285lect29 - 1 Lecture 30 Example 1 x 00 3 x 5 x =& 4 cos 5 t Complimentary equation x 00 3 x 5 x = 0 Characteristic equation r 2 3 r 5 = 0

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 Lecture 30 Example 1 x 00 + 3 x + 5 x = & 4 cos 5 t: Complimentary equation: x 00 + 3 x + 5 x = 0 : Characteristic equation: r 2 + 3 r + 5 = 0 Solution is: r = & 3 2 + 1 2 i p 11 ; r = & 3 2 & 1 2 i p 11 : Transient solution: x tr ( t ) = e & 3 2 t C 1 cos p 11 2 t + C 2 sin p 11 2 t: ! Steady periodic solution: x sp ( t ) = A cos 5 t + B sin 5 t ; x sp ( t ) = & 5 A sin 5 t + 5 B cos 5 t ; x 00 sp ( t ) = & 25 ( A cos 5 t + B sin 5 t ) : x 00 + 3 x + 5 x = & 25 ( A cos 5 t + B sin 5 t ) + 3 ( & 5 A sin 5 t + 5 B cos 5 t ) + 5 ( A cos 5 t + B sin 5 t ) = cos 5 t ( & 25 A + 15 B + 5 A ) + sin t ( & 25 B & 15 A + 5 B ) = & 4 cos 5 t ) & 20 A + 15 B = & 4 & 15 A & 20 B = 0 : ) & 20 A + 15 B = & 4 3 A + 4 B = 0 ) B = & 3 4 A ) & 20 A + 15 & & 3 4 ¡ A = & 4 ) A = 16 125 ) B = & 3 4 A = & 3 4 16 125 = & 12 125 : 1 Steady periodic solution: x sp ( t ) = 16 125 cos 5 t & 12 125 sin 5 t: s & 16 125 ¡ 2 + & 12 125 ¡ 2 = 4 25 : So, x sp ( t ) = 4 25 & 25 4 16 125 cos 5 t & 25 4 12 125 sin 5 t ¡ = 4 25 & 4 5 cos 5...
View Full Document

This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

Page1 / 5

285lect29 - 1 Lecture 30 Example 1 x 00 3 x 5 x =& 4 cos 5 t Complimentary equation x 00 3 x 5 x = 0 Characteristic equation r 2 3 r 5 = 0

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online