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# lecture10 - Section 4.2(then 4.1 2nd order DFQ with...

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Section 4.2 (then 4.1) 2nd order DFQ with constant coefficients Feb 19, 2009 Characteristic Method

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Today’s Session Next Time: Section 4.1 (Motivation/Particular Case of 4.2) A Summary of This Session: (1) Solving constant coefficient homogeneous DFQ’s (1st order, second order, 3rd order, etc); (2) Characteristic Equation with real or repeated roots; (3) Concepts of linearity and independence; (4) nonhomogeneous DFQ with particular and general solutions. Characteristic Method
First order, constant coefficients Problem: Solve ay + by = 0, a negationslash = 0, and b are constants. Try y = e r x . So: y = r e r x . The equation becomes: ( ar + b ) e r x = 0 This leads to the characteristic equation : ar + b = 0. So: r = b a and y = e b a x The general solution is given by: y = C e b a x Solution set: 1st order = family of curves which depends on one parameter The solution set is generated by the function y 1 = e b a x since all solutions look like y = C y 1 . Characteristic Method

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Second order homogeneous with constant coefficients Problem: Solve ay ′′ + by + cy = 0, a negationslash = 0, b , and c are constants. Try y = e r x . So: y = r e r x . Also y ′′ = r 2 e r x . The equation becomes: ( ar 2 + br + c ) e r x = 0 This leads to the characteristic equation : ar 2 + br + c = 0. The solutions to this algebraic equation are given by r = b ± b 2 4 ac 2 a Let’s call these two roots r 1 and r 2 . So y 1 = e r 1 x and y 2 = e r 2 x are both solutions. In fact any linear combination of these two solutions is also a solution: y = C 1 e r 1 x + C 2 e r 2 x This is the general solution to the homogeneous problem at hand.
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lecture10 - Section 4.2(then 4.1 2nd order DFQ with...

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