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Homework Help: 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


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    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


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    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)
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