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1st order linear diff eq. problem

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of

    [tex]\frac{dy}{dt}[/tex] = 2y +sin(2t)

    2. Relevant equations

    The general solution of a nonhomogeneous ode is the particular solution of the nonhomo plus the solution of the homogeneous ode.: y(t)= y[tex]_{p}[/tex](t)+y[tex]_{h}[/tex](t)

    3. The attempt at a solution

    [tex]\frac{dy}{dt}[/tex] - 2y = sin(2t)


    Initial guess, is that y[tex]_{p}[/tex] = a*Sin(2t)+b*Cos(2t), where a and b are unknown constants later to be found

    Then, the d.e. becomes:

    [tex]\frac{d}{dt}[/tex] ( a*Sin(2t)+b*Cos(2t) ) - 2( a*Sin(2t)+b*Cos(2t) ) = Sin2t

    =2aCos(2t)-2bSin(2t) - 2aSin(2t)+2bCos(2t) = Sin2t

    =(2a+2b)*Cos(2t) + (-2b-2a)*Sin(2t) = Sin2t

    Using the fact that polynomial coefficients must be equal:

    2a+2b=0
    -2b-2a=1

    And here is my issue. These simultaneous equations are false, and thus I cannot possibly solve the d.e.

    Can someone tell me where I messed up? Thanks

    PS. I know this can also be found using integrating factors, but i need help on the section with "guessing" the solution
     
  2. jcsd
  3. Jul 11, 2010 #2

    vela

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    You got a sign wrong when you went from this line

    [tex]\frac{d}{dt} (a \sin 2t + b \cos 2t ) - 2(a \sin 2t + b \cos 2t ) = \sin 2t[/tex]

    to your next one. You should have gotten

    [tex]2a\cos 2t - 2b \sin 2t - 2a \sin 2t - 2b \cos 2t = \sin 2t[/tex]
     
  4. Jul 11, 2010 #3
    Hey thanks. You hw helpers save my life.
     
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