c18 - 18.03 Class 18, March 20, 2006 Review of constant...

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Unformatted text preview: 18.03 Class 18, March 20, 2006 Review of constant coefficient linear equations: Big example, superposition, and Frequency Response [1] Example. x" + 4x = 0 PLEASE KNOW the solution to the homogeneous harmonic oscillator x" + omega^2 x = 0 are sinusoids of circular frequency omega ! Here, a cos(2t) + b sin(2t). In the real example I drive it: x" + 4x = t cos(2t) . The complex equation is z" + 4z = t e^{2it} . If it weren't for the t we could try to apply ERF: p(s) = s^2 + 4, p(2i) = -4 + 4 = 0 , though, so it doesn't apply; we do have the resonance response formula, which gives z_p = t e^{2it}/p'(2i) = -(it/4) e^{2it} so x_p = (t/4) sin(2t) . But there is a t there. We should then use "Variation of Parameters": Look for solutions of the form z = e^{2it} u for u an Unknown function. 4] z = e^{2it} u 0] z' = e^{2it} ( u' + 2i u ) 1] z" = e^{2it} ( u" + 2i u' + 2i u' + (2i)^2 u )------------------------------------------------------ e^{2it} t = e^{2it} ( u" + 4i u' + (4-4) u ) so u" + 4i u' = t Reduction of order: v = u' , v' + 4i v = t ; Use undetermined coefficients: v = at + b 4i] v = at + b v' = a----------------...
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This note was uploaded on 01/31/2011 for the course MAT 17A taught by Professor Staff during the Winter '08 term at UC Davis.

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c18 - 18.03 Class 18, March 20, 2006 Review of constant...

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