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Homework Help: Check DE general solution with complex roots

  1. Aug 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve [itex]y''+4y'+5y=0[/itex]

    find solutions for y(0)=1 and y'(0)=0

    2. Relevant equations

    Quadratic equation

    3. The attempt at a solution

    Hows this look ?

    assume solution is in the form of [itex]y=ce^{rx}[/itex]

    substitute [itex]y=ce^{rx}[/itex] into the equation.

    [itex]cr^2e^{rx}+4cre^{rx}+5ce^{rx}=0[/itex]

    we then divide by [itex]ce^{rx}[/itex] to give

    [itex]r^2+4r+5=0[/itex]

    So we than find the roots which are [itex]r_1=-2+i,r_2=-2-i[/itex]

    which gives the solution of

    [itex]y=c_1e^{-2+ix}+c_2e^{-2-ix}[/itex].

    but as they are complex roots we use the equation

    [itex]y=c_1e^{\lambda x}cos \mu x +c_2e^{\lambda x}sin \mu x }[/itex].


    where [itex]\mu = 1 , \lambda = -2[/itex]

    so

    [itex]y(x)=c_1e^{-2x}cos x +c_2e^{-2x}sin x [/itex].

    and

    [itex]y'(x)=e^{-2x}(sin (x)(-c_1-sc_2)+cos(x)(c_2-2c_1)) [/itex].


    to find [itex]c_1[/itex] and [itex]c_2[/itex] apply the initial conditions.

    [itex]y(0) = c_1 = 1[/itex]
    and
    [itex]y'(0) = c_2-2= 0 [/itex] so [itex]c_2 = 2[/itex]

    so that makes the general solution

    [itex]y(x)=e^{-2x}cos x +2e^{-2x}sin x [/itex].
     
    Last edited: Aug 13, 2010
  2. jcsd
  3. Aug 13, 2010 #2

    Mark44

    Staff: Mentor

    It's easy to check.
    1) Verify that y(0) = 1 and y'(0) = 2
    2) Verify that your solution satisfies y'' + 4y' + 5y = 0.
     
  4. Aug 15, 2010 #3
    It's easy to check.
    2) Verify that your solution satisfies y'' + 4y' + 5y = 0.

    I got
    y'' = -5
    y' = 0
    y =1

    so its the solution thanks alot
     
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