3 complex roots to the characteristic equation mean

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Unformatted text preview: i)t =e e αt iβ t = eαt (cos(β t) + i sin(β t)) y2 (t) = e(α−β i)t =e e αt −iβ t Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: y1 (t) = e(α+β i)t =e e αt iβ t = eαt (cos(β t) + i sin(β t)) y2 (t) = e(α−β i)t =e e αt −iβ t = e (cos(−β t) + i sin(−β t)) αt Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: y1 (t) = e(α+β i)t =e e αt iβ t = eαt (cos(β t) + i sin(β t)) y2 (t) = e(α−β i)t =e e αt −iβ t = e (cos(−β t) + i sin(−β t)) αt = e (cos(β t) − i sin(β t)) αt Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: y1 (t) = eαt (cos(β t) + i sin(β t)) y2 (t) = eαt (cos(β t) − i sin(β t)) Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: y1 (t) = eαt (cos(β t) + i sin(β t)) y2 (t) = eαt (cos(β t) − i sin(β t)) • Instead of using these to form the general solution, let’s use them to find two real valued solutions: Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: y1 (t) = eαt (cos(β t) + i sin(β t)) y2 (t) = eαt (cos(β t) − i sin(β t)) • Instead of using these to form the general solution, let’s use them to find two real valued solutions: 1 1 αt y1 (t) + y2 (t) = e cos(β t) 2 2 Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: y1 (t) = eαt (cos(β t) + i sin(β t)) y2 (t) = eαt (cos(β t) − i sin(β t)) • Instead of using these to form the general solution, let’s use them to find two real valued solutions: 1 1 αt y1 (t) + y2 (t) = e cos(β t) 2 2 1 1 y1 (t) − y2 (t) = eαt sin(β t) 2i 2i Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: y1 (t) = eαt (cos(β t) + i sin(β t)) y2 (t) = eαt (cos(β t) − i sin(β t)) • Instead of using these to form the general solution, let’s use them to find two real valued solutions: 1 1 αt y1 (t) + y2 (t) = e cos(β t) 2 2 1 1 y1 (t) − y2 (t) = eαt sin(β t) 2i 2i • General solution: y (t) = C1 eαt cos(β t) + C2 eαt sin(β t) Complex roots (Section 3.3) • To be sure this is a general solution, we must check the Wronskian: W (e αt cos(β t), e αt sin(β t))(t) = (for you to fill in later - is it non-zero?) ￿ ￿ W (y1 , y2 )(t) = y1 (t)y2 (t) − y1 (t)y2 (t) Recall: Complex roots (Section 3.3) • Example: Find the (real valued) general solution to the equation y + 2y + 5y = 0 ￿￿ ￿ • Step 1: Assume y (t) = e , plug this into the equation and find values of r that make it work. rt (A) r1 = 1+2i, r2 = 1-2i (D) r1 = 2+4i, r2 = 2-4i (B) r1 = -1+2i, r2 = -1-2i...
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This note was uploaded on 02/12/2014 for the course MATH 256 taught by Professor Ericcytrynbaum during the Spring '13 term at The University of British Columbia.

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