐2 = ๐ธ๐ธ cos2๐ฅ๐ฅ + ๐น๐น sin 2๐ฅ๐ฅ 7๐ฅ๐ฅ๐๐6๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐3 = ๐บ๐บ๐ฅ๐ฅ + ๐ป๐ป ๐๐6๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐ = ๐ฆ๐ฆ๐๐1 + ๐ฆ๐ฆ๐๐2 + ๐ฆ๐ฆ๐๐3 โ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ2 + ๐ต๐ต๐ฅ๐ฅ + ๐ถ๐ถ + ๐ธ๐ธ cos2๐ฅ๐ฅ + ๐น๐น sin 2๐ฅ๐ฅ + ๐บ๐บ๐ฅ๐ฅ + ๐ป๐ป ๐๐6๐ฅ๐ฅ Note: No duplication of terms between ๐ฆ๐ฆ๐๐ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 100
E.g. 7: ๐ฆ๐ฆโฒโฒ โ 2๐ฆ๐ฆโฒ + ๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ With ๐ฆ๐ฆ๐๐ = ๐๐1๐๐๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ๐๐๐ฅ๐ฅ (computed earlier) What do we assume for ๐ฆ๐ฆ๐๐? โข ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐๐๐ฅ๐ฅ โ will fail since ๐๐๐ฅ๐ฅ is part of ๐ฆ๐ฆ๐๐
โข ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ will fail since ๐ฅ๐ฅ๐๐๐ฅ๐ฅ is part of ๐ฆ๐ฆ๐๐
โ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ2๐๐๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐โฒ = 2๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ2๐๐๐ฅ๐ฅ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐โฒโฒ = 2๐ด๐ด๐๐๐ฅ๐ฅ + 2๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ + 2๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ2๐๐๐ฅ๐ฅ
Case II: A function in the potential particular solution is also a solution of the associated Homogeneous Differential Equation.
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 101
โ 2๐ด๐ด๐๐๐ฅ๐ฅ + 4๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ2๐๐๐ฅ๐ฅ โ 4๐ด๐ด๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ 2๐ด๐ด๐ฅ๐ฅ2๐๐๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ2๐๐๐ฅ๐ฅ = ๐๐๐ฅ๐ฅ
โ 2๐ด๐ด๐๐๐ฅ๐ฅ = ๐๐๐ฅ๐ฅ โ 2๐ด๐ด = 1 โ ๐ด๐ด =12
โ ๐๐๐๐ =๐๐๐๐๐๐
๐๐๐๐๐๐
โ ๐๐๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ +๐๐๐๐๐๐
๐๐๐๐๐๐
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 102
Hence if ๐๐ ๐ฅ๐ฅ consists of no terms similar to Table 3.4.1 and that:
๐ฆ๐ฆ๐๐ = ๐ฆ๐ฆ๐๐1 + ๐ฆ๐ฆ๐๐2 + โฏ+ ๐ฆ๐ฆ๐๐๐๐ (assumption) Where ๐ฆ๐ฆ๐๐๐๐ , ๐ ๐ = 1, 2, 3, โฆ โฆ ,๐๐ are potential particular solution Multiplication rule: If any ๐๐๐๐๐๐ contains terms that duplicate terms in ๐๐๐๐, then that ๐๐๐๐๐๐ must be multiplied by ๐๐๐๐, where n is the smallest positive integer that eliminates that duplication
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 103
E.g. 8: ๐ฆ๐ฆโฒโฒ + ๐ฆ๐ฆ = 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ Initial conditions: ๐ฆ๐ฆ ๐๐ = 0;๐ฆ๐ฆโฒ ๐๐ = 2 Step 1: ๐๐๐๐๐๐๐ ๐ ๐๐ ๐ฆ๐ฆโฒโฒ + ๐ฆ๐ฆ = 0 ๐๐2 + 1 = 0 โ ๐๐2 = โ1 = ๐๐2 โ ๐๐ = ยฑ๐๐ โ ๐ฆ๐ฆ๐๐ = ๐๐1๐๐๐๐๐ฅ๐ฅ + ๐๐2๐๐โ๐๐๐ฅ๐ฅ = ๐๐1 cos๐ฅ๐ฅ + ๐๐๐๐1 sin ๐ฅ๐ฅ + ๐๐2 cos ๐ฅ๐ฅ โ ๐๐๐๐2 sin ๐ฅ๐ฅ = ๐๐1 + ๐๐2 cos๐ฅ๐ฅ + ๐๐ ๐๐1 โ ๐๐2 sin ๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐ = ๐ผ๐ผ1 cos ๐ฅ๐ฅ + ๐ผ๐ผ2 sin ๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 104
๐๐ ๐ฅ๐ฅ = 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ
โข 4๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐๐ด๐ด๐ฅ๐ฅ+๐ต๐ต
โข 10 sin ๐ฅ๐ฅ โ ๐ถ๐ถ cos ๐ฅ๐ฅ + ๐ธ๐ธ sin ๐ฅ๐ฅ
(but these are part of ๐ฆ๐ฆ๐๐)
= ๐ถ๐ถ๐ฅ๐ฅ cos๐ฅ๐ฅ + ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ โ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต + ๐ถ๐ถ๐ฅ๐ฅ cos ๐ฅ๐ฅ + ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ โข ๐ฆ๐ฆ๐๐โฒ = ๐ด๐ด + ๐ถ๐ถ cos ๐ฅ๐ฅ + ๐ฅ๐ฅ โ sin ๐ฅ๐ฅ + ๐ธ๐ธ sin ๐ฅ๐ฅ + ๐ฅ๐ฅ cos๐ฅ๐ฅ = ๐ด๐ด + ๐ถ๐ถ cos๐ฅ๐ฅ โ ๐ถ๐ถ๐ฅ๐ฅ sin ๐ฅ๐ฅ + ๐ธ๐ธ sin ๐ฅ๐ฅ + ๐ธ๐ธ๐ฅ๐ฅ cos ๐ฅ๐ฅ
๐ฆ๐ฆโฒโฒ + ๐ฆ๐ฆ = 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 105
โข ๐ฆ๐ฆ๐๐โฒโฒ = โ๐ถ๐ถ sin ๐ฅ๐ฅ โ ๐ถ๐ถ sin ๐ฅ๐ฅ + ๐ฅ๐ฅ cos๐ฅ๐ฅ + ๐ธ๐ธ cos ๐ฅ๐ฅ + ๐ธ๐ธ cos๐ฅ๐ฅ โ ๐ฅ๐ฅ sin ๐ฅ๐ฅ = โ๐ถ๐ถ sin ๐ฅ๐ฅ โ ๐ถ๐ถ sin ๐ฅ๐ฅ โ ๐ถ๐ถ๐ฅ๐ฅ cos ๐ฅ๐ฅ + ๐ธ๐ธ cos ๐ฅ๐ฅ + ๐ธ๐ธ cos ๐ฅ๐ฅ โ ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ = โ2๐ถ๐ถ sin ๐ฅ๐ฅ + 2๐ธ๐ธ cos ๐ฅ๐ฅ โ ๐ถ๐ถ๐ฅ๐ฅ cos ๐ฅ๐ฅ โ ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ We have: ๐ฆ๐ฆโฒโฒ + ๐ฆ๐ฆ = 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ โ โ2๐ถ๐ถ sin ๐ฅ๐ฅ + 2๐ธ๐ธ cos ๐ฅ๐ฅ โ ๐ถ๐ถ๐ฅ๐ฅ cos๐ฅ๐ฅ โ ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต + ๐ถ๐ถ๐ฅ๐ฅ cos ๐ฅ๐ฅ + ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ
= 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ
๐ฆ๐ฆโฒโฒ + ๐ฆ๐ฆ = 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 106
โ โ2๐ถ๐ถ sin ๐ฅ๐ฅ + 2๐ธ๐ธ cos ๐ฅ๐ฅ โ ๐ถ๐ถ๐ฅ๐ฅ cos ๐ฅ๐ฅ โ ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ + ๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต + ๐ถ๐ถ๐ฅ๐ฅ cos ๐ฅ๐ฅ + ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ
= 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ โ ๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต โ 2๐ถ๐ถ sin ๐ฅ๐ฅ โ ๐ถ๐ถ๐ฅ๐ฅ cos ๐ฅ๐ฅ + ๐ถ๐ถ๐ฅ๐ฅ cos ๐ฅ๐ฅ + 2๐ธ๐ธ cos ๐ฅ๐ฅ โ ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ + ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ
= 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ โ ๐จ๐จ๐๐ + ๐ฉ๐ฉ + โ๐๐๐ช๐ช๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐ + ๐๐๐๐๐๐๐๐๐๐๐๐ = ๐๐๐๐ + ๐๐ + ๐๐๐๐๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐ + ๐๐ โ ๐ด๐ด๐ฅ๐ฅ = 4๐ฅ๐ฅ โ ๐ด๐ด = 4 โ ๐ต๐ต = 0 โ โ2๐ถ๐ถ sin ๐ฅ๐ฅ = 10 sin ๐ฅ๐ฅ โ โ2๐ถ๐ถ = 10 โ ๐ถ๐ถ = โ5 โ 2๐ธ๐ธ cos ๐ฅ๐ฅ = 0 โ ๐ธ๐ธ = 0
๐ฆ๐ฆโฒโฒ + ๐ฆ๐ฆ = 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 107
WE know, ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ฅ๐ฅ + ๐ต๐ต + ๐ถ๐ถ๐ฅ๐ฅ cos๐ฅ๐ฅ + ๐ธ๐ธ๐ฅ๐ฅ sin ๐ฅ๐ฅ & ๐ด๐ด = 4,๐ต๐ต = 0,๐ถ๐ถ = โ5,๐ท๐ท = 0 โ ๐๐๐๐ = ๐๐๐๐ โ ๐๐๐๐๐๐๐๐๐๐ ๐๐ We know: ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐ โ ๐ฆ๐ฆ = ๐ผ๐ผ1 cos ๐ฅ๐ฅ + ๐ผ๐ผ2 sin ๐ฅ๐ฅ + 4๐ฅ๐ฅ โ 5๐ฅ๐ฅ cos๐ฅ๐ฅ Initial conditions: ๐ฆ๐ฆ ๐ฅ๐ฅ = 0 ๐๐๐๐๐๐ ๐ฆ๐ฆโฒ ๐ฅ๐ฅ = 2 โข ๐ฆ๐ฆ ๐๐ = 0 โ 0 = ๐ผ๐ผ1 cos๐๐ + ๐ผ๐ผ2 sin๐๐ + 4๐๐ โ 5๐๐ cos๐๐
โ 0 = โ๐ผ๐ผ1 + 0 + 4๐๐ + 5๐๐ โ ๐ผ๐ผ1 = 9๐๐
๐ฆ๐ฆโฒโฒ + ๐ฆ๐ฆ = 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 108
โข ๐ฆ๐ฆโฒ ๐๐ = 2
โ ๐ฆ๐ฆโฒ = โ๐ผ๐ผ1 sin ๐ฅ๐ฅ + ๐ผ๐ผ2 cos ๐ฅ๐ฅ + 4 โ 5 cos๐ฅ๐ฅ โ ๐ฅ๐ฅ sin ๐ฅ๐ฅ โ 2 = โ9๐๐ sin๐๐ + ๐ผ๐ผ2 cos๐๐ + 4 โ 5 cos๐๐ + 5๐๐ sin๐๐ โ 2 = โ๐ผ๐ผ2 + 4 + 5 โ ๐ผ๐ผ2 = 9 โ 2 โ ๐ผ๐ผ2 = 7 Therefore:
๐๐ = ๐๐๐๐๐๐๐๐๐๐๐๐ + ๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ ๐๐ + ๐๐๐๐ โ ๐๐๐๐๐๐๐๐๐๐๐๐
๐ฆ๐ฆโฒโฒ + ๐ฆ๐ฆ = 4๐ฅ๐ฅ + 10 sin ๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = 4๐ฅ๐ฅ โ 5๐ฅ๐ฅ ๐๐๐๐๐ ๐ ๐ฅ๐ฅ
Undetermined Coefficients
9/30/2014 Dr. Eli Saber 109
โข To solve a non-Homogeneous D.E.
๐๐๐๐๐ฆ๐ฆ ๐๐ + ๐๐๐๐โ1๐ฆ๐ฆ ๐๐โ1 + โฏ+ ๐๐1๐ฆ๐ฆ1 + ๐๐0๐ฆ๐ฆ0= ๐๐ ๐ฅ๐ฅ
Step1: Finding a complementary solution ๐ฆ๐ฆ๐๐ by equating it to 0. Step2: Finding a particular solution ๐ฆ๐ฆ๐๐. Step3: ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โข In case of multiple additive terms in the right
hand side that constitute ๐๐(๐ฅ๐ฅ), take into account all factors contributing to ๐ฆ๐ฆ๐๐
โข Multiplication rule: If any ๐๐๐๐๐๐ contains terms
that duplicate terms in ๐๐๐๐, then that ๐๐๐๐๐๐ must be multiplied by ๐๐๐๐, where n is the smallest positive integer that eliminates that duplication
Summary
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.
Section 3.5 Variation of Parameters
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Variation of Parameters
9/30/2014 Dr. Eli Saber 111
โข See also section 2.3 for first order differential equations
Advantages: โข Always yields a particular solution ๐ฆ๐ฆ๐๐ assuming ๐ฆ๐ฆ๐๐ can be found. โข Not limited to cases such as the described in Table 3.4.1 (slide 109) โข Not limited to differential equation with constant coefficients.
Variation of Parameters
9/30/2014 Dr. Eli Saber 112
Given ๐๐2 ๐ฅ๐ฅ ๐ฆ๐ฆโฒโฒ + ๐๐1 ๐ฅ๐ฅ ๐ฆ๐ฆโฒ + ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ
Divide by ๐๐2 ๐ฅ๐ฅ
โน ๐ฆ๐ฆโฒโฒ +๐๐1๐๐2
๐ฅ๐ฅ ๐ฆ๐ฆโฒ +๐๐0๐๐2
๐ฅ๐ฅ ๐ฆ๐ฆ =๐๐ ๐ฅ๐ฅ๐๐2
โน ๐ฆ๐ฆโฒโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ (similar to ๐ฆ๐ฆ๐ฆ + ๐๐(๐ฅ๐ฅ)๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)) Assumptions: โข ๐๐(๐ฅ๐ฅ),๐๐(๐ฅ๐ฅ), ๐๐(๐ฅ๐ฅ) are continuous on some interval ๐ผ๐ผ โข ๐ฆ๐ฆ๐๐ can be found
๐ท๐ท(๐๐) ๐ธ๐ธ(๐๐) ๐๐(๐๐)
Variation of Parameters
9/30/2014 Dr. Eli Saber 113
Method: โข For first order differential equation ๐ฆ๐ฆโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ , seek a solution ๐๐๐๐ = ๐๐๐๐(๐๐)๐๐๐๐(๐๐) ๐๐๐๐(๐๐): fundamental solution for homogeneous D.E โข For second order D.E ๐ฆ๐ฆโฒโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ , seek a solution ๐๐๐๐ = ๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐ + ๐๐๐๐(๐๐)๐๐๐๐(๐๐)
๐๐๐๐ ๐๐ , ๐๐๐๐(๐๐): fundamental solution for homogeneous D.E
Variation of Parameters
9/30/2014 Dr. Eli Saber 114
๐ฆ๐ฆ๐๐ = ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2 โน ๐ฆ๐ฆ๐๐โฒ = ๐๐1๐ฆ๐ฆ1โฒ + ๐๐1โฒ ๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2โฒ + ๐๐2โฒ ๐ฆ๐ฆ2 โ ๐ฆ๐ฆ๐๐๐ฆ๐ฆ = ๐๐1๐ฆ๐ฆ1โฒโฒ + ๐๐1โฒ ๐ฆ๐ฆ1โฒ + ๐๐1โฒ ๐ฆ๐ฆ1โฒ + ๐๐1โฒโฒ๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2โฒโฒ + ๐๐2โฒ ๐ฆ๐ฆ2โฒ + ๐๐2โฒ ๐ฆ๐ฆ2โฒ + ๐๐2โฒโฒ๐ฆ๐ฆ2 Substitute into D.E: ๐ฆ๐ฆ๐๐โฒโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ๐๐โฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐๐ ๐ฅ๐ฅ โน ๐๐1๐ฆ๐ฆ1โฒโฒ + ๐๐1โฒ ๐ฆ๐ฆ1โฒ + ๐๐1โฒ ๐ฆ๐ฆ1โฒ + ๐๐1โฒโฒ๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2โฒโฒ + ๐๐2โฒ ๐ฆ๐ฆ2โฒ + ๐๐2โฒ ๐ฆ๐ฆ2โฒ + ๐๐2โฒโฒ๐ฆ๐ฆ2 ๐๐ ๐ฅ๐ฅ ๐๐1๐ฆ๐ฆ1โฒ + ๐๐1โฒ ๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2โฒ + ๐๐2โฒ ๐ฆ๐ฆ2 +๐๐ ๐ฅ๐ฅ ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2 = ๐๐(๐ฅ๐ฅ)
Variation of Parameters
9/30/2014 Dr. Eli Saber 115
Rearranging the equations, ๐๐1 ๐ฆ๐ฆ1โฒโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ1โฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ1 + ๐๐2 ๐ฆ๐ฆ2โฒโฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ2โฒ + ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ2 +๐๐1โฒโฒ๐ฆ๐ฆ1 + ๐๐1โฒ ๐ฆ๐ฆ1โฒ + ๐๐2โฒโฒ๐ฆ๐ฆ2 + ๐๐2โฒ ๐ฆ๐ฆ2โฒ + ๐๐ ๐๐1โฒ ๐ฆ๐ฆ1 + ๐๐2โฒ ๐ฆ๐ฆ2 + ๐๐1โฒ ๐ฆ๐ฆ1โฒ + ๐๐2โฒ ๐ฆ๐ฆ2โฒ = ๐๐(๐ฅ๐ฅ)
= ๐๐ = ๐๐
Since ๐ฆ๐ฆ1 & ๐ฆ๐ฆ2 are the solutions to the homogeneous equation
Variation of Parameters
9/30/2014 Dr. Eli Saber 116
๐๐1โฒโฒ๐ฆ๐ฆ1 + ๐๐1
โฒ๐ฆ๐ฆ1โฒ + ๐๐2
โฒโฒ๐ฆ๐ฆ2 + ๐๐2โฒ๐ฆ๐ฆ2
โฒ + ๐๐ ๐๐1โฒ๐ฆ๐ฆ1 + ๐๐2
โฒ๐ฆ๐ฆ2 + ๐๐1โฒ๐ฆ๐ฆ1
โฒ + ๐๐2โฒ๐ฆ๐ฆ2
โฒ = ๐๐(๐ฅ๐ฅ)
โน๐ ๐ ๐ ๐ ๐๐
๐๐๐๐โฒ๐๐๐๐ +๐ ๐ ๐ ๐ ๐๐
๐๐๐๐โฒ๐๐๐๐ + ๐๐ ๐๐1โฒ๐ฆ๐ฆ1 + ๐๐2
โฒ๐ฆ๐ฆ2 + ๐๐1โฒ๐ฆ๐ฆ1
โฒ + ๐๐2โฒ๐ฆ๐ฆ2
โฒ = ๐๐(๐ฅ๐ฅ)
โน๐๐๐๐๐ฅ๐ฅ ๐๐1
โฒ๐ฆ๐ฆ1 + ๐๐2โฒ๐ฆ๐ฆ2
+ ๐๐[๐๐1โฒ๐ฆ๐ฆ1 + ๐๐2
โฒ๐ฆ๐ฆ2] + ๐๐1โฒ๐ฆ๐ฆ1
โฒ + ๐๐2โฒ๐ฆ๐ฆ2
โฒ = ๐๐(๐ฅ๐ฅ)
โข Have two unknown functions ๐๐1 & ๐๐2 โน Need two equations โน make further assumption that ๐๐๐๐โ๐๐๐๐ + ๐๐๐๐โ๐๐๐๐ = ๐๐ โน ๐๐๐๐โ๐๐๐๐โ + ๐๐๐๐โ๐๐๐๐โ = ๐๐(๐๐)
Variation of Parameters
9/30/2014 Dr. Eli Saber 117
โข Hence, we have two equations with two unknowns:
๐ฆ๐ฆ1๐๐1โ + ๐ฆ๐ฆ2๐๐2โ = 0
๐ฆ๐ฆ1๐ฆ๐๐1โ + ๐ฆ๐ฆ2โฒ๐๐2โ = ๐๐(๐ฅ๐ฅ)
โข Solve for ๐๐1โฒ & ๐๐2โฒ & then integrate to get ๐๐1 & ๐๐2
โข Using Cramerโs rule: ๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2๐ฆ
๐๐1๐ฆ๐๐2๐ฆ
= 0๐๐(๐ฅ๐ฅ)
(1)
(2)
๐๐ ๐จ๐จ ๐๐
Variation of Parameters
9/30/2014 Dr. Eli Saber 118
We have, ๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2๐ฆ
๐๐1๐ฆ๐๐2๐ฆ
= 0๐๐(๐ฅ๐ฅ)
๐๐๐๐โฒ =
๐๐ ๐๐๐๐๐๐(๐๐) ๐๐๐๐๐ฆ๐๐๐๐ ๐๐๐๐๐๐๐๐๐ฆ ๐๐๐๐๐ฆ
& ๐๐๐๐โฒ =
๐๐๐๐ ๐๐๐๐๐๐๐ฆ ๐๐(๐๐)๐๐๐๐ ๐๐๐๐๐๐๐๐๐ฆ ๐๐๐๐๐ฆ
Note:
Wโก๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1โฒ ๐ฆ๐ฆ2โฒ
โ ๐ก๐กโ๐๐ ๐พ๐พ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐ฆ๐ฆ1 & ๐ฆ๐ฆ2
Hence, Since y1 & y2 are independent โน Wโ ๐๐ ๐๐๐๐๐๐ โ๐๐ โ ๐ฐ๐ฐ
Variation of Parameters
9/30/2014 Dr. Eli Saber 119
Summary: Given ๐๐2๐ฆ๐ฆ๐ฆ๐ฆ + ๐๐1๐ฆ๐ฆ๐ฆ + ๐๐0๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)
1. Put Eq. into standard form by dividing throughout by a2(x)
๐ฆ๐ฆโฒโฒ +๐๐1๐๐2
๐ฆ๐ฆโฒ +๐๐0๐๐2๐ฆ๐ฆ =
๐๐ ๐ฅ๐ฅ๐๐2(๐ฅ๐ฅ)
2. Find the complementary solution ๐ฆ๐ฆ๐๐
= ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2
3. Compute Wronskian of ๐ฆ๐ฆ1 & ๐ฆ๐ฆ2 ๐๐ =๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1โฒ ๐ฆ๐ฆ2โฒ
๐ท๐ท(๐๐) ๐ธ๐ธ(๐๐) ๐๐(๐๐)
Variation of Parameters
9/30/2014 Dr. Eli Saber 120
4. Compute ๐๐1โ & ๐๐2โ using:
๐๐1โฒ =
0 ๐ฆ๐ฆ2๐๐(๐ฅ๐ฅ) ๐ฆ๐ฆ2๐ฆ๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2๐ฆ
;๐๐2โฒ =
๐ฆ๐ฆ1 0๐ฆ๐ฆ1๐ฆ ๐๐(๐ฅ๐ฅ)๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2๐ฆ
5. Find ๐๐1 & ๐๐2 by integrating ๐๐1๐ฆ & ๐๐2๐ฆ respectively. 6. Form ๐ฆ๐ฆ๐๐ = ๐๐1(๐ฅ๐ฅ)๐ฆ๐ฆ1(๐ฅ๐ฅ) + ๐๐2(๐ฅ๐ฅ)๐ฆ๐ฆ2(๐ฅ๐ฅ) 7. General Solution: ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
Variation of Parameters
9/30/2014 Dr. Eli Saber 121
Note: When integrating ๐๐1โ & ๐๐2โ, you donโt need to introduce any constants because:
โข ๐ฆ๐ฆ๐๐ = ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2
๏ฟฝ๐๐1๐ฆ ๐๐๐ฅ๐ฅ = ๐๐1 + ๐๐1,๏ฟฝ๐๐2๐ฆ
๐๐๐ฅ๐ฅ = ๐๐2 + ๐๐2
โน ๐ฆ๐ฆ๐๐ = (๐๐1 + ๐๐1)๐ฆ๐ฆ1 + (๐๐2 + ๐๐2)๐ฆ๐ฆ2 โน ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐ = ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2 + ๐๐1๐ฆ๐ฆ1 + ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2 + ๐๐2๐ฆ๐ฆ2
Rearranging, ๐ฆ๐ฆ = (๐๐1 + ๐๐1)๐ฆ๐ฆ1 + (๐๐2 + ๐๐2)๐ฆ๐ฆ2 + ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2
โน ๐ฆ๐ฆ = ๐ถ๐ถ๐๐๐ฆ๐ฆ1 + ๐ถ๐ถ๐๐๐ฆ๐ฆ2 + ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2
Where, ๐ผ๐ผ1&๐ผ๐ผ2: constants computed using initial conditions or boundary conditions
๐๐1, ๐๐2 are constants
Variation of Parameters
9/30/2014 Dr. Eli Saber 122
E.g.1: ๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = ๐ฅ๐ฅ + 1 ๐๐2๐ฅ๐ฅ
1. Equation is already in standard form: ๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = ๐ฅ๐ฅ + 1 ๐๐2๐ฅ๐ฅ 2. Find ๐ฆ๐ฆ๐๐: โ ๐๐2 โ 4๐๐ + 4 = 0 โ ๐๐โ 2 2 = 0 โ ๐๐1 = ๐๐2 = 2 ๐ฆ๐ฆ๐๐ = ๐๐1๐๐2๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ๐๐2๐ฅ๐ฅ 3. Compute ๐๐
๐๐ =๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2๐ฆ
= ๐๐2๐ฅ๐ฅ ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ2๐๐2๐ฅ๐ฅ ๐๐2๐ฅ๐ฅ + 2๐ฅ๐ฅ ๐๐2๐ฅ๐ฅ
= ๐๐2๐ฅ๐ฅ ๐๐2๐ฅ๐ฅ + 2๐ฅ๐ฅ ๐๐2๐ฅ๐ฅ โ ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ 2๐๐2๐ฅ๐ฅ = ๐๐4๐ฅ๐ฅ + 2๐ฅ๐ฅ ๐๐4๐ฅ๐ฅ โ 2๐ฅ๐ฅ ๐๐4๐ฅ๐ฅ โ ๐๐ = ๐๐4๐ฅ๐ฅ
๐ท๐ท(๐๐) ๐ธ๐ธ(๐๐) ๐๐(๐๐)
๐๐๐๐ ๐๐๐๐
Variation of Parameters
9/30/2014 Dr. Eli Saber 123
4. Compute ๐๐1โฒ& ๐๐2๐ฆ
๐๐1 = 0 ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ๐ฅ๐ฅ + 1 ๐๐2๐ฅ๐ฅ ๐๐2๐ฅ๐ฅ + 2๐ฅ๐ฅ๐๐2๐ฅ๐ฅ = โ๐๐ ๐๐ + ๐๐ ๐๐๐๐๐๐
๐๐2 = ๐๐2๐ฅ๐ฅ 02๐๐2๐ฅ๐ฅ ๐ฅ๐ฅ + 1 ๐๐2๐ฅ๐ฅ = ๐๐ + ๐๐ ๐๐๐๐๐๐
Now,
๐๐1โฒ =๐๐1๐๐ = โ
๐ฅ๐ฅ + 1 ๐ฅ๐ฅ๐๐4๐ฅ๐ฅ
๐๐4๐ฅ๐ฅ = โ๐ฅ๐ฅ ๐ฅ๐ฅ + 1 โ ๐๐1โฒ = โ๐ฅ๐ฅ2 โ ๐ฅ๐ฅ
๐๐1 = ๏ฟฝ โ๐ฅ๐ฅ2 โ ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
๐๐1 = โ13 ๐ฅ๐ฅ
3 โ๐ฅ๐ฅ2
2
๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = ๐ฅ๐ฅ + 1 ๐๐2๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐๐1๐๐2๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
๐๐ = ๐๐4๐ฅ๐ฅ
Variation of Parameters
9/30/2014 Dr. Eli Saber 124
Now,
๐๐2โฒ =๐๐2๐๐ =
๐ฅ๐ฅ + 1 ๐๐4๐ฅ๐ฅ
๐๐4๐ฅ๐ฅ = ๐ฅ๐ฅ + 1 โ ๐๐2๐ฆ
๐๐2 = ๏ฟฝ ๐ฅ๐ฅ + 1 ๐๐๐ฅ๐ฅ
๐๐2 =๐ฅ๐ฅ2
2 + ๐ฅ๐ฅ
Hence, ๐ฆ๐ฆ๐๐ = ๐๐1 ๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐ฅ๐ฅ + ๐๐2 ๐ฅ๐ฅ ๐ฆ๐ฆ2 ๐ฅ๐ฅ
๐ฆ๐ฆ๐๐ ๐ฅ๐ฅ = โ13๐ฅ๐ฅ3 โ
๐ฅ๐ฅ2
2๐๐2๐ฅ๐ฅ +
๐ฅ๐ฅ2
๐ฅ๐ฅ+ ๐ฅ๐ฅ ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = ๐ฅ๐ฅ + 1 ๐๐2๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐๐1๐๐2๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
๐๐ = ๐๐4๐ฅ๐ฅ ๐๐2 = ๐ฅ๐ฅ + 1 ๐๐4๐ฅ๐ฅ
๐๐1 = โ13๐ฅ๐ฅ3 โ
๐ฅ๐ฅ2
2
Variation of Parameters
9/30/2014 Dr. Eli Saber 125
๐ฆ๐ฆ๐๐ ๐ฅ๐ฅ = โ13๐ฅ๐ฅ3 โ
๐ฅ๐ฅ2
2๐๐2๐ฅ๐ฅ +
๐ฅ๐ฅ2
๐ฅ๐ฅ+ ๐ฅ๐ฅ ๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
= โ13
๐ฅ๐ฅ3๐๐2๐ฅ๐ฅ โ12๐ฅ๐ฅ2๐๐2๐ฅ๐ฅ +
12๐ฅ๐ฅ3๐๐2๐ฅ๐ฅ + ๐ฅ๐ฅ2๐๐2๐ฅ๐ฅ
๐ฆ๐ฆ๐๐ =16 ๐ฅ๐ฅ3๐๐2๐ฅ๐ฅ +
12 ๐ฅ๐ฅ2๐๐2๐ฅ๐ฅ
And, ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โ ๐๐ = ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐๐๐ +๐๐๐๐๐๐
๐๐๐๐๐๐๐๐ +๐๐๐๐๐๐
๐๐๐๐๐๐๐๐
๐ฆ๐ฆโฒโฒ โ 4๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = ๐ฅ๐ฅ + 1 ๐๐2๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐๐1๐๐2๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ๐๐2๐ฅ๐ฅ
๐๐ = ๐๐4๐ฅ๐ฅ ๐๐2 = ๐ฅ๐ฅ + 1 ๐๐4๐ฅ๐ฅ
๐๐1 = โ13๐ฅ๐ฅ3 โ
๐ฅ๐ฅ2
2
๐๐2 =๐ฅ๐ฅ2
2+ ๐ฅ๐ฅ
Variation of Parameters
9/30/2014 Dr. Eli Saber 126
E.g.2: ๐ฆ๐ฆโฒโฒ โ 5๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 8๐๐๐ฅ๐ฅ
1. Equation is already in standard form: ๐ฆ๐ฆโฒโฒ โ 5๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 8๐๐๐ฅ๐ฅ 2. Find ๐ฆ๐ฆ๐๐: โ ๐๐2 โ 5๐๐ + 4 = 0 โ (๐๐โ 1)(๐๐โ 4) = 0 โ ๐๐1 = 1&๐๐2 = 4 ๐ฆ๐ฆ๐๐ = ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐4๐ฅ๐ฅ 3. Compute ๐๐
๐๐ =๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2๐ฆ
= ๐๐๐ฅ๐ฅ ๐๐4๐ฅ๐ฅ๐๐๐ฅ๐ฅ 4๐๐4๐ฅ๐ฅ
= ๐๐๐ฅ๐ฅ 4๐๐4๐ฅ๐ฅ โ ๐๐๐ฅ๐ฅ๐๐4๐ฅ๐ฅ = 4๐๐5๐ฅ๐ฅ โ ๐๐5๐ฅ๐ฅ โ ๐๐ = 3๐๐5๐ฅ๐ฅ
๐ท๐ท(๐๐) ๐ธ๐ธ(๐๐) ๐๐(๐๐)
๐๐๐๐ ๐๐๐๐
Sec. 3.4. E.g. 4 (slide 94-95)
Variation of Parameters
9/30/2014 Dr. Eli Saber 127
4. Compute ๐๐1โฒ& ๐๐2๐ฆ
๐๐1โฒ =
0 ๐ฆ๐ฆ2๐๐(๐ฅ๐ฅ) ๐ฆ๐ฆ2๐ฆ
๐๐ =0 ๐๐4๐ฅ๐ฅ
8๐๐๐ฅ๐ฅ 4๐๐4๐ฅ๐ฅ3๐๐5๐ฅ๐ฅ = โ
8๐๐5๐ฅ๐ฅ
3๐๐5๐ฅ๐ฅ = โ83
๐๐2โฒ =
๐ฆ๐ฆ1 0๐ฆ๐ฆ1๐ฆ ๐๐(๐ฅ๐ฅ)
๐๐ =๐๐๐ฅ๐ฅ 0๐๐๐ฅ๐ฅ 8๐๐๐ฅ๐ฅ
3๐๐5๐ฅ๐ฅ =8๐๐2๐ฅ๐ฅ
3๐๐5๐ฅ๐ฅ =83 ๐๐โ3๐ฅ๐ฅ
5. Find ๐๐1 ๐๐๐๐๐๐ ๐๐2
๐๐1 = ๏ฟฝโ83๐๐๐ฅ๐ฅ = โ
83 ๐ฅ๐ฅ
๐๐2 = ๏ฟฝ83 ๐๐
โ3๐ฅ๐ฅ ๐๐๐ฅ๐ฅ = โ89 ๐๐โ3๐ฅ๐ฅ
๐ฆ๐ฆโฒโฒ โ 5๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 8๐๐๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐4๐ฅ๐ฅ
๐๐ = 3๐๐5๐ฅ๐ฅ
Variation of Parameters
9/30/2014 Dr. Eli Saber 128
Hence, ๐ฆ๐ฆ๐๐ = ๐๐1 ๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐ฅ๐ฅ + ๐๐2 ๐ฅ๐ฅ ๐ฆ๐ฆ2 ๐ฅ๐ฅ
๐ฆ๐ฆ๐๐ = โ83๐ฅ๐ฅ ๐๐๐ฅ๐ฅ + โ
89
๐๐โ3๐ฅ๐ฅ ๐๐4๐ฅ๐ฅ
๐ฆ๐ฆ๐๐ = โ83 ๐ฅ๐ฅ ๐๐๐ฅ๐ฅ โ
89 ๐๐
๐ฅ๐ฅ
And, ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โ ๐๐ = ๐ถ๐ถ๐๐๐๐๐๐ + ๐ถ๐ถ๐๐๐๐๐๐๐๐ โ๐๐๐๐๐๐๐๐
๐๐ โ๐๐๐๐๐๐
๐๐
๐ฆ๐ฆโฒโฒ โ 5๐ฆ๐ฆโฒ + 4๐ฆ๐ฆ = 8๐๐๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐4๐ฅ๐ฅ
๐๐ = 3๐๐5๐ฅ๐ฅ
๐๐1 = โ83๐ฅ๐ฅ & ๐๐2 = โ
89๐๐โ3๐ฅ๐ฅ
Variation of Parameters
9/30/2014 Dr. Eli Saber 129
We got,
โ ๐๐ = ๐ถ๐ถ๐๐๐๐๐๐ + ๐ถ๐ถ๐๐๐๐๐๐๐๐ โ๐๐๐๐๐๐๐๐๐๐ โ
๐๐๐๐๐๐๐๐
From Sec. 3.4. E.g. 4 (slide 94-95), we have:
๐๐ = ๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ โ๐๐๐๐๐๐๐๐๐๐
as the solution.
Notice, ๐ฆ๐ฆ = ๐ผ๐ผ1๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐4๐ฅ๐ฅ โ83๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ 8
9๐๐๐ฅ๐ฅ = ๐ผ๐ผ1๐๐๐ฅ๐ฅ โ
89๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐4๐ฅ๐ฅ โ
83๐ฅ๐ฅ๐๐๐ฅ๐ฅ
= ๐ผ๐ผ1 โ89
๐๐๐ฅ๐ฅ + ๐ผ๐ผ2๐๐4๐ฅ๐ฅ โ83๐ฅ๐ฅ๐๐๐ฅ๐ฅ
= ๐๐๐๐๐๐๐ฅ๐ฅ + ๐๐๐๐๐๐4๐ฅ๐ฅ โ83๐ฅ๐ฅ๐๐๐ฅ๐ฅ
Variation of Parameters
9/30/2014 Dr. Eli Saber 130
โข Higher Order Equations Generalize method to linear nth order D.E.
๐ฆ๐ฆ(๐๐) + ๐๐๐๐โ1 ๐ฅ๐ฅ ๐ฆ๐ฆ(๐๐โ1) + โฏ+ ๐๐1 ๐ฅ๐ฅ ๐ฆ๐ฆโฒ + ๐๐0 ๐ฅ๐ฅ ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ
If ๐ฆ๐ฆc = ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2+. . . +๐๐๐๐๐ฆ๐ฆ๐๐ is the complementary function , then a particular solution is:
๐ฆ๐ฆ๐๐ = ๐๐1(๐ฅ๐ฅ)๐ฆ๐ฆ1(๐ฅ๐ฅ) + ๐๐2(๐ฅ๐ฅ)๐ฆ๐ฆ2(๐ฅ๐ฅ)+. . . +๐๐๐๐(๐ฅ๐ฅ)๐ฆ๐ฆ๐๐ (๐ฅ๐ฅ)
,where the ๐๐๐๐โฒ , ๐๐ = 1,2, โฆ ,๐๐ are determined by the ๐๐ eqn.
๐ฆ๐ฆ1๐๐1โฒ + ๐ฆ๐ฆ2๐๐2โฒ + โฏ+ ๐ฆ๐ฆ๐๐๐๐๐๐โฒ = 0 ๐ฆ๐ฆ1โฒ๐๐1โฒ + ๐ฆ๐ฆ2โฒ๐๐2โฒ + โฏ+ ๐ฆ๐ฆ๐๐โฒ๐๐๐๐โฒ = 0
โฎ โฎ
๐ฆ๐ฆ1(๐๐โ1)๐๐1โฒ + ๐ฆ๐ฆ2
(๐๐โ1)๐๐2โฒ + โฏ+ ๐ฆ๐ฆ๐๐๐๐โ1 ๐๐๐๐โฒ = ๐๐(๐ฅ๐ฅ)
Variation of Parameters
9/30/2014 Dr. Eli Saber 131
โ
๐ฆ๐ฆ1 ๐ฆ๐ฆ2 โฆ ๐ฆ๐ฆ๐๐๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2โฒ โฆ ๐ฆ๐ฆ๐๐๐ฆโฎ โฎ โฎ โฎ
๐ฆ๐ฆ1๐๐โ1 ๐ฆ๐ฆ2
๐๐โ1 โฆ ๐ฆ๐ฆ๐๐๐๐โ1
๐๐1โฒ
๐๐2โฒโฎ๐๐๐๐โฒ
=
00โฎ
๐๐(๐ฅ๐ฅ)
And, ๐๐๐๐โฒ =๐๐๐๐๐๐ ; ๐๐ = 1,2, โฆ ,๐๐
Where, ๐๐1 =
0 ๐ฆ๐ฆ2 โฏ ๐ฆ๐ฆ๐๐0 ๐ฆ๐ฆ2โฒ โฏ ๐ฆ๐ฆ๐๐โฒโฎ โฎ โฎ โฎ
๐๐(๐ฅ๐ฅ) ๐ฆ๐ฆ2(๐๐โ1) โฏ ๐ฆ๐ฆ๐๐ (๐๐โ1)
๐๐๐๐ can be computed by integrating ๐๐๐๐โฒ ;๐๐ = 1,2, โฆ ,๐๐ ๐ฆ๐ฆ๐๐ = ๐๐1(๐ฅ๐ฅ)๐ฆ๐ฆ1(๐ฅ๐ฅ) + ๐๐2(๐ฅ๐ฅ)๐ฆ๐ฆ2(๐ฅ๐ฅ)+. . . +๐๐๐๐(๐ฅ๐ฅ)๐ฆ๐ฆ๐๐
(๐ฅ๐ฅ)
Variation of Parameters
9/30/2014 Dr. Eli Saber 132
Summary
Given ๐๐2๐ฆ๐ฆ๐ฆ๐ฆ + ๐๐1๐ฆ๐ฆ๐ฆ + ๐๐0๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ) 1. Put Eq. into standard form by dividing throughout by a2(x)
๐ฆ๐ฆโฒโฒ +๐๐1๐๐2
๐ฆ๐ฆโฒ +๐๐0๐๐2๐ฆ๐ฆ =
๐๐ ๐ฅ๐ฅ๐๐2(๐ฅ๐ฅ)
2. Find ๐ฆ๐ฆ๐๐
= ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2
Verify ๐ฆ๐ฆ๐๐
for the D.E.
3. Compute ๐๐ =๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1โฒ ๐ฆ๐ฆ2โฒ
4. Compute ๐๐1โ & ๐๐2โ using:
๐๐1โฒ =
0 ๐ฆ๐ฆ2๐๐(๐ฅ๐ฅ) ๐ฆ๐ฆ2๐ฆ๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2๐ฆ
;๐๐2โฒ =
๐ฆ๐ฆ1 0๐ฆ๐ฆ1๐ฆ ๐๐(๐ฅ๐ฅ)๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2๐ฆ
5. Find ๐๐1 & ๐๐2 by integrating ๐๐1โฒ & ๐๐2โฒ respectively. 6. Form ๐ฆ๐ฆ๐๐ = ๐๐1(๐ฅ๐ฅ)๐ฆ๐ฆ1(๐ฅ๐ฅ) + ๐๐2(๐ฅ๐ฅ)๐ฆ๐ฆ2(๐ฅ๐ฅ) 7. General Solution: ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐ 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:
๐๐๐๐๐ฅ๐ฅ๐๐๐๐๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐
+ ๐๐๐๐โ1๐ฅ๐ฅ๐๐โ1 ๐๐๐๐โ1๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐โ1
+ โฆ + ๐๐1๐ฅ๐ฅ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐0 ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)
, where ๐๐๐๐,๐๐๐๐โ1, โฆ ,๐๐1,๐๐0 are constants
And the degree ๐๐ at ๐ฅ๐ฅ๐๐ matches the order ๐๐ of the differentiation ๐๐๐๐๐ฆ๐ฆ
๐๐๐ฅ๐ฅ๐๐
is called a Cauchy-Euler Equation E.g.
1) ๐ฅ๐ฅ2๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
โ 2๐ฅ๐ฅ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
โ 4๐ฆ๐ฆ = 0
2) ๐ฅ๐ฅ2๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
โ 3๐ฅ๐ฅ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ 3๐ฆ๐ฆ = 2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ
same same
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 135
General 2nd order: ๐๐๐ฅ๐ฅ2 ๐๐2๐ฆ๐ฆ
๐๐๐ฅ๐ฅ2+ ๐๐๐ฅ๐ฅ ๐๐๐ฆ๐ฆ
๐๐๐ฅ๐ฅ+ ๐๐๐ฆ๐ฆ = 0
Proceed to develop solution for 2nd order and then generalize.
๐๐๐ฅ๐ฅ2๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 + ๐๐๐ฅ๐ฅ
๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ + ๐๐๐ฆ๐ฆ = 0
Note: ๐๐๐ฅ๐ฅ2 = 0 @ ๐ฅ๐ฅ = 0 confine attention to interval ๐ผ๐ผ โก 0,โ For (โโ, 0), let ๐ก๐ก = โ๐ฅ๐ฅ
Homogeneous
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 136
Try a solution of the form ๐ฆ๐ฆ = ๐ฅ๐ฅ๐๐
โ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
= ๐๐๐ฅ๐ฅ๐๐โ1 &๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
= ๐๐ ๐๐โ 1 ๐ฅ๐ฅ๐๐โ2
โ ๐๐๐ฅ๐ฅ2 ๐๐ ๐๐โ 1 ๐ฅ๐ฅ๐๐โ2 + ๐๐๐ฅ๐ฅ ๐๐๐ฅ๐ฅ๐๐โ1 + ๐๐๐ฅ๐ฅ๐๐ = 0 โ ๐๐๐๐ ๐๐โ 1 ๐ฅ๐ฅ๐๐ + ๐๐๐๐๐ฅ๐ฅ๐๐ + ๐๐๐ฅ๐ฅ๐๐ = 0 โ ๐๐๐๐ ๐๐ โ 1 + ๐๐๐๐ + ๐๐ ๐ฅ๐ฅ๐๐ = 0 โ ๐๐๐๐2 โ ๐๐๐๐ + ๐๐๐๐ + ๐๐ ๐ฅ๐ฅ๐๐ = 0 โ ๐๐๐๐2 + ๐๐ โ ๐๐ ๐๐ + ๐๐ ๐ฅ๐ฅ๐๐ = 0 Thus, ๐๐ = ๐๐๐๐ is a solution of the D.E. whenever ๐๐ is a solution to the auxiliary equation
๐๐๐๐๐๐ + ๐๐ โ ๐๐ ๐๐ + ๐๐ = ๐๐
๐๐๐ฅ๐ฅ2๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
+ ๐๐๐ฅ๐ฅ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐๐ฆ๐ฆ = 0
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 137
Case 1: Distinct Real Roots If ๐๐1 & ๐๐2 are the real roots of ๐๐๐๐2 + ๐๐ โ ๐๐ ๐๐ + ๐๐ = 0 with ๐๐1 โ ๐๐2 โ ๐ฆ๐ฆ1 = ๐ฅ๐ฅ๐๐1 & ๐ฆ๐ฆ2 = ๐ฅ๐ฅ๐๐2 form a fundamental set of solutions and the general solution is ๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ๐๐1 + ๐๐2๐ฅ๐ฅ๐๐2 General case: ๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ๐๐1 + ๐๐2๐ฅ๐ฅ๐๐2 + โฏ+ ๐๐๐๐๐ฅ๐ฅ๐๐๐๐ ๐๐๐ก๐ก๐ก order
Case 1: Distinct Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 138
E.g. ๐ฅ๐ฅ2 ๐๐2๐ฆ๐ฆ
๐๐๐ฅ๐ฅ2โ 2๐ฅ๐ฅ ๐๐๐ฆ๐ฆ
๐๐๐ฅ๐ฅ โ 4๐ฆ๐ฆ = 0
Assume ๐ฆ๐ฆ = ๐ฅ๐ฅ๐๐ as the solution. โ ๐ฅ๐ฅ2 ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐โ2 โ 2๐ฅ๐ฅ ๐๐๐ฅ๐ฅ๐๐โ1 โ 4๐ฅ๐ฅ๐๐ = 0 โ ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐ โ 2๐๐ ๐ฅ๐ฅ๐๐ โ 4๐ฅ๐ฅ๐๐ = 0 โ ๐๐ ๐๐ โ 1 โ 2๐๐ โ 4 ๐ฅ๐ฅ๐๐ = 0 โ ๐๐2 โ๐๐ โ 2๐๐ โ 4 ๐ฅ๐ฅ๐๐ = 0 โ ๐๐2 โ 3๐๐ โ 4 ๐ฅ๐ฅ๐๐ = 0
๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ = ๐๐๐ฅ๐ฅ๐๐โ1
๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 = ๐๐ ๐๐โ 1 ๐ฅ๐ฅ๐๐โ2
Auxiliary Equation
Case 1: Distinct Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 139
๐๐2 โ 3๐๐ โ 4 = 0 โ ๐๐ + 1 ๐๐ โ 4 = 0 โ ๐๐ = โ1 ๐๐๐๐ ๐๐ = 4 Hence,
๐๐ = ๐๐๐๐๐๐โ๐๐ + ๐๐๐๐๐๐๐๐
Case 1: Distinct Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 140
Case 2: Repeated Real Roots If the roots are repeated i.e. ๐๐1 = ๐๐2, only one solution: ๐๐ = ๐๐๐๐๐๐ โ ๐๐๐๐2 + ๐๐ โ ๐๐ ๐๐ + ๐๐ = 0
โ ๐๐ =โ ๐๐ โ ๐๐ ยฑ ๐๐ โ ๐๐ 2 โ 4๐๐๐๐
2๐๐
For ๐๐1 = ๐๐2,โ ๐๐ โ ๐๐ 2 โ 4๐๐๐๐ = 0 โ ๐๐ โ ๐๐ 2 = 4๐๐๐๐
Hence, ๐๐1 = ๐๐2 = โ (๐๐โ๐๐)2๐๐
Case 2: Repeated Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 141
Construct a second solution like Section 3.2.
๐๐๐ฅ๐ฅ2๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
+ ๐๐๐ฅ๐ฅ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐๐ฆ๐ฆ = 0
โ๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 +
๐๐๐ฅ๐ฅ๐๐๐ฅ๐ฅ2
๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ +
๐๐๐๐๐ฅ๐ฅ2 ๐ฆ๐ฆ = 0
โ๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2 +
๐๐๐๐๐ฅ๐ฅ
๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ +
๐๐๐๐๐ฅ๐ฅ2 ๐ฆ๐ฆ = 0
Let ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐ฅ๐ฅ โ ๐ฆ๐ฆโฒ = ๐๐๐ฆ๐ฆ1โฒ + ๐๐โฒ๐ฆ๐ฆ1 & ๐ฆ๐ฆโฒโฒ = ๐๐๐ฆ๐ฆโฒโฒ + 2๐๐โฒ๐ฆ๐ฆ1โฒ + ๐๐โฒโฒ๐ฆ๐ฆ1
Case 2: Repeated Real Roots
๐ท๐ท(๐๐) Q(๐๐)
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 142
๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
+๐๐๐๐๐ฅ๐ฅ
๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+๐๐๐๐๐ฅ๐ฅ2
๐ฆ๐ฆ = 0
Replace ๐ฆ๐ฆ1,๐ฆ๐ฆ1โฒ ,๐ฆ๐ฆ1โฒ ๐ฆ
โ ๐๐๐ฆ๐ฆ1โฒโฒ + 2๐๐โฒ๐ฆ๐ฆ1โฒ + ๐๐โฒโฒ๐ฆ๐ฆ1 +๐๐๐๐๐ฅ๐ฅ ๐๐๐ฆ๐ฆ1โฒ + ๐๐โฒ๐ฆ๐ฆ1 +
๐๐๐๐๐ฅ๐ฅ2 ๐๐๐ฆ๐ฆ1 = 0
โ ๐๐ ๐ฆ๐ฆ1โฒโฒ +๐๐๐๐๐ฅ๐ฅ ๐ฆ๐ฆ1
โฒ +๐๐๐๐๐ฅ๐ฅ2 ๐ฆ๐ฆ1 + ๐ฆ๐ฆ1๐๐โฒโฒ + 2๐ฆ๐ฆ1โฒ +
๐๐๐๐๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐๐โฒ = 0
= ๐ฆ๐ฆ1๐๐โฒโฒ + 2๐ฆ๐ฆโฒ +๐๐๐๐๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐๐โฒ = 0
Case 2: Repeated Real Roots
๐ฆ๐ฆ1 = ๐ฅ๐ฅ๐๐1 ๐ฆ๐ฆ = ๐๐๐ฆ๐ฆ1
๐ฆ๐ฆโฒ = ๐๐๐ฆ๐ฆ1โฒ + ๐๐โฒ๐ฆ๐ฆ1 ๐ฆ๐ฆโฒโฒ = ๐๐๐ฆ๐ฆโฒโฒ + 2๐๐โฒ๐ฆ๐ฆ1โฒ + ๐๐โฒโฒ๐ฆ๐ฆ1
=0 since ๐๐๐๐ = ๐๐๐๐ is a solution
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 143
๐ฆ๐ฆ1๐๐โฒโฒ + 2๐ฆ๐ฆโฒ +๐๐๐๐๐ฅ๐ฅ
๐ฆ๐ฆ1 ๐๐โฒ = 0
Let ๐ค๐ค = ๐๐โฒ โ ๐ฆ๐ฆ1๐ค๐คโฒ + 2๐ฆ๐ฆ1โฒ + ๐๐๐๐๐ฅ๐ฅ
๐ฆ๐ฆ1 ๐ค๐ค = 0
โ ๐ฆ๐ฆ1๐๐๐ค๐ค๐๐๐ฅ๐ฅ = โ 2 ๐ฆ๐ฆ1โฒ +
๐๐๐๐๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐ค๐ค
โ๐๐๐ค๐ค๐ค๐ค = โ
1๐ฆ๐ฆ1
2๐ฆ๐ฆ1โฒ +๐๐๐๐๐ฅ๐ฅ ๐ฆ๐ฆ1 ๐๐๐ฅ๐ฅ
โ๐๐๐ค๐ค๐ค๐ค = โ2
๐ฆ๐ฆ1โฒ
๐ฆ๐ฆ1 ๐๐๐ฅ๐ฅ โ
๐๐๐๐๐ฅ๐ฅ ๐๐๐ฅ๐ฅ
๐ฆ๐ฆ1 = ๐ฅ๐ฅ๐๐1 ๐ฆ๐ฆ = ๐๐๐ฆ๐ฆ1
๐ฆ๐ฆโฒ = ๐๐๐ฆ๐ฆ1โฒ + ๐๐โฒ๐ฆ๐ฆ1 ๐ฆ๐ฆโฒโฒ = ๐๐๐ฆ๐ฆโฒโฒ + 2๐๐โฒ๐ฆ๐ฆ1โฒ + ๐๐โฒโฒ๐ฆ๐ฆ1
Case 2: Repeated Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 144
โ๐๐๐ค๐ค๐ค๐ค
= โ2๐๐1๐ฅ๐ฅ๐๐1โ1
๐ฅ๐ฅ๐๐1 ๐๐๐ฅ๐ฅ โ
๐๐๐๐๐ฅ๐ฅ
๐๐๐ฅ๐ฅ
โ ๏ฟฝ๐๐๐ค๐ค๐ค๐ค
= ๏ฟฝโ2๐๐1๐ฅ๐ฅ ๐๐๐ฅ๐ฅ โ ๏ฟฝ
๐๐๐๐
1๐ฅ๐ฅ
๐๐๐ฅ๐ฅ
โ ln |๐ค๐ค| = โ2 ๐๐1 ln ๐ฅ๐ฅ โ๐๐๐๐ ln ๐ฅ๐ฅ + ๐๐
โ ln |๐ค๐ค| + 2 ๐๐1 ln ๐ฅ๐ฅ +๐๐๐๐ ln ๐ฅ๐ฅ = ๐๐
โ ln ๐ค๐ค + ln ๐ฅ๐ฅ 2๐๐1 + ln ๐ฅ๐ฅ๐๐๐๐ = ๐๐
โ ln ๐ค๐ค๐ฅ๐ฅ2๐๐1๐ฅ๐ฅ๐๐๐๐ = ๐๐
Case 2: Repeated Real Roots
๐๐๐ค๐ค๐ค๐ค
= โ2๐ฆ๐ฆ1โฒ
๐ฆ๐ฆ1 ๐๐๐ฅ๐ฅ โ
๐๐๐๐๐ฅ๐ฅ
๐๐๐ฅ๐ฅ
๐ฆ๐ฆ1 = ๐ฅ๐ฅ๐๐1
โ ๐ฆ๐ฆ1โฒ = ๐๐1๐ฅ๐ฅ๐๐1โ1
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 145
โ ln ๐ค๐ค๐ฅ๐ฅ2๐๐1๐ฅ๐ฅ๐๐๐๐ = ๐๐
โ ๐ค๐ค๐ฅ๐ฅ2๐๐1๐ฅ๐ฅ๐๐๐๐ = ๐๐๐๐
But ๐ค๐ค = ๐๐โฒ โ ๐๐โฒ๐ฅ๐ฅ2๐๐1๐ฅ๐ฅ๐๐๐๐ = ๐๐๐๐
โ ๐๐โฒ = ๐๐๐๐๐ฅ๐ฅโ2๐๐1๐ฅ๐ฅโ ๐๐๐๐
โ ๐๐ = ๏ฟฝ๐๐๐๐๐ฅ๐ฅโ2๐๐1๐ฅ๐ฅโ ๐๐๐๐ ๐๐๐ฅ๐ฅ
Now, ๐ฆ๐ฆ2 = ๐๐๐ฆ๐ฆ1 = ๐ฅ๐ฅ๐๐1 โซ ๐๐๐๐๐ฅ๐ฅ๐๐โ๐๐๐๐ ๐ฅ๐ฅโ ๐๐๐๐ ๐๐๐ฅ๐ฅ
= ๐ฅ๐ฅ๐๐1 ๏ฟฝ๐๐๐๐ ๐ฅ๐ฅ๐๐๐๐โ1โ
๐๐๐๐ ๐๐๐ฅ๐ฅ = ๐ฅ๐ฅ๐๐1 ๏ฟฝ๐๐๐๐ ๐ฅ๐ฅโ1๐๐๐ฅ๐ฅ
Case 2: Repeated Real Roots
๐๐1 = โ๐๐ โ ๐๐2๐๐
โ 2๐๐1 = โ๐๐ โ ๐๐๐๐
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 146
๐ฆ๐ฆ2 = ๐ฅ๐ฅ๐๐1 ๏ฟฝ๐๐๐๐ ๐ฅ๐ฅโ1๐๐๐ฅ๐ฅ
๐ฆ๐ฆ2 = ๐๐๐๐๐ฅ๐ฅ๐๐1 ln ๐ฅ๐ฅ = ๐๐2๐ฅ๐ฅ๐๐1 ln ๐ฅ๐ฅ General solution: ๐๐ = ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐
Case 2: Repeated Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 147
E.g. 4๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ + 8๐ฅ๐ฅ๐ฆ๐ฆโฒ + ๐ฆ๐ฆ = 0
Let ๐ฆ๐ฆ = ๐ฅ๐ฅ๐๐ โ ๐ฆ๐ฆโฒ = ๐๐๐ฅ๐ฅ๐๐โ1 โ ๐ฆ๐ฆโฒโฒ = ๐๐ ๐๐โ 1 ๐ฅ๐ฅ๐๐โ2 โ 4๐ฅ๐ฅ2 ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐โ2 + 8๐ฅ๐ฅ ๐๐๐ฅ๐ฅ๐๐โ1 + ๐ฅ๐ฅ๐๐ = 0 โ 4๐๐ ๐๐โ 1 ๐ฅ๐ฅ๐๐ + 8๐๐ ๐ฅ๐ฅ๐๐ + ๐ฅ๐ฅ๐๐ = 0 โ 4๐๐2 โ 4๐๐ + 8๐๐ + 1 ๐ฅ๐ฅ๐๐ = 0 โ 4๐๐2 + 4๐๐ + 1 = 0
โ 2๐๐ + 1 2 = 0 โ ๐๐ = โ12
โ ๐๐ = ๐๐๐๐๐๐โ ๐๐๐๐ + ๐๐๐๐๐๐
โ ๐๐๐๐ ๐ฅ๐ฅ๐ฅ๐ฅ ๐๐
Case 2: Repeated Real Roots
๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ๐๐1 + ๐๐2๐ฅ๐ฅ๐๐1 ๐๐๐๐ ๐ฅ๐ฅ
Repeated roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 148
Note: For higher order equations, if ๐๐1 is a root of multiplicity ๐พ๐พ
โ ๐ฅ๐ฅ๐๐1 , ๐ฅ๐ฅ๐๐1 ln๐ฅ๐ฅ , ๐ฅ๐ฅ๐๐1 ln๐ฅ๐ฅ 2, โฆ , ๐ฅ๐ฅ๐๐1 ln ๐ฅ๐ฅ ๐๐โ1 are ๐พ๐พ linearly independent solutions
โ ๐๐ = ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐ ๐๐ + โฏ+ ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐ ๐๐โ๐๐
Case 2: Repeated Real Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 149
Case 3: Conjugate Complex Roots If the roots are conjugate pairs i.e. ๐๐1 = ๐ผ๐ผ + ๐๐๐ฝ๐ฝ & ๐๐2 = ๐ผ๐ผ โ ๐๐๐ฝ๐ฝ (๐ผ๐ผ,๐ฝ๐ฝ > 0)
โ ๐๐ = ๐๐๐๐๐๐๐ถ๐ถ+๐๐๐ท๐ท + ๐๐๐๐๐๐๐ถ๐ถโ๐๐๐ท๐ท We can rewrite that in terms of ๐๐๐๐๐ ๐ & ๐ ๐ ๐ ๐ ๐๐ as:
๐ฅ๐ฅ๐๐๐๐ = ๐๐ln ๐ฅ๐ฅ ๐๐๐๐ = ๐๐ln ๐ฅ๐ฅ๐๐๐๐ = cos ๐ฝ๐ฝ ln ๐ฅ๐ฅ + ๐๐ sin(๐ฝ๐ฝ ln ๐ฅ๐ฅ)
๐ฅ๐ฅโ๐๐๐๐ = ๐๐ln ๐ฅ๐ฅ โ๐๐๐๐ = ๐๐โ ln ๐ฅ๐ฅ๐๐๐๐ = cos ๐ฝ๐ฝ ln ๐ฅ๐ฅ โ ๐๐ sin(๐ฝ๐ฝ ln ๐ฅ๐ฅ)
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 150
We have, ๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ๐ผ๐ผ+๐๐๐๐ + ๐๐2๐ฅ๐ฅ๐ผ๐ผโ๐๐๐๐ โ ๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ๐ผ๐ผ ๐๐๐๐๐ ๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ + ๐๐ ๐ ๐ ๐ ๐ ๐๐(๐ฝ๐ฝ ln ๐ฅ๐ฅ) + ๐๐2๐ฅ๐ฅ๐ผ๐ผ ๐๐๐๐๐ ๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ โ ๐๐ ๐ ๐ ๐ ๐ ๐๐(๐ฝ๐ฝ ln ๐ฅ๐ฅ)
= ๐๐1๐ฅ๐ฅ๐ผ๐ผ ๐๐๐๐๐ ๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ + ๐๐๐๐1๐ฅ๐ฅ๐ผ๐ผ ๐ ๐ ๐ ๐ ๐๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ๐ผ๐ผ ๐๐๐๐๐ ๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ โ ๐๐๐๐2๐ฅ๐ฅ๐ผ๐ผ ๐ ๐ ๐ ๐ ๐๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ = ๐ฅ๐ฅ๐ผ๐ผ ๐๐1 ๐๐๐๐๐ ๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ + ๐๐2 ๐๐๐๐๐ ๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ + ๐๐๐๐1 ๐ ๐ ๐ ๐ ๐๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ โ ๐๐๐๐2 ๐ ๐ ๐ ๐ ๐๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ = ๐ฅ๐ฅ๐ผ๐ผ ๐๐๐๐๐ ๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ {๐๐1 + ๐๐2} + ๐๐{๐๐1 โ ๐๐2} ๐ ๐ ๐ ๐ ๐๐ ๐ฝ๐ฝ ln ๐ฅ๐ฅ ๐๐ = ๐๐๐ถ๐ถ โ๐๐ ๐๐๐๐๐๐ ๐ท๐ท ๐๐๐๐๐๐ +โ๐๐ ๐๐๐๐๐๐ ๐ท๐ท ๐๐๐๐๐๐
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 151
E.g.
4๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ + 17๐ฆ๐ฆ = 0 with I.C. ๐ฆ๐ฆ 1 = โ1; ๐ฆ๐ฆโฒ 1 = โ12
Let ๐ฆ๐ฆ = ๐ฅ๐ฅ๐๐ โ ๐ฆ๐ฆโฒ = ๐๐๐ฅ๐ฅ๐๐โ1 โ ๐ฆ๐ฆโฒโฒ = ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐โ2 โ 4๐ฅ๐ฅ2 ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐โ2 + 17๐ฅ๐ฅ๐๐ = 0 โ 4๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐ + 17๐ฅ๐ฅ๐๐ = 0 โ ๐ฅ๐ฅ๐๐ 4๐๐2 โ 4๐๐ + 17 = 0 Auxiliary Eqn. : ๐๐๐๐๐๐ โ ๐๐๐๐ + ๐๐๐๐ = ๐๐
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 152
Auxiliary Eqn. : 4๐๐2 โ 4๐๐ + 17 = 0
๐๐ =โ โ4 ยฑ 16 โ 4(4)(17)
8=
4 ยฑ 16 โ 2728
=4 ยฑ 256 ๐๐2
8
๐๐ =4 ยฑ 4 โ 64 ๐๐2
8 =4 ยฑ ๐๐2(8)
8 =12 ยฑ 2๐๐
โ ๐๐๐๐ =๐๐๐๐ + ๐๐๐๐ & ๐๐๐๐ =
๐๐๐๐ โ ๐๐๐๐
๐ผ๐ผ =12 & ๐ฝ๐ฝ = 2
โ ๐๐ = ๐๐๐๐๐๐๐๐๐๐+๐๐๐๐ + ๐๐๐๐๐๐
๐๐๐๐โ๐๐๐๐ ๐๐๐๐ ๐๐ = ๐๐
๐๐๐๐ โ๐๐ ๐๐๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐ +โ๐๐ ๐๐๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 153
Now,
I.C. ๐ฆ๐ฆ 1 = โ1;๐ฆ๐ฆโฒ 1 = โ12
๐ฆ๐ฆ 1 = โ1 โ โ1 = 112 โ1 ๐๐๐๐๐ ๐ 2 ๐๐๐๐ 1 +โ2 ๐ ๐ ๐ ๐ ๐๐ 2 ๐๐๐๐ 1
โ โ1 = 1 โ1 1 +โ2 0 โ โ1= โ1
๐ฆ๐ฆโฒ 1 = โ12
๐ฆ๐ฆโฒ =โ112 ๐ฅ๐ฅ
โ12 cos 2 ln ๐ฅ๐ฅ + ๐ฅ๐ฅ12 โ sin 2 ln ๐ฅ๐ฅ
2๐ฅ๐ฅ +
โ212๐ฅ๐ฅโ
12 sin 2 ln ๐ฅ๐ฅ + ๐ฅ๐ฅ
12 cos 2 ln ๐ฅ๐ฅ 2
๐ฅ๐ฅ
๐ฆ๐ฆ = ๐ฅ๐ฅ12 โ1 ๐๐๐๐๐ ๐ 2 ๐๐๐๐ ๐ฅ๐ฅ +โ2 ๐ ๐ ๐ ๐ ๐๐ 2 ๐๐๐๐ ๐ฅ๐ฅ
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 154
๐ฆ๐ฆโฒ =โ112๐ฅ๐ฅโ
12 cos 2 ln ๐ฅ๐ฅ + ๐ฅ๐ฅ
12 โ sin 2 ln ๐ฅ๐ฅ
2๐ฅ๐ฅ
+
โ212๐ฅ๐ฅโ
12 sin 2 ln ๐ฅ๐ฅ + ๐ฅ๐ฅ
12 cos 2 ln ๐ฅ๐ฅ 2
๐ฅ๐ฅ
โ โ12
= โ1 1
2 112
cos 0 + 112 โ sin 0
21
+
+โ2 1
2 112
sin 0 + 112 cos 0 2
1
โ โ12
= โ112
+โ2โโ2= โ12
+12
= 0 โโ2= 0
โ ๐๐ = โ๐๐๐๐๐๐๐๐๐๐๐๐ (๐๐ ๐ฅ๐ฅ๐ฅ๐ฅ ๐๐)
๐ฆ๐ฆ = ๐ฅ๐ฅ12 โ1 ๐๐๐๐๐ ๐ 2 ๐๐๐๐ ๐ฅ๐ฅ +โ2 ๐ ๐ ๐ ๐ ๐๐ 2 ๐๐๐๐ ๐ฅ๐ฅ
โ1= โ1
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 155
E.g.
๐ฅ๐ฅ3๐๐3๐ฆ๐ฆ๐๐๐ฅ๐ฅ3
+ 5๐ฅ๐ฅ2๐๐2๐ฆ๐ฆ๐๐๐ฅ๐ฅ2
+ 7๐ฅ๐ฅ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ 8๐ฆ๐ฆ = 0
Assume ๐ฆ๐ฆ = ๐ฅ๐ฅ๐๐ โ ๐ฆ๐ฆโฒ = ๐๐๐ฅ๐ฅ๐๐โ1 โ ๐ฆ๐ฆโฒโฒ = ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐โ2 โ ๐ฆ๐ฆโฒโฒโฒ = ๐๐ ๐๐โ 1 ๐๐โ 2 ๐ฅ๐ฅ๐๐โ3 โ ๐ฅ๐ฅ3 ๐๐ ๐๐ โ 1 ๐๐โ 2 ๐ฅ๐ฅ๐๐โ3 + 5๐ฅ๐ฅ2 ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐โ2 +7๐ฅ๐ฅ ๐๐๐ฅ๐ฅ๐๐โ1 + 8๐ฅ๐ฅ๐๐ = 0 โ ๐ฅ๐ฅ๐๐ ๐๐ ๐๐ โ 1 ๐๐โ 2 + 5๐๐ ๐๐ โ 1 + 7๐๐ + 8 = 0 โ ๐ฅ๐ฅ๐๐ ๐๐ ๐๐2 โ 3๐๐ + 2 + 5๐๐2 โ 5๐๐ + 7๐๐ + 8 = 0 โ ๐ฅ๐ฅ๐๐ ๐๐3 โ 3๐๐2 + 2๐๐ + 5๐๐2 + 2๐๐ + 8 = 0 โ ๐ฅ๐ฅ๐๐ ๐๐3 + 2๐๐2 + 4๐๐ + 8 = 0
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 156
โ ๐ฅ๐ฅ๐๐ ๐๐3 + 2๐๐2 + 4๐๐ + 8 = 0 โ ๐ฅ๐ฅ๐๐ ๐๐ + 2 ๐๐2 + 4 = 0 ๐๐ + 2 ๐๐2 + 4 = 0
โ ๐๐ + 2 ๐๐ + 2๐๐ ๐๐ โ 2๐๐ = 0 โ ๐๐1 = โ2,๐๐2 = โ2๐๐,๐๐3 = 2๐๐ Solution: ๐๐ = ๐๐๐๐๐๐โ๐๐ + ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐โ๐๐๐๐ or ๐๐ = ๐๐๐๐๐๐โ๐๐ + ๐๐๐๐ ๐๐๐๐๐๐(๐๐ ๐ฅ๐ฅ๐ฅ๐ฅ ๐๐) + ๐๐๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ(๐๐ ๐ฅ๐ฅ๐ฅ๐ฅ ๐๐)
Case 3: Conjugate Complex Roots
Auxiliary Equation
๐๐2 + 4 = 0 โ ๐๐2 = โ4 โ ๐๐2 = 4๐๐2 โ ๐๐ = ยฑ2๐๐
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 157
E.g. ๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 3๐ฆ๐ฆ = 2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ
โข Non Homogeneous Eqn. solve associated Homogeneous Eqn.
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 3๐ฆ๐ฆ = 0 Assume ๐ฆ๐ฆ = ๐ฅ๐ฅ๐๐ โ ๐ฆ๐ฆโฒ = ๐๐๐ฅ๐ฅ๐๐โ1 โ ๐ฆ๐ฆโฒโฒ = ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐โ2 โ ๐ฅ๐ฅ2 ๐๐ ๐๐ โ 1 ๐ฅ๐ฅ๐๐โ2 โ 3๐ฅ๐ฅ ๐๐๐ฅ๐ฅ๐๐โ1 + 3๐ฅ๐ฅ๐๐ = 0 โ ๐๐2 โ๐๐ ๐ฅ๐ฅ๐๐ โ 3๐๐๐ฅ๐ฅ๐๐ + 3๐ฅ๐ฅ๐๐ = 0 โ ๐ฅ๐ฅ๐๐ ๐๐2 โ๐๐ โ 3๐๐ + 3 = 0 โ ๐ฅ๐ฅ๐๐ ๐๐2 โ 4๐๐ + 3 = 0
Case 3: Conjugate Complex Roots
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 158
โ ๐ฅ๐ฅ๐๐ ๐๐2 โ 4๐๐ + 3 = 0 Auxiliary Eqn. ๐๐2 โ 4๐๐ + 3 = 0 โ ๐๐ โ 1 ๐๐ โ 3 = 0 โ ๐๐1 = 1 & ๐๐2 = 3
โ ๐๐๐๐ = ๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐ โข Utilize Variation of Parameters to solve for particular solution ๐ฆ๐ฆ๐๐
๐ฆ๐ฆ๐๐ = ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2
,where ๐ฆ๐ฆ1 = ๐ฅ๐ฅ & ๐ฆ๐ฆ2 = ๐ฅ๐ฅ3
Case 3: Conjugate Complex Roots
Auxiliary Equation
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 3๐ฆ๐ฆ = 2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 159
Note: To use Variation of Parameters must transform the equation
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 3๐ฆ๐ฆ = 2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ
Divide by ๐ฅ๐ฅ2,
๐ฆ๐ฆโฒโฒ โ3๐ฅ๐ฅ๐ฅ๐ฅ2 ๐ฆ๐ฆโฒ +
3๐ฅ๐ฅ2 ๐ฆ๐ฆ =
2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ
๐ฅ๐ฅ2
โ ๐ฆ๐ฆโฒโฒ โ3๐ฅ๐ฅ ๐ฆ๐ฆโฒ +
3๐ฅ๐ฅ2 ๐ฆ๐ฆ = 2๐ฅ๐ฅ2๐๐๐ฅ๐ฅ
Case 3: Conjugate Complex Roots
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 3๐ฆ๐ฆ = 2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐๐1๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ3 ๐ฆ๐ฆ1 = ๐ฅ๐ฅ &๐ฆ๐ฆ2 = ๐ฅ๐ฅ3
๐ท๐ท(๐๐) ๐ธ๐ธ(๐๐) ๐๐(๐๐)
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 160
Form all Wronskians: ๐๐ =
๐ฆ๐ฆ1 ๐ฆ๐ฆ2๐ฆ๐ฆ1๐ฆ ๐ฆ๐ฆ2โฒ
= ๐ฅ๐ฅ ๐ฅ๐ฅ31 3๐ฅ๐ฅ2
= 3๐ฅ๐ฅ3 โ ๐ฅ๐ฅ3 = 2๐ฅ๐ฅ3
๐๐1 =0 ๐ฆ๐ฆ2
๐๐(๐ฅ๐ฅ) ๐ฆ๐ฆ2โฒ= 0 ๐ฅ๐ฅ3
2๐ฅ๐ฅ2๐๐๐ฅ๐ฅ 3๐ฅ๐ฅ2= โ2๐ฅ๐ฅ5๐๐๐ฅ๐ฅ
๐๐2 = ๐ฅ๐ฅ 01 2๐ฅ๐ฅ2๐๐๐ฅ๐ฅ = 2๐ฅ๐ฅ3๐๐๐ฅ๐ฅ
โ ๐๐1โฒ =๐๐1๐๐ = โ
2๐ฅ๐ฅ5๐๐๐ฅ๐ฅ
2๐ฅ๐ฅ3 = โ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ
โ ๐๐2โฒ =๐๐2๐๐ =
2๐ฅ๐ฅ3๐๐๐ฅ๐ฅ
2๐ฅ๐ฅ3 = ๐๐๐ฅ๐ฅ
Case 3: Conjugate Complex Roots
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 3๐ฆ๐ฆ = 2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐๐1๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ3 ๐ฆ๐ฆ1 = ๐ฅ๐ฅ &๐ฆ๐ฆ2 = ๐ฅ๐ฅ3 ๐๐ ๐ฅ๐ฅ = 2๐ฅ๐ฅ2๐๐๐ฅ๐ฅ
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 161
Integrate ๐๐1โฒ & ๐๐2โฒ to get ๐๐1&๐๐2 :
๐๐2 = ๏ฟฝ๐๐2โฒ ๐๐๐ฅ๐ฅ = ๏ฟฝ๐๐๐ฅ๐ฅ๐๐๐ฅ๐ฅ = ๐๐๐๐
๐๐1 = ๏ฟฝ๐๐1โฒ ๐๐๐ฅ๐ฅ = โ๏ฟฝ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ๐๐๐ฅ๐ฅ
Let ๐ผ๐ผ = ๐ฅ๐ฅ2 โ ๐๐๐ผ๐ผ = 2๐ฅ๐ฅ ๐๐๐ฅ๐ฅ ;๐๐๐ฝ๐ฝ = ๐๐๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ ๐ฝ๐ฝ = ๐๐๐ฅ๐ฅ
โ ๐๐1 = โ ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ โ ๏ฟฝ๐๐๐ฅ๐ฅ2๐ฅ๐ฅ๐๐๐ฅ๐ฅ
โ ๐๐1 = โ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ + 2๏ฟฝ๐๐๐ฅ๐ฅ๐ฅ๐ฅ๐๐๐ฅ๐ฅ
Case 3: Conjugate Complex Roots
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 3๐ฆ๐ฆ = 2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐๐1๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ3 ๐ฆ๐ฆ1 = ๐ฅ๐ฅ &๐ฆ๐ฆ2 = ๐ฅ๐ฅ3 ๐๐ ๐ฅ๐ฅ = 2๐ฅ๐ฅ2๐๐๐ฅ๐ฅ ๐๐1โฒ = โ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ
๐๐2โฒ = ๐๐๐ฅ๐ฅ
Integration by parts
Cauchy-Euler Equations
9/30/2014 Dr. Eli Saber 162
โ ๐๐1 = โ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ + 2๏ฟฝ๐๐๐ฅ๐ฅ๐ฅ๐ฅ๐๐๐ฅ๐ฅ
โ ๐๐1 = โ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ + 2 ๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ ๐๐๐ฅ๐ฅ โ ๐๐๐๐ = โ๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐ โ ๐๐๐๐๐๐ Now, ๐ฆ๐ฆ๐๐ = ๐๐1๐ฆ๐ฆ1 + ๐๐2๐ฆ๐ฆ2 = โ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ + 2๐ฅ๐ฅ๐๐๐ฅ๐ฅ โ 2๐๐๐ฅ๐ฅ ๐ฅ๐ฅ + ๐๐๐ฅ๐ฅ๐ฅ๐ฅ3 = โ๐ฅ๐ฅ3๐๐๐ฅ๐ฅ + 2๐ฅ๐ฅ2๐๐๐ฅ๐ฅ โ 2๐๐๐ฅ๐ฅ๐ฅ๐ฅ + ๐๐๐ฅ๐ฅ๐ฅ๐ฅ3 = ๐๐๐๐๐๐๐๐๐๐ โ ๐๐๐๐๐๐๐๐ โ ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โ ๐๐ = ๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ โ ๐๐๐๐๐๐๐๐
Case 3: Conjugate Complex Roots
๐ฅ๐ฅ2๐ฆ๐ฆโฒโฒ โ 3๐ฅ๐ฅ๐ฆ๐ฆโฒ + 3๐ฆ๐ฆ = 2๐ฅ๐ฅ4๐๐๐ฅ๐ฅ ๐ฆ๐ฆ๐๐ = ๐๐1๐ฅ๐ฅ + ๐๐2๐ฅ๐ฅ3 ๐ฆ๐ฆ1 = ๐ฅ๐ฅ &๐ฆ๐ฆ2 = ๐ฅ๐ฅ3 ๐๐ ๐ฅ๐ฅ = 2๐ฅ๐ฅ2๐๐๐ฅ๐ฅ ๐๐1โฒ = โ๐ฅ๐ฅ2๐๐๐ฅ๐ฅ
๐๐2 = ๐๐๐ฅ๐ฅ
Cauchy-Euler Equations
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Summary
โข Identified when ๐๐๐๐ matches the order of the differentiation ๐ ๐
๐๐๐๐๐ ๐ ๐๐๐๐
๐๐๐๐๐ฅ๐ฅ๐๐๐๐๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ๐๐
+ โฆ + ๐๐1๐ฅ๐ฅ๐๐๐ฆ๐ฆ๐๐๐ฅ๐ฅ
+ ๐๐0 ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ)
Step1: Obtain Complementary Solution(๐ฆ๐ฆ๐๐) โข Consider ๐๐ ๐๐ = ๐๐
โข Try the form ๐๐ = ๐๐๐๐
โข Auxiliary Equation:
๐๐๐๐๐๐ + ๐๐ โ ๐๐ ๐๐ + ๐๐ = ๐๐
โข Obtain roots for the equation โ Case 1: Distinct Real Roots โ ๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ๐๐1 + ๐๐2๐ฅ๐ฅ๐๐2 + โฏ+ ๐๐๐๐๐ฅ๐ฅ๐๐๐๐
โ Case 2: Repeated Real Roots โ ๐ฆ๐ฆ = ๐๐1๐ฅ๐ฅ๐๐1 + ๐๐2๐ฅ๐ฅ๐๐1 ๐๐๐๐ ๐ฅ๐ฅ +
๐๐3๐ฅ๐ฅ๐๐1 ๐๐๐๐ ๐ฅ๐ฅ 2 + โฏ+๐๐๐๐๐ฅ๐ฅ๐๐1 ๐๐๐๐ ๐ฅ๐ฅ ๐๐โ1
โ Case 3: Conjugate Complex Roots โ ๐ฆ๐ฆ =
๐ฅ๐ฅ๐ผ๐ผ โ1 ๐๐๐๐๐ ๐ ๐ฝ๐ฝ ๐๐๐๐ ๐ฅ๐ฅ +โ2 ๐ ๐ ๐ ๐ ๐๐ ๐ฝ๐ฝ ๐๐๐๐ ๐ฅ๐ฅ
Step2: Obtain Particular Solution (๐ฆ๐ฆ๐๐) โข Use either Undetermined Coefficients
(3.4) or Variation of Parameters (3.5)
Step3: Combine to obtain general solution โข ๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
Step4: Verify the solution
Section 3.8 Linear Models
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Linear Models
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Note: ๐๐๐ ๐ = ๐ ๐ ๐ ๐
๐๐๐ฟ๐ฟ = ๐ฟ๐ฟ๐๐๐ ๐ ๐๐๐ก๐ก
๐ ๐ ๐๐ = ๐๐๐๐๐๐๐๐๐๐๐ก๐ก
; ๐ ๐ =๐๐๐๐๐๐๐ก๐ก
โ๐๐๐ ๐ ๐๐๐ก๐ก
=๐๐2๐๐๐๐๐ก๐ก2
๐๐๐๐ = ๐ ๐ ๐๐๐ก๐ก โ ๐๐ = โซ ๐ ๐ ๐๐๐ก๐ก ๐๐ charge
Kirchoffโs Voltage Law:
๐ธ๐ธ = ๐ ๐ ๐ ๐ + ๐ฟ๐ฟ๐๐๐ ๐ ๐๐๐ก๐ก
+1๐ถ๐ถ
๏ฟฝ๐ ๐ ๐๐๐ก๐ก
C
L
R
E
3.8.4. : Series Circuit (LRC)
๐ ๐ (๐ก๐ก)
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๐ธ๐ธ = ๐ ๐ ๐ ๐ + ๐ฟ๐ฟ๐๐๐ ๐ ๐๐๐ก๐ก
+1๐ถ๐ถ
๏ฟฝ๐ ๐ ๐๐๐ก๐ก
โ ๐ธ๐ธ = ๐ ๐ ๐๐๐๐๐๐๐ก๐ก
+ ๐ฟ๐ฟ๐๐2๐๐๐๐๐ก๐ก2
+๐๐๐ถ๐ถ
โ ๐น๐น๐๐๐๐๐๐๐ก๐ก + ๐ณ๐ณ
๐๐2๐๐๐๐๐ก๐ก2 +
๐๐๐ช๐ช๐๐ = ๐ธ๐ธ
โข ๐ ๐ = ๐๐๐๐๐๐๐ก๐ก
;๐๐๐ ๐ = ๐ ๐ ๐ ๐
โข ๐๐๐ฟ๐ฟ = ๐ฟ๐ฟ ๐๐๐๐๐๐๐ก๐ก
โข ๐๐๐ถ๐ถ = 1๐ถ๐ถ โซ ๐ ๐ ๐๐๐ก๐ก
โข ๐ธ๐ธ(๐ก๐ก): forcing function
C
L
R
E
3.8.4. : Series Circuit (LRC)
๐ ๐ (๐ก๐ก)
Linear Models
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๐ ๐ ๐๐๐๐๐๐๐ก๐ก
+ ๐ฟ๐ฟ๐๐2๐๐๐๐๐ก๐ก2
+๐๐๐ถ๐ถ
= ๐ธ๐ธ
Rearranging the equation, we get:
๐ฟ๐ฟ๐๐2๐๐๐๐๐ก๐ก2
+ ๐ ๐ ๐๐๐๐๐๐๐ก๐ก
+1๐ถ๐ถ๐๐ = ๐ธ๐ธ
Auxiliary Eqn: ๐ฟ๐ฟ๐๐2 + ๐ ๐ ๐๐ + 1๐ถ๐ถ
= 0
โ ๐๐ =โ๐ ๐ ยฑ ๐ ๐ 2 โ 4 ๐ฟ๐ฟ 1
๐ถ๐ถ2๐ฟ๐ฟ
โ ๐๐ =โ๐ ๐ ยฑ ๐ ๐ 2 โ 4๐ฟ๐ฟ
๐ถ๐ถ2๐ฟ๐ฟ
C
L
R
E
3.8.4. : Series Circuit (LRC)
๐ ๐ (๐ก๐ก)
(Assume ๐ธ๐ธ ๐ก๐ก = 0)
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๐๐ =โ๐ ๐ ยฑ ๐ ๐ 2 โ 4๐ฟ๐ฟ
๐ถ๐ถ2๐ฟ๐ฟ
โข If ๐น๐น๐๐ โ ๐๐๐ณ๐ณ๐ช๐ช
> ๐๐ over damped
โข If ๐น๐น๐๐ โ ๐๐๐ณ๐ณ๐ช๐ช
= ๐๐ critically damped
โข If ๐น๐น๐๐ โ ๐๐๐ณ๐ณ๐ช๐ช
< ๐๐ under damped
Now,
๐๐ =โ๐ ๐ ยฑ ๐ ๐ 2 โ 4๐ฟ๐ฟ
๐ถ๐ถ2๐ฟ๐ฟ = โ
๐ ๐ 2๐ฟ๐ฟ ยฑ
๐ ๐ 2 โ 4๐ฟ๐ฟ๐ถ๐ถ
2๐ฟ๐ฟ
3.8.4. : Series Circuit (LRC)
๐ถ๐ถ ๐ท๐ท
C
L
R
E ๐ ๐ (๐ก๐ก)
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E.g. ๐ฟ๐ฟ = 0.25๐ป๐ป;๐ ๐ = 10ฮฉ;๐ถ๐ถ = 0.001๐น๐น ๐ธ๐ธ ๐ก๐ก = 0๐๐; ๐๐ 0 = ๐๐0; ๐ ๐ 0 = 0๐ด๐ด Solution:
๐ฟ๐ฟ๐๐2๐๐๐๐๐ก๐ก2 + ๐ ๐
๐๐๐๐๐๐๐ก๐ก +
1๐ถ๐ถ ๐๐ = 0 โ 0.25๐๐โฒโฒ + 10๐๐โฒ + 1000๐๐ = 0
โ ๐๐โฒโฒ + 40๐๐โฒ + 4000๐๐ = 0 Aux. Eq.: ๐๐2 + 40๐๐ + 4000 = 0
๐๐ =โ40 ยฑ 1600 โ 4(1)(4000)
2 =โ40 ยฑ 1600 โ 16000
2
3.8.4. : Series Circuit (LRC)
C= 0.001๐น๐น
L= 0.25๐ป๐ป
R=10 ฮฉ
E=0V ๐ ๐ (๐ก๐ก)
Linear Models
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๐๐ =โ40 ยฑ โ14400
2
๐๐ =โ40 ยฑ 14400๐๐2
2 =โ40 ยฑ 16(900)๐๐2
2
=โ40 ยฑ 4 30 ๐๐
2 = โ๐๐๐๐ ยฑ ๐๐๐๐๐๐
๐๐1 = โ20 + 60๐๐ & ๐๐2 = โ20 โ 60๐๐ Hence: ๐ผ๐ผ = โ20 & ๐ฝ๐ฝ = 60 โ ๐๐ ๐ก๐ก = ๐๐โ20๐ก๐ก ๐๐1 cos60๐ก๐ก + ๐๐2 sin60๐ก๐ก
3.8.4. : Series Circuit (LRC) ๐ฟ๐ฟ = 0.25๐ป๐ป; ๐ ๐ = 10ฮฉ; ๐ถ๐ถ = 0.001๐น๐น ๐ธ๐ธ ๐ก๐ก = 0๐๐; ๐๐ 0 = ๐๐0; ๐ ๐ 0 = 0๐ด๐ด Aux. Eqn. ๐๐2 + 40๐๐ + 4000 = 0
Linear Models
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โข ๐๐ 0 = ๐๐0
โ ๐๐0 = ๐๐0 ๐๐1 cos 0 + ๐๐2 sin 0 โ ๐๐0 = 1 ๐๐1 + 0 โ ๐๐1 = ๐๐0 Hence, we now have:
๐๐ ๐๐ = ๐๐โ๐๐๐๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐
โ ๐ ๐ ๐ก๐ก =๐๐๐๐๐๐๐ก๐ก = โ20๐๐โ20๐ก๐ก[๐๐0 cos60๐ก๐ก + ๐๐2 sin 60๐ก๐ก]
+๐๐โ20๐ก๐ก[โ60๐๐0 sin 60๐ก๐ก + 60๐๐2 cos60๐ก๐ก]
3.8.4. : Series Circuit (LRC) ๐ฟ๐ฟ = 0.25๐ป๐ป; ๐ ๐ = 10ฮฉ; ๐ถ๐ถ = 0.001๐น๐น ๐ธ๐ธ ๐ก๐ก = 0๐๐; ๐๐ 0 = ๐๐0; ๐ ๐ 0 = 0๐ด๐ด Aux. Eqn. ๐๐2 + 40๐๐ + 4000 = 0
๐๐ ๐ก๐ก = ๐๐โ20๐ก๐ก ๐๐1 cos60๐ก๐ก + ๐๐2 sin 60๐ก๐ก
Linear Models
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๐ ๐ ๐ก๐ก =โ20๐๐โ20๐ก๐ก[๐๐0 cos60๐ก๐ก + ๐๐2 sin 60๐ก๐ก]
+๐๐โ20๐ก๐ก[โ60๐๐0 sin 60๐ก๐ก + 60๐๐2 cos60๐ก๐ก]
But ๐ ๐ 0 = 0 โ 0 = โ20 ๐๐0 + 0 + 1 0 + 60 ๐๐2 โ 0 = โ20 ๐๐0 + 60๐๐2 โ 60๐๐2 = 20๐๐0
โ ๐๐2 =2060 ๐๐0 โ ๐๐2 =
13 ๐๐0
โ ๐๐ ๐ก๐ก = ๐๐โ20๐ก๐ก ๐๐0 ๐๐๐๐๐ ๐ 60๐ก๐ก +๐๐03 ๐ ๐ ๐ ๐ ๐๐ 60๐ก๐ก
โ ๐๐ ๐ก๐ก = ๐๐0๐๐โ20๐ก๐ก ๐๐๐๐๐ ๐ 60๐ก๐ก +13 ๐ ๐ ๐ ๐ ๐๐ 60๐ก๐ก
3.8.4. : Series Circuit (LRC) ๐ฟ๐ฟ = 0.25๐ป๐ป; ๐ ๐ = 10ฮฉ; ๐ถ๐ถ = 0.001๐น๐น ๐ธ๐ธ ๐ก๐ก = 0๐๐; ๐๐ 0 = ๐๐0; ๐ ๐ 0 = 0๐ด๐ด Aux. Eqn. ๐๐2 + 40๐๐ + 4000 = 0
๐๐ ๐ก๐ก = ๐๐โ20๐ก๐ก ๐๐0 ๐๐๐๐๐ ๐ 60๐ก๐ก + ๐๐2 ๐ ๐ ๐ ๐ ๐๐ 60๐ก๐ก
๐๐ ๐ก๐ก = ๐๐0๐๐โ20๐ก๐ก ๐๐๐๐๐ ๐ 60๐ก๐ก +13๐ ๐ ๐ ๐ ๐๐ 60๐ก๐ก
We know, sin ๐ด๐ด + ๐ต๐ต = sin๐ด๐ด cos๐ต๐ต + cos๐ด๐ด sin๐ต๐ต We can transform ๐๐(๐ก๐ก) into an alternate form:
๐๐ ๐ก๐ก = ๐๐0๐๐โ20๐ก๐ก (1) ๐๐๐๐๐ ๐ 60๐ก๐ก +13 ๐ ๐ ๐ ๐ ๐๐ 60๐ก๐ก
โ ๐๐ ๐ก๐ก = ๐๐0๐๐โ20๐ก๐ก103
1103
๐๐๐๐๐ ๐ 60๐ก๐ก +13103
๐ ๐ ๐ ๐ ๐๐ 60๐ก๐ก
Linear Models
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3.8.4. : Series Circuit (LRC)
๐๐ 13
1 1 2 +
13
2
=๐๐๐๐๐๐
sin๐๐ =1103
; cos๐๐ =13103
๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐ ๐๐๐๐๐๐๐๐
โ ๐๐ ๐ก๐ก = ๐๐0103
๐๐โ20๐ก๐ก sin[60๐ก๐ก + ๐๐]
Note: sin๐๐ = 310โ ๐๐ = sinโ1 3
10= 1.249 rad
โ ๐๐ ๐ก๐ก = ๐๐0103 ๐๐โ20๐ก๐ก sin[60๐ก๐ก + 1.249]
Linear Models
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3.8.4. : Series Circuit (LRC)
๐๐ 13
1 1 2 +
13
2
=๐๐๐๐๐๐
sin๐๐ =1103
; cos๐๐ =13103
Note: โข ๐๐๐๐(๐ก๐ก): solution to the homogeneous equation is called the transient solution
โข ๐๐๐๐ ๐ก๐ก : solution to the non-homogeneous equation (i.e. ๐ธ๐ธ(๐ก๐ก) โ 0) is called the
steady-state solution
Linear Models
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3.8.4. : Series Circuit (LRC)
E.g. ๐ฟ๐ฟ = 1๐ป๐ป;๐ ๐ = 2ฮฉ;๐ถ๐ถ = 0.25๐น๐น; ๐ธ๐ธ ๐ก๐ก = 50 cos ๐ก๐ก ๐ ๐ ๐๐๐๐๐ก๐ก๐ ๐ Find the steady-state charge and the steady-state current in the LRC Circuit (Advanced Eng. Mathematics โ 5th Edition โ Ex. 3.8 Prob. 49) Solution:
๐ธ๐ธ ๐ก๐ก = ๐ฟ๐ฟ๐๐๐ ๐ ๐๐๐ก๐ก
+ ๐ ๐ ๐ ๐ +๐๐๐ถ๐ถ
โ ๐ฟ๐ฟ๐๐2๐๐๐๐๐ก๐ก2
+ ๐ ๐ ๐๐๐๐๐๐๐ก๐ก
+๐๐๐ถ๐ถ
= ๐ธ๐ธ ๐ก๐ก
โ 1๐๐2๐๐๐๐๐ก๐ก2
+ 2๐๐๐๐๐๐๐ก๐ก
+๐๐
0.25= 50 cos ๐ก๐ก
Linear Models
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3.8.4. : Series Circuit (LRC)
C= 0.25๐น๐น
L= 1๐ป๐ป
R=2 ฮฉ
E=50 cos (t) V ๐ ๐ (๐ก๐ก)
โ 1๐๐2๐๐๐๐๐ก๐ก2
+ 2๐๐๐๐๐๐๐ก๐ก
+ 4 ๐๐ = 50 cos ๐ก๐ก
Homogeneous Eqn. ๐๐2๐๐๐๐๐ก๐ก2
+ 2 ๐๐๐๐๐๐๐ก๐ก
+ 4 ๐๐ = 0 โ ๐๐2 + 2๐๐ + 4 = 0
โ ๐๐ =โ2 ยฑ 4 โ 4(1)(4)
2 =โ2 ยฑ 12
2 =โ2 ยฑ 4 3 ๐๐2
2
โ ๐๐ = โ1 ยฑ ๐๐ 3 โ ๐ผ๐ผ = โ1 & ๐ฝ๐ฝ = 3
๐๐๐๐ ๐ก๐ก = ๐๐1๐๐ โ1+๐๐ 3 ๐ก๐ก + ๐๐2๐๐ โ1โ๐๐ 3 ๐ก๐ก or
๐๐๐๐ ๐ก๐ก = ๐๐โ๐ก๐ก โ1 cos 3๐ก๐ก+โ2 sin 3๐ก๐ก
Linear Models
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3.8.4. : Series Circuit (LRC)
From table 3.4.1., we can write: ๐๐๐๐ ๐ก๐ก = ๐ด๐ด cos ๐ก๐ก + ๐ต๐ต sin ๐ก๐ก ๐๐โฒ ๐ก๐ก = โ๐ด๐ด sin ๐ก๐ก + ๐ต๐ต cos ๐ก๐ก ๐๐โฒโฒ ๐ก๐ก = โ๐ด๐ด cos ๐ก๐ก โ ๐ต๐ต sin ๐ก๐ก ReโSubstituting back in eqn. โ โ๐ด๐ด cos ๐ก๐ก โ ๐ต๐ต sin ๐ก๐ก + 2 โ๐ด๐ด sin ๐ก๐ก + ๐ต๐ต cos ๐ก๐ก + 4 ๐ด๐ด cos ๐ก๐ก + ๐ต๐ต sin ๐ก๐ก = 50 cos ๐ก๐ก โ โ๐ด๐ด cos ๐ก๐ก โ ๐ต๐ต sin ๐ก๐ก โ 2๐ด๐ด sin ๐ก๐ก + 2๐ต๐ต cos ๐ก๐ก + 4๐ด๐ด cos ๐ก๐ก + 4๐ต๐ต sin ๐ก๐ก = 50 cos ๐ก๐ก โ cos ๐ก๐ก 3๐ด๐ด + 2๐ต๐ต + sin ๐ก๐ก โ2๐ด๐ด + 3๐ต๐ต = 50 cos ๐ก๐ก โ 3๐ด๐ด + 2๐ต๐ต = 50
โ โ2๐ด๐ด + 3๐ต๐ต = 0 โ 2๐ด๐ด = 3๐ต๐ต โ ๐ด๐ด =32๐ต๐ต
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3.8.4. : Series Circuit (LRC)
1๐๐2๐๐๐๐๐ก๐ก2 + 2
๐๐๐๐๐๐๐ก๐ก + 4 ๐๐ = 50 cos ๐ก๐ก
3๐ด๐ด + 2๐ต๐ต = 50
332๐ต๐ต + 2๐ต๐ต = 50 โ
92๐ต๐ต + 2๐ต๐ต = 50 โ
132๐ต๐ต = 50 โ ๐ต๐ต =
10013
๐ด๐ด =32๐ต๐ต =
32โ
10013
=30026
โ ๐ด๐ด =15013
๐๐๐๐ ๐ก๐ก =15013
cos ๐ก๐ก +10013
sin ๐ก๐ก
We already have: ๐๐๐๐ ๐ก๐ก = ๐๐โ๐ก๐ก โ1 cos 3๐ก๐ก+โ2 sin 3๐ก๐ก Hence,
๐๐ ๐ก๐ก = ๐๐โ๐ก๐ก ๐๐1 cos 3๐ก๐ก + ๐๐2 sin 3๐ก๐ก +15013
cos ๐ก๐ก +10013
sin ๐ก๐ก
Linear Models
9/30/2014 Dr. Eli Saber 179
3.8.4. : Series Circuit (LRC)
1๐๐2๐๐๐๐๐ก๐ก2 + 2
๐๐๐๐๐๐๐ก๐ก + 4 ๐๐ = 50 cos ๐ก๐ก
3๐ด๐ด + 2๐ต๐ต = 50
๐ด๐ด =32๐ต๐ต
Transient Solution Steady-State Solution
๐๐๐ ๐ ๐ ๐ ๐ก๐ก = ๐๐๐๐ ๐ก๐ก โ ๐๐๐ ๐ ๐ ๐ ๐ก๐ก =15013
cos ๐ก๐ก +10013
sin ๐ก๐ก
๐ ๐ ๐ก๐ก =๐๐๐๐๐๐๐ก๐ก
= โ๐๐โ๐ก๐ก ๐๐1 cos 3๐ก๐ก + ๐๐2 sin 3๐ก๐ก + ๐๐โ๐ก๐ก โ๐๐1 3 sin 3๐ก๐ก + ๐๐2 3 cos 3๐ก๐ก โ15013
sin ๐ก๐ก +10013
cos ๐ก๐ก
โ ๐ ๐ ๐ก๐ก = โ๐๐1๐๐โ๐ก๐ก cos 3๐ก๐ก โ ๐๐2๐๐โ๐ก๐ก sin 3๐ก๐ก โ ๐๐1๐๐โ๐ก๐ก 3 sin 3๐ก๐ก + ๐๐2๐๐โ๐ก๐ก 3 cos 3๐ก๐ก โ15013
sin ๐ก๐ก
+10013
cos ๐ก๐ก
โ ๐ ๐ ๐ก๐ก = ๐๐โ๐ก๐ก โ๐๐1 + 3๐๐2 cos 3๐ก๐ก + ๐๐โ๐ก๐ก โ๐๐2 โ 3๐๐1 sin 3๐ก๐ก โ15013
sin ๐ก๐ก +10013
cos ๐ก๐ก
โ ๐ ๐ ๐ ๐ ๐ ๐ ๐ก๐ก = โ15013
sin ๐ก๐ก + 10013
cos ๐ก๐ก
Linear Models
9/30/2014 Dr. Eli Saber 180
3.8.4. : Series Circuit (LRC)
๐๐ ๐ก๐ก = ๐๐โ๐ก๐ก ๐๐1 cos 3๐ก๐ก + ๐๐2 sin 3๐ก๐ก +15013
cos ๐ก๐ก +10013
sin ๐ก๐ก
Linear Models
9/30/2014 Dr. Eli Saber 181
Summary
๐ฟ๐ฟ๐๐๐ ๐ ๐๐๐ก๐ก
+ ๐ ๐ ๐ ๐ +1๐ถ๐ถ๏ฟฝ ๐ ๐ ๐๐๐ก๐ก = ๐ธ๐ธ(๐ก๐ก)
โ ๐ฟ๐ฟ๐๐2๐๐๐๐๐ก๐ก2
+ ๐ ๐ ๐๐๐๐๐๐๐ก๐ก
+๐๐๐ถ๐ถ
= ๐ธ๐ธ ๐ก๐ก
Auxiliary Eqn: ๐ฟ๐ฟ๐๐2 + ๐ ๐ ๐๐ + 1๐ถ๐ถ
= 0
๐๐ =โ๐ ๐ ยฑ ๐ ๐ 2 โ 4๐ฟ๐ฟ
๐ถ๐ถ2๐ฟ๐ฟ
= โ๐ ๐ 2๐ฟ๐ฟ
ยฑ๐ ๐ 2 โ 4๐ฟ๐ฟ
๐ถ๐ถ2๐ฟ๐ฟ
Use already known methods to obtain ๐ฆ๐ฆ๐๐
๐๐ = ๐๐๐๐ + ๐๐๐๐
C
L
R
E ๐ ๐ (๐ก๐ก)
๐ถ๐ถ ๐ท๐ท
โ obtain ๐ฆ๐ฆ๐๐
Section 3.12 Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 182
Newtonโs 2nd Law:
๐๐1๐๐2๐ฅ๐ฅ1๐๐๐ก๐ก2 = โ๐๐1๐ฅ๐ฅ1 + ๐๐2 ๐ฅ๐ฅ2 โ ๐ฅ๐ฅ1
๐๐2๐๐2๐ฅ๐ฅ2๐๐๐ก๐ก2 = โ๐๐2 ๐ฅ๐ฅ2 โ ๐ฅ๐ฅ1
Also can be written as: ๐๐1๐ฅ๐ฅ1โฒโฒ = โ๐๐1๐ฅ๐ฅ1 + ๐๐2 ๐ฅ๐ฅ2 โ ๐ฅ๐ฅ1 ๐๐2๐ฅ๐ฅ2โฒโฒ = โ๐๐2 ๐ฅ๐ฅ2 โ ๐ฅ๐ฅ1
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 183
Coupled Spring/Mass Systems
A coupled system of Differential Equations
Ref. D. Zill & W. Wright, Advanced Engineering Mathematics. 5th Ed.
Given ๐๐๐๐๐ฆ๐ฆ ๐๐ + ๐๐๐๐โ1๐ฆ๐ฆ ๐๐โ1 + โฏ+ ๐๐1๐ฆ๐ฆโฒ + ๐๐0๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ ,where ๐๐๐๐ , ๐ ๐ = 0,1,2,3, โฆ ,๐๐ are constants Rewrite as: ๐๐๐๐๐ท๐ท๐๐ + ๐๐๐๐โ1๐ท๐ท๐๐โ1 + โฏ+ ๐๐1๐ท๐ท + ๐๐0 ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ Then group like terms for solving.
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 184
Systematic Elimination
Given: ๐ฅ๐ฅโฒโฒ + 2๐ฅ๐ฅโฒ + ๐ฆ๐ฆโฒโฒ = ๐ฅ๐ฅ + 3๐ฆ๐ฆ + sin ๐ก๐ก ๐ฅ๐ฅโฒ + ๐ฆ๐ฆโฒ = โ4๐ฅ๐ฅ + 2๐ฆ๐ฆ + ๐๐โ๐ก๐ก โ ๐ฅ๐ฅโฒโฒ + 2๐ฅ๐ฅโฒ + ๐ฆ๐ฆโฒโฒ โ ๐ฅ๐ฅ โ 3๐ฆ๐ฆ = sin ๐ก๐ก โ ๐ฅ๐ฅโฒ + ๐ฆ๐ฆโฒ + 4๐ฅ๐ฅ โ 2๐ฆ๐ฆ = ๐๐โ๐ก๐ก โ ๐ท๐ท2๐ฅ๐ฅ + 2๐ท๐ท๐ฅ๐ฅ + ๐ท๐ท2๐ฆ๐ฆ โ ๐ฅ๐ฅ โ 3๐ฆ๐ฆ = sin ๐ก๐ก โ ๐ท๐ท๐ฅ๐ฅ + ๐ท๐ท๐ฆ๐ฆ + 4๐ฅ๐ฅ โ 2๐ฆ๐ฆ = ๐๐โ๐ก๐ก โ ๐ซ๐ซ๐๐ + ๐๐๐ซ๐ซ โ ๐๐ ๐ฅ๐ฅ + ๐ซ๐ซ๐๐ โ ๐๐ ๐ฆ๐ฆ = sin ๐ก๐ก โ ๐ซ๐ซ + ๐๐ ๐ฅ๐ฅ + ๐ซ๐ซโ ๐๐ ๐ฆ๐ฆ = ๐๐โ๐ก๐ก
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 185
A solution of a system of D.E. is a set of sufficiently differentiable functions
๐ฅ๐ฅ = โ 1 ๐ก๐ก ๐ฆ๐ฆ = โ 2 ๐ก๐ก ๐ง๐ง = โ 3 ๐ก๐ก
โฎ ๐๐๐๐๐๐ ๐ ๐ ๐๐ ๐๐๐๐
that satisfies each equation in the system on some common interval ๐ผ๐ผ
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 186
Solution of System
E.g. Linear 1st order equations: ๐๐๐ฅ๐ฅ๐๐๐ก๐ก
= 3๐ฆ๐ฆ
๐๐๐ฆ๐ฆ๐๐๐ก๐ก
= 2๐ฅ๐ฅ
Solution: ๐๐๐ฅ๐ฅ๐๐๐ก๐ก = 3๐ฆ๐ฆ โ ๐ท๐ท๐ฅ๐ฅ โ 3๐ฆ๐ฆ = 0
๐๐๐ฆ๐ฆ๐๐๐ก๐ก = 2๐ฅ๐ฅ โ ๐ท๐ท๐ฆ๐ฆ โ 2๐ฅ๐ฅ = 0
๐ท๐ท๐ฅ๐ฅ โ 3๐ฆ๐ฆ = 0 โ ๐๐๐ ๐ ๐๐๐๐๐๐๐ก๐ก๐๐ ๐ค๐ค๐ ๐ ๐ก๐กโ ๐ท๐ท โ ๐ท๐ท2๐ฅ๐ฅ โ 3๐ท๐ท๐ฆ๐ฆ = 0 ๐ท๐ท๐ฆ๐ฆ โ 2๐ฅ๐ฅ = 0 โ ๐๐๐๐๐๐๐ก๐ก๐ ๐ ๐ ๐ ๐๐๐ฆ๐ฆ ๐๐๐ฆ๐ฆ 3 โ +3๐ท๐ท๐ฆ๐ฆ โ 6๐ฅ๐ฅ = 0
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 187
๐ซ๐ซ๐๐๐๐ โ ๐๐๐๐ = ๐๐
โ ๐ท๐ท2๐ฅ๐ฅ โ 6๐ฅ๐ฅ = 0 Auxiliary Equation: ๐๐2 โ 6 = 0 โ ๐๐2 = 6 โ ๐๐ = ยฑ 6
๐๐ ๐๐ = ๐๐๐๐๐๐โ ๐๐๐๐ + ๐๐๐๐๐๐ ๐๐๐๐ Now, to obtain ๐ฆ๐ฆ(๐ก๐ก): ๐ท๐ท๐ฅ๐ฅ โ 3๐ฆ๐ฆ = 0 โ ๐๐๐๐๐๐๐ก๐ก๐ ๐ ๐ ๐ ๐๐๐ฆ๐ฆ ๐๐๐ฆ๐ฆ 2 โ 2๐ท๐ท๐ฅ๐ฅ โ 6๐ฆ๐ฆ = 0 ๐ท๐ท๐ฆ๐ฆ โ 2๐ฅ๐ฅ = 0 โ ๐๐๐ ๐ ๐๐๐๐๐๐๐ก๐ก๐๐ ๐ค๐ค๐ ๐ ๐ก๐กโ ๐ท๐ท โ ๐ท๐ท2๐ฆ๐ฆ โ 2๐ท๐ท๐ฅ๐ฅ = 0 Auxiliary Equation: ๐๐2 โ 6 = 0 โ ๐๐2 = 6 โ ๐๐ = ยฑ 6
๐๐ ๐๐ = ๐๐๐๐๐๐โ ๐๐๐๐ + ๐๐๐๐๐๐ ๐๐๐๐
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 188
๐ซ๐ซ๐๐๐๐ โ ๐๐๐๐ = ๐๐
๐ฅ๐ฅ๐ฆ ๐ก๐ก = โ 6๐๐1๐๐โ 6๐ก๐ก + 6๐๐2๐๐ 6๐ก๐ก
We know: ๐๐๐ฅ๐ฅ๐๐๐ก๐ก
= 3๐ฆ๐ฆ
โ โ 6๐๐1๐๐โ 6๐ก๐ก + 6๐๐2๐๐ 6๐ก๐ก = 3๐๐3๐๐โ 6๐ก๐ก + 3๐๐4๐๐ 6๐ก๐ก โ โ 6๐๐1 โ 3๐๐3 ๐๐โ 6๐ก๐ก + 6๐๐2 โ 3๐๐4 ๐๐ 6๐ก๐ก = 0 โ๐ก๐ก
โ โ 6๐๐1 โ 3๐๐3 = 0 โ ๐๐3 = โ6
3๐๐1
โ 6๐๐2 โ 3๐๐4 = 0 โ ๐๐4 =6
3c2
๐๐ ๐๐ = ๐๐๐๐๐๐โ ๐๐๐๐ + ๐๐๐๐๐๐ ๐๐๐๐ & ๐๐ ๐๐ = โ๐๐๐๐๐๐๐๐๐๐โ ๐๐๐๐ +
๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 189
๐ฅ๐ฅ ๐ก๐ก = ๐๐1๐๐โ 6๐ก๐ก + ๐๐2๐๐ 6๐ก๐ก ๐ฆ๐ฆ ๐ก๐ก = ๐๐3๐๐โ 6๐ก๐ก + ๐๐4๐๐ 6๐ก๐ก
E.g. ๐ฅ๐ฅโฒ โ 4๐ฅ๐ฅ + ๐ฆ๐ฆโฒโฒ = ๐ก๐ก2 ๐ฅ๐ฅโฒ + ๐ฅ๐ฅ + ๐ฆ๐ฆโฒ = 0
Solution: ๐ท๐ท๐ฅ๐ฅ โ 4๐ฅ๐ฅ + ๐ท๐ท2๐ฆ๐ฆ = ๐ก๐ก2 โ ๐ท๐ท โ 4 ๐ฅ๐ฅ + ๐ท๐ท2๐ฆ๐ฆ = ๐ก๐ก2 ๐ท๐ท๐ฅ๐ฅ + ๐ฅ๐ฅ + ๐ท๐ท๐ฆ๐ฆ = 0 โ ๐ท๐ท + 1 ๐ฅ๐ฅ + ๐ท๐ท๐ฆ๐ฆ = 0 Solving for ๐ฆ๐ฆ first: 1 โ ๐ท๐ท + 1 โ ๐ท๐ท โ 4 ๐ท๐ท + 1 ๐ฅ๐ฅ + ๐ท๐ท2 ๐ท๐ท + 1 ๐ฆ๐ฆ = ๐ท๐ท + 1 ๐ก๐ก2
2 โ ๐ท๐ท โ 4 โ ๐ท๐ท + 1 ๐ท๐ท โ 4 ๐ฅ๐ฅ + ๐ท๐ท ๐ท๐ท โ 4 ๐ฆ๐ฆ = 0
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 190
(๐๐)
(๐๐)
๐ซ๐ซ๐๐ ๐ซ๐ซ + ๐๐ ๐๐ โ ๐ซ๐ซ ๐ซ๐ซโ ๐๐ ๐๐ = ๐ซ๐ซ + ๐๐ ๐๐๐๐ (โ) (โ) (โ)
โ ๐ท๐ท2 ๐ท๐ท + 1 ๐ฆ๐ฆ โ ๐ท๐ท ๐ท๐ท โ 4 ๐ฆ๐ฆ = ๐ท๐ท + 1 ๐ก๐ก2 โ ๐ท๐ท3 + ๐ท๐ท2 โ ๐ท๐ท2 + 4๐ท๐ท ๐ฆ๐ฆ = ๐ท๐ท๐ก๐ก2 + ๐ก๐ก2 โ ๐ท๐ท3 + 4๐ท๐ท ๐ฆ๐ฆ = 2๐ก๐ก + ๐ก๐ก2
โ๐๐3๐ฆ๐ฆ๐๐๐ก๐ก3 + 4
๐๐๐ฆ๐ฆ๐๐๐ก๐ก = 2๐ก๐ก + ๐ก๐ก2
โ ๐ท๐ท3 + 4๐ท๐ท ๐ฆ๐ฆ = 2๐ก๐ก + ๐ก๐ก2 Aux. Equation: ๐๐3 + 4๐๐ = 0 โ ๐๐ ๐๐2 + 4 = 0 โ ๐๐1 = 0;๐๐2 = 2๐๐;๐๐3 = โ2๐๐
๐ฆ๐ฆ๐๐ = ๐๐1๐๐0๐ก๐ก + ๐๐2 cos2๐ก๐ก + ๐๐3 sin 2๐ก๐ก
๐ฆ๐ฆ๐๐ = ๐๐1 + ๐๐2 ๐๐๐๐๐ ๐ 2๐ก๐ก + ๐๐3 ๐ ๐ ๐ ๐ ๐๐ 2๐ก๐ก
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 191
Here, ๐ท๐ท๐ก๐ก2 = ๐๐๐๐๐ก๐ก
๐ก๐ก2 = 2๐ก๐ก
Particular Solution ๐๐: use undetermined coefficient โ ๐ด๐ด๐ ๐ ๐ ๐ ๐๐๐๐๐๐ ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ก๐ก3 + ๐ต๐ต๐ก๐ก2 + ๐ถ๐ถ๐ก๐ก โ ๐ฆ๐ฆ๐๐โฒ = 3๐ด๐ด๐ก๐ก2 + 2๐ต๐ต๐ก๐ก + ๐ถ๐ถ; ๐ฆ๐ฆ๐๐โฒโฒ = 6๐ด๐ด๐ก๐ก + 2๐ต๐ต; ๐ฆ๐ฆ๐๐โฒโฒโฒ = 6๐ด๐ด โ 6๐ด๐ด + 4 3๐ด๐ด๐ก๐ก2 + 2๐ต๐ต๐ก๐ก + ๐ถ๐ถ = ๐ก๐ก2 + 2๐ก๐ก โ 6๐ด๐ด + 12๐ด๐ด๐ก๐ก2 + 8๐ต๐ต๐ก๐ก + 4๐ถ๐ถ = ๐ก๐ก2 + 2๐ก๐ก โ 12๐ด๐ด๐ก๐ก2 + 8๐ต๐ต๐ก๐ก + 6๐ด๐ด + 4๐ถ๐ถ = ๐ก๐ก2 + 2๐ก๐ก
โ 12๐ด๐ด = 1 โ ๐ด๐ด =1
12
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 192
๐๐3๐ฆ๐ฆ๐๐๐ก๐ก3
+ 4๐๐๐ฆ๐ฆ๐๐๐ก๐ก
= ๐ก๐ก2 + 2๐ก๐ก ๐ฆ๐ฆ๐๐ = ๐๐1 + ๐๐2 ๐๐๐๐๐ ๐ 2๐ก๐ก + ๐๐3 ๐ ๐ ๐ ๐ ๐๐ 2๐ก๐ก
Note: ๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ก๐ก3 + ๐ต๐ต๐ก๐ก2 + ๐ถ๐ถ๐ก๐ก Here, +๐ท๐ท is not considered since ๐ฆ๐ฆ๐๐ already has a constant term
โ 8๐ต๐ต = 2 โ ๐ต๐ต =14
โ 6๐ด๐ด + 4๐ถ๐ถ = 0 โ 4๐ถ๐ถ = โ6๐ด๐ด = โ61
12
โ ๐ถ๐ถ = โ18
Hence,
๐ฆ๐ฆ๐๐ =1
12 ๐ก๐ก3 +
14 ๐ก๐ก
2 โ18 ๐ก๐ก
๐ฆ๐ฆ = ๐ฆ๐ฆ๐๐ + ๐ฆ๐ฆ๐๐
โ ๐๐ = ๐๐๐๐ + ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐๐๐ +๐๐๐๐๐๐ ๐๐
๐๐ +๐๐๐๐ ๐๐
๐๐ โ๐๐๐๐ ๐๐
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 193
๐๐3๐ฆ๐ฆ๐๐๐ก๐ก3
+ 4๐๐๐ฆ๐ฆ๐๐๐ก๐ก
= ๐ก๐ก2 + 2๐ก๐ก ๐ฆ๐ฆ๐๐ = ๐๐1 + ๐๐2 ๐๐๐๐๐ ๐ 2๐ก๐ก + ๐๐3 ๐ ๐ ๐ ๐ ๐๐ 2๐ก๐ก
๐ฆ๐ฆ๐๐ = ๐ด๐ด๐ก๐ก3 + ๐ต๐ต๐ก๐ก2 + ๐ถ๐ถ๐ก๐ก
๐ด๐ด =1
12
We have: (1) โก ๐ท๐ท๐ฅ๐ฅ โ 4๐ฅ๐ฅ + ๐ท๐ท2๐ฆ๐ฆ = ๐ก๐ก2 โ ๐ท๐ท โ 4 ๐ฅ๐ฅ + ๐ท๐ท2๐ฆ๐ฆ = ๐ก๐ก2 2 โก ๐ท๐ท๐ฅ๐ฅ + ๐ฅ๐ฅ + ๐ท๐ท๐ฆ๐ฆ = 0 โ ๐ท๐ท + 1 ๐ฅ๐ฅ + ๐ท๐ท๐ฆ๐ฆ = 0
Solving for ๐ฅ๐ฅ now: 1 โ ๐ท๐ท โ 4 ๐ฅ๐ฅ + ๐ท๐ท2๐ฆ๐ฆ = ๐ก๐ก2
2 โ ๐ท๐ท โ ๐ท๐ท ๐ท๐ท + 1 ๐ฅ๐ฅ + ๐ท๐ท2๐ฆ๐ฆ = 0
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 194
๐ซ๐ซโ ๐๐ โ ๐ซ๐ซ ๐ซ๐ซ + ๐๐ ๐๐ = ๐๐๐๐ (โ) (โ) (โ)
โ ๐ท๐ท โ 4 โ ๐ท๐ท ๐ท๐ท + 1 ๐ฅ๐ฅ = ๐ก๐ก2 โ ๐ท๐ท โ 4 โ ๐ท๐ท2 โ ๐ท๐ท ๐ฅ๐ฅ = ๐ก๐ก2 โ โ 4 + ๐ท๐ท2 = ๐ก๐ก2 โ ๐ท๐ท2 + 4 ๐ฅ๐ฅ = โ๐ก๐ก2 Aux. Equation: ๐๐2 + 4 = 0 โ ๐๐1 = 2๐๐;๐๐2 = โ2๐๐
๐ฅ๐ฅ๐๐ = ๐๐4 cos2๐ก๐ก + ๐๐5 sin2๐ก๐ก
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 195
Particular Solution ๐๐: use undetermined coefficient ๐ด๐ด๐ ๐ ๐ ๐ ๐๐๐๐๐๐ ๐ฅ๐ฅ๐๐ = ๐ด๐ด๐ก๐ก2 + ๐ต๐ต๐ก๐ก + ๐ถ๐ถ (Table 3.4.1) โ ๐ฅ๐ฅ๐๐โฒ = 2๐ด๐ด๐ก๐ก + ๐ต๐ต; ๐ฅ๐ฅ๐๐โฒโฒ = 2๐ด๐ด
๐ท๐ท2๐ฅ๐ฅ + 4๐ฅ๐ฅ = โ๐ก๐ก2 โ๐๐2๐ฅ๐ฅ๐๐๐๐๐ก๐ก2 + 4๐ฅ๐ฅ๐๐ = โ๐ก๐ก2
โ 2๐ด๐ด + 4 ๐ด๐ด๐ก๐ก2 + ๐ต๐ต๐ก๐ก + ๐ถ๐ถ = โ๐ก๐ก2 โ 2๐ด๐ด + 4๐ด๐ด๐ก๐ก2 + 4๐ต๐ต๐ก๐ก + 4๐ถ๐ถ = โ๐ก๐ก2 4๐ด๐ด๐ก๐ก2 + 4๐ต๐ต๐ก๐ก + 2๐ด๐ด + 4๐ถ๐ถ = โ๐ก๐ก2
Solving Systems of Linear Equations
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๐ท๐ท2 + 4 ๐ฅ๐ฅ = โ๐ก๐ก2 ๐ฅ๐ฅ๐๐ = ๐๐4 cos2๐ก๐ก + ๐๐5 sin 2๐ก๐ก
4๐ด๐ด๐ก๐ก2 + 4๐ต๐ต๐ก๐ก + 2๐ด๐ด + 4๐ถ๐ถ = โ๐ก๐ก2
4๐ด๐ด = โ1 โ ๐ด๐ด = โ14
4๐ต๐ต = 0 โ ๐ต๐ต = 0
2๐ด๐ด + 4๐ถ๐ถ = 0 โ ๐ถ๐ถ = โ12๐ด๐ด = โ
12
โ14
=18โ ๐ถ๐ถ =
18
๐ฅ๐ฅ๐๐ = โ14๐ก๐ก2 +
18
๐ฅ๐ฅ = ๐ฅ๐ฅ๐๐ + ๐ฅ๐ฅ๐๐
โ ๐๐ = ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐๐๐ โ๐๐๐๐๐๐๐๐ +
๐๐๐๐
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 197
๐ท๐ท2 + 4 ๐ฅ๐ฅ = โ๐ก๐ก2 ๐ฅ๐ฅ๐๐ = ๐๐4 cos2๐ก๐ก + ๐๐5 sin 2๐ก๐ก
๐ฅ๐ฅ = ๐๐4 cos2๐ก๐ก + ๐๐5 sin 2๐ก๐ก โ14๐ก๐ก2 +
18
๐ฆ๐ฆ = ๐๐1 + ๐๐2 cos2๐ก๐ก + ๐๐3 sin2๐ก๐ก +1
12๐ก๐ก3 +
14๐ก๐ก2 โ
18๐ก๐ก
Re-substituting ๐ฅ๐ฅ, ๐ฆ๐ฆ in ๐๐โฒ + ๐๐ + ๐๐โฒ = ๐๐
โ โ2๐๐4 cos2๐ก๐ก + 2๐๐5 cos2๐ก๐ก โ12๐ก๐ก + ๐๐4 cos2๐ก๐ก + ๐๐5 sin 2๐ก๐ก โ
14๐ก๐ก2 +
18
+ โ2๐๐2 cos2๐ก๐ก + 2๐๐3 cos2๐ก๐ก +14๐ก๐ก3 +
12๐ก๐ก โ
18
= 0
โ sin 2๐ก๐ก โ2 ๐๐4 + ๐๐5 โ 2๐๐2 + cos2๐ก๐ก 2๐๐5 + ๐๐4 + 2๐๐3 = 0 โ โ2 ๐๐4 + ๐๐5 โ 2๐๐2 = 0 & 2๐๐5 + ๐๐4 + 2๐๐3 = 0
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 198
โ ๐๐5 โ 2๐๐4 = 2๐๐2 โ 2๐๐5 + ๐๐4 = โ2๐๐3
โ ๐๐4 = โ 15
4๐๐2 + 2๐๐3
โ ๐๐5 =15 2๐๐2 โ 4๐๐3
โ ๐๐ = โ ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ + ๐๐
๐๐๐๐๐๐๐๐ โ ๐๐๐๐๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐๐๐ โ ๐๐
๐๐๐๐๐๐ + ๐๐
๐๐
โ ๐๐ = ๐๐๐๐ + ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ + ๐๐๐๐ ๐๐๐ฌ๐ฌ๐ฅ๐ฅ๐๐๐๐ + ๐๐๐๐๐๐๐๐๐๐ + ๐๐
๐๐๐๐๐๐ โ ๐๐
๐๐๐๐
Solving Systems of Linear Equations
9/30/2014 Dr. Eli Saber 199
๐ฅ๐ฅ = ๐๐4 ๐๐๐๐๐ ๐ 2๐ก๐ก + ๐๐5 ๐ ๐ ๐ ๐ ๐๐ 2๐ก๐ก โ14๐ก๐ก2 +
18
๐ฆ๐ฆ = ๐๐1 + ๐๐2 ๐๐๐๐๐ ๐ 2๐ก๐ก + ๐๐3 ๐ ๐ ๐ ๐ ๐๐ 2๐ก๐ก +1
12๐ก๐ก3 +
14๐ก๐ก2 โ
18๐ก๐ก
9/30/2014 Dr. Eli Saber 200
End of Chapter 3