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Solve differential equation with variation of parameters

  1. Sep 6, 2005 #1
    solve using method of variation of parameters

    y''-y = 2/(1+e^x)

    y'' ==> second order
     
  2. jcsd
  3. Sep 6, 2005 #2

    HallsofIvy

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    Staff Emeritus
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    And your question is?

    It should be easy to see that ex and e-x are two independent solutions to y"- y= 0.

    The "variation of parameters" is to look for a solution of the form
    y(x)= u(x)ex+ v(x)e-x
    In fact, there are an infinite number of u(x), v(x) that would work.

    Differentiating, y'= u'(x)ex+ u(x)ex+ v'(x)e-x- v(x)e-x.

    ASSUME that u'(x)ex+ v'(x)e-x= 0.
    (Since there are an infinite number of u(x), v(x) that would work above, this is just "narrowing the search".)

    With that assumption y'= u(x)ex- v(x)e-x.

    Differentiating again, y"= u'(x)ex+ u(x)ex- v'(x)e-x+ ve-x.

    Plug that into the original equation and, because ex and e-x satisfy the original homogeneous equation, the "u(x)" and "v(x)" terms cancel leaving just u'ex- v'e-x= 2/(1+ex). Treat that, along with
    u'ex+ v'e-x= 0 (above) as two linear equations for u', v'.
    Integrate those solutions to find u and v and plug into y(x)= u(x)ex+ v'(x)e-x.
     
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