Chapter 4 Part II

Chapter 4 Part II - 304 CHAPTER 4 Second-Order Linear...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
304 CHAPTER 4 Second-Order Linear Differential Equations 4.4 Undetermined Coefficients ! Inspection First 1. ′′ −=⇒ ()=− yy ty t t p 2. ′′ + =⇒ () = y t t p 22 3. t t p 2 2 4. ty y t y t t p ′′ + 4 2 5. ′′ − += y y t p 4 2 6. −= t t p 2cos cos 7. ′′ − += ⇒ y e y te t p t 8. ′′′ + += +⇒ y t y tt p 2 ! Educated Prediction The homogeneous equation ′′ + y 2 5 0 has characteristic equation rr 2 250 ++= , which has complex roots −± 12 i . Hence, yt c e t ce t h () =+ −− sin cos , so for the right-hand sides ft , we try the following: 9. f t t t y t At Bt Ct D p =− =++ + 23 33 2 10. ft t e y t A t Be t p t 11. t A t B t p 2sin sin cos 12. e t y t e A t B t t p t 2 sin cos sin ! Guess Again The homogeneous equation ′′ − yyy 690 has characteristic equation 2 −+= , which has a double root 3, 3. Hence, e c t e h 1 3 2 3 . We try particular solutions of the form: 13. t A t B t C t D t p () ++ cos sin cos 2 14. f t te y t At Bt e t p t 2 3 bg (We can’t have any terms here dependent on terms in the homogeneous solution.) 15. ft e t A e B t C t t p t + sin sin cos 16. f t t t y t At Bt Ct Dt E p =−+⇒ =+++ + 42 4 3 2 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
SECTION 4.4 Undetermined Coefficients 305 ! Determining the Undetermined 17. ′ = y 1. The homogeneous solution is yt c h () = , where c is any constant. By simple inspection we observe that yt t p is a solution of the nonhomogeneous equation. Hence, the general solution is yt t c () =+ . 18. += yy 1. The homogeneous solution is e h t where c is any constant. By simple inspec- tion we observe that p 1 is a solution of the nonhomogeneous equation. Hence, the general solution is ce t 1. 19. t . e h t , yA t B p , ′ = p . Substituting into DE, AA t Bt ++ = . Coeffi- cient of t : A = 1. Coefficient of 1: AB 0 . Hence, A = 1, B =− 1. p 1, yc e t t 20. ′′ = y 1. The homogeneous solution of the equation is h 12 , where c 1 , c 2 are arbitrary constants. By inspection, we note that p = 1 2 2 is a particular solution. Hence, the solution of the homogeneous equation is t ct c + 1 2 2 . If you could not find a particular solution by inspection, you could try a solution of the form yt A t p 2 . 21. ′′ + ′ = 4 1. The characteristic equation is rr 2 40 , which has roots 0, –4. Hence, the homo- geneous solution is yt c ce h t 4 . The constant on the right-hand side of the differential equation indicates we seek a particular solution of the form p , except that the homogeneous solution has a constant solution; thus we seek a solution of the form t p . Substituting this expression into the differential equation yields 4 1 A = , or A = 1 4 . Hence, we have a particular solution t p 1 4 ,
Background image of page 2
306 CHAPTER 4 Second-Order Linear Differential Equations so the general solution is yt c ce t t () =+ + 12 4 1 4 . 22. ′′ += yy 4 1. The characteristic equation is r 2 40 , which has roots ± 2 i . Hence, the homoge- neous solution is y t ct h 22 cos sin .
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This lab report was uploaded on 04/04/2008 for the course APPM 2360 taught by Professor Williamheuett during the Fall '07 term at Colorado.

Page1 / 47

Chapter 4 Part II - 304 CHAPTER 4 Second-Order Linear...

This preview shows document pages 1 - 4. Sign up to view the full document.

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