1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second-order Linear Equations

  1. Apr 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose y(t) is a complex-valued solution of y'' +py' + qy=0 where p and q are real numbers. Show that if y(t)=yre(t) + iyim(t), where yre(t) and yim(t) are real valued, then both yre(t) and yim(t) are solutions of the second-order equation.

    2. Relevant equations
    We can use the idea that a complex numbers is zero as long as it's real and imaginary parts are equal to zero.

    3. The attempt at a solution
    I do not have an attempt at this solution. I do not know where to begin.

    thanks for your help :)
     
  2. jcsd
  3. Apr 17, 2016 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Begin by plugging y(t)=yre(t) + iyim(t) into your DE and see what happens.
     
  4. Apr 17, 2016 #3
    then I would get y''re(t) + py're(t) +qyre(t) +y''im(t) +py'im(t) +qyim(t)=0

    so the real part would add up too zero and the imaginary part would add up to zero.
     
  5. Apr 17, 2016 #4
    would this be correct?
     
  6. Apr 17, 2016 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Where'd the factor of ##i## go?
     
  7. Apr 17, 2016 #6
    oops so it would be y''re(t) + py're(t) +qyre(t) +iy''im(t) +ipy'im(t) +iqyim(t)=0
     
  8. Apr 17, 2016 #7
    lets take yre as a solution, y''re + py're+qyre=0
    then take yim as a solution, y''im+py'im +qyim=0
    the multiply with i both sides of the 2nd eqn,
    i(y''im+py'im +qyim)=0

    linear combination of particular solutions isalso a solution,

    general solution y(t)=yre(t) + i yim(t)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Second-order Linear Equations
Loading...