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How to get a particular solution.

  1. Apr 13, 2005 #1
    I have a "Solve the initial value problem" It is:

    y'' + y = 2cosx - 3sinx

    I know how to do everything except for get yp. I know it has to be something so when i substitute yp into the left hand side of the equation, i get the right hand side, 2cosx - 3sinx.

    By this definition i would think yp = Acosx - Bsinx, where A = 2 and B = 3. But that doesn't seem right.

    Can anyone help me out?

    Thank you.
     
  2. jcsd
  3. Apr 13, 2005 #2

    dextercioby

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    Yeah,use Lagrange's method of variation of constants...Solve the homogenous ODE and then make the 2 constants (associated with the 2 independent solution spanning the solutions space) become functions of "x"...


    Daniel.
     
  4. Apr 13, 2005 #3

    saltydog

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    For the equation:

    [tex]y^{''}+y=2Cos[x]-3Sin[x][/tex]

    Whenever the right-hand side is a particular solution of a linear homogeneous equation, then you can apply that differential operator to both sides of the above equation to collapse it into a homogeneous one:

    Since RHS is a particular solution to homogeneous equation (in operator notation):

    [tex](D^2+1)y=0[/tex]

    Applying this opearator to both sides of the original equation leads to:

    [tex](D^2+1)(D^2+1)R=0[/tex]

    Solution of this one is:

    [tex]y(x)=c_1Sin[x]+c_2Cos[x]+AxSin[x]+BxCos[x][/tex]

    but:

    Well, read the book to fill in the details, but then just substitute:

    [tex]y_p(x)=AxSin[x]+BxCos[x][/tex]

    into the original equation, equate coefficients. You're done.
     
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