User:Egm6322.s12.team2.steele.m2/Mtg6

=EGM6321 - Principles of Engineering Analysis 1, Fall 2011=

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L2-ODE-VC: 2nd order Need 2 conditions to solve for 2 constants 2 types of problems: BVP and IVP 1.	Boundary Value Problem (BVP) Solution domain: Closed interval [a,b] Closed = End points a and b are included in domain. Boundary conditions: Prescribe known values

2.	Initial Value Problem (IVP) Solution domain: (Semi) Open interval $$[a, +\infty) \equiv [a,+\infty[$$ .	Open = End point $$+\infty$$ included in domain. Initial conditions: Prescribe known values $$y(a) \alpha, \ y\prime (a) = \beta$$

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For BVPs and IVPs: Two facts A.	Existence and uniqueness of solution Solution exists and is unique; no proof here; accept this result. B.	Superposition of solutions L2-ODE-VC [[media:Pea1.f11.mtg5.djvu|Eqn(1) p.5-5:]] (2)

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$$L_2(\cdot)$$ is linear, thus superposition is applicable. R1.7: Show that $$L_2(\cdot)$$ is linear. 2 types of solutions: Homogeneous and particular

B1. Homogeneous solution

B2. Particular solution $$y_P(x)$$

Solution:

$$L_2(\cdot)$$ is linear; hence

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Since the solution for (1) p.6-2 exists and is unique, then, due to (4) – (5) p.6-3, the solution for (1) p.6-2 should be Question: $$L_2(\cdot)$$ is 2nd-order differential operator; there are 2 integration constants, thus we need 2 boundary conditions or 2 initial conditions to solve for these 2 constants; where are those 2 constants? Answer: There are actually 2 linearly independent homogeneous solutions: $$y^1_H(x)$$ and $$y^2_H(x)$$. The homogeneous solution $$y_H(x)$$ is a linear combination of $$y^1_H(x)$$ and $$y^2_H(x)$$: 2 integration constants

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The 2 integration constants are determined by requiring the solution y(x) to satisfy the boundary conditions (1) p.6-1 or the initial conditions (2) p.6-1: Ref: cf. King 2003 p.4 R1.8: Find the integration constants in terms of The boundary conditions (1) p.6-1	-> F10 The initial conditions (2) p.6-1 		-> F11 Example: Legendre differential operator/equation King 2003 p. 31