{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture 5 Notes

Lecture 5 Notes - Today Solving a second order linear...

This preview shows pages 1–16. Sign up to view the full content.

Today • Solving a second order linear homogeneous equation with constant coefFcients • complex roots to the characteristic equation, • repeated roots to the characteristic equation (Reduction of Order). • Connections to matrix algebra. • Solving a second order linear non homogeneous equation.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Reminder: Euler’s formula e i θ = cos θ + i sin θ
Complex roots (Section 3.3) • For the general case, , by assuming we get the characteristic equation : ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Complex roots (Section 3.3) • For the general case, , by assuming we get the characteristic equation : • When b 2 - 4ac < 0, we get complex roots: ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c r 1 , 2 = b ± b 2 4 ac 2 a
Complex roots (Section 3.3) • For the general case, , by assuming we get the characteristic equation : • When b 2 - 4ac < 0, we get complex roots: ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c r 1 , 2 = b ± b 2 4 ac 2 a = b ± 1 4 ac b 2 2 a

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Complex roots (Section 3.3) • For the general case, , by assuming we get the characteristic equation : • When b 2 - 4ac < 0, we get complex roots: ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c r 1 , 2 = b ± b 2 4 ac 2 a = b ± 1 4 ac b 2 2 a = b ± i 4 ac b 2 2 a
Complex roots (Section 3.3) • For the general case, , by assuming we get the characteristic equation : • When b 2 - 4ac < 0, we get complex roots: ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c r 1 , 2 = b ± b 2 4 ac 2 a = b ± 1 4 ac b 2 2 a = b ± i 4 ac b 2 2 a = b 2 a ± 4 ac b 2 2 a i

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Complex roots (Section 3.3) • For the general case, , by assuming we get the characteristic equation : • When b 2 - 4ac < 0, we get complex roots: ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c r 1 , 2 = b ± b 2 4 ac 2 a = b ± 1 4 ac b 2 2 a = b ± i 4 ac b 2 2 a = α ± β i = b 2 a ± 4 ac b 2 2 a i
Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: y 1 ( t ) = e ( α + β i ) t
Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: = e α t e i β t y 1 ( t ) = e ( α + β i ) t

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: = e α t e i β t = e α t (cos( β t )+ i sin( β t )) y 1 ( t ) = e ( α + β i ) t
Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: = e α t e i β t = e α t (cos( β t )+ i sin( β t )) y 1 ( t ) = e ( α + β i ) t y 2 ( t ) = e ( α β i ) t

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: = e α t e i β t = e α t (cos( β t )+ i sin( β t )) = e α t e i β t y 1 ( t ) = e ( α + β i ) t y 2 ( t ) = e ( α β i ) t
Complex roots (Section 3.3) • Complex roots to the characteristic equation mean complex valued solution to the ODE: = e α t e i β t = e α t (cos( β t )+ i sin(

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 108

Lecture 5 Notes - Today Solving a second order linear...

This preview shows document pages 1 - 16. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online