This preview shows page 1. Sign up to view the full content.
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 nonzero?)
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 = 12i
(D) r1 = 2+4i, r2 = 24i
(B) r1 = 1+2i, r2 = 12i...
View
Full
Document
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.
 Spring '13
 EricCytrynbaum
 Differential Equations, Algebra, Equations

Click to edit the document details