Elementary Differential Equations 8th edition by Boyce ch09

Elementary Differential Equations, with ODE Architect CD

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
—————————————————————————— —— CHAPTER 9. ________________________________________________________________________ page 498 Chapter Nine Section 9.1 2 Setting results in the algebraic equations a b + Þ œ / x 0 <> Œ Œ Œ &  <  " $ "  < œ ! ! 0 0 " # . For a nonzero solution, we must have . The roots of ./>  < œ <  ' <  ) œ ! a b A I # the characteristic equation are and . For , the system of equations < œ # < œ % < œ # " # reduces to . The corresponding eigenvector is $ œ 0 0 " # 0 a b " X œ " ß $ Þ a b Substitution of < œ % œ results in the single equation . A corresponding eigenvector is 0 0 " # 0 a b # X œ " ß " Þ a b a b , . The eigenvalues are and , hence the critical point is an . real positive unstable node a b -ß . .
Image of page 1

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

View Full Document Right Arrow Icon
—————————————————————————— —— CHAPTER 9. ________________________________________________________________________ page 499 3 . a b + Solution of the ODE requires analysis of the algebraic equations Œ Œ Œ #  <  " $  #  < œ ! ! 0 0 " # . For a nonzero solution, we must have . The roots of the ./>  < œ <  " œ ! a b A I # characteristic equation are and . For , the system of equations < œ " < œ  " < œ " " # reduces to . The corresponding eigenvector is 0 0 " # œ 0 a b " X œ " ß " Þ a b Substitution of < œ  " $ œ ! results in the single equation . A corresponding eigenvector is 0 0 " # 0 a b # X œ " ß $ Þ a b a b , . The eigenvalues are , with . real saddle < <  ! " # . Hence the critical point is a a b -ß . .
Image of page 2
—————————————————————————— —— CHAPTER 9. ________________________________________________________________________ page 500 5 . The characteristic equation is given by a b + º º "  <  & "  $  < œ <  # <  # œ ! Þ # The equation has roots and . complex < œ  "  3 < œ  "  3 " # For , < œ  "  3 the components of the solution vector must satisfy . Thus the 0 0 " #  #  3 œ ! a b corresponding eigenvector is 0 a b " X œ #  3 ß " Þ < œ  "  3 a b Substitution of results in the single equation . A corresponding eigenvector is 0 0 " #  #  3 œ ! a b 0 a b # X œ #  3 ß " Þ a b a b , . The eigenvalues are , with negative real part. Hence the origin complex conjugates is a . stable spiral
Image of page 3

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

View Full Document Right Arrow Icon
—————————————————————————— —— CHAPTER 9. ________________________________________________________________________ page 501 a b -ß . . 6 . Solution of the ODEs is based on the analysis of the algebraic equations a b + Œ Œ Œ #  <  & "  #  < œ ! ! 0 0 " # . For a nonzero solution, we require that . The roots of the ./>  < œ <  " œ ! a b A I # characteristic equation are . Setting , the equations are equivalent to < œ „3 < œ 3 0 0 " #  #  3 œ ! a b . The eigenvectors are 0 0 a b a b " # X X œ #  3 ß " œ #  3 ß " Þ a b a b and a b , . The eigenvalues are . Hence the critical point is a . purely imaginary center
Image of page 4
—————————————————————————— —— CHAPTER 9. ________________________________________________________________________ page 502 a b -ß . . 7 . a b + Setting results in the algebraic equations x œ / 0 <> Œ Œ Œ $  <  # %  "  < œ ! ! 0 0 " # . For a nonzero solution, we require that . The roots ./>  < œ <  #<  & œ ! a b A I # of the characteristic equation are . Substituting , the two < œ " „ #3 < œ "  #3 equations reduce to . The two eigenvectors are a b "  3 œ ! 0 0
Image of page 5

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

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern