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General solution of first-order differential equation

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of the first-order differential equation,

    y' + 3y = 2xe^(-3x)


    2. Relevant equations

    y' + P(x)y = Q(x)

    Integrating factor = e^(∫P(x) dx)

    3. The attempt at a solution

    Since it's already in the form y' + P(x)y = Q(x),

    the integrating factor is I(x) = e^(∫3 dx) = e^(3x)

    Now multiplying both sides by the integrating factor,

    e^(3x)*y' + 3ye^(3x) = 2xe^(-3x)e^(3x)

    d/dx (e^(3x)y) = 2x

    Finally, integrating both sides and solving for y,

    e^(3x)y = x^2 + C

    General solution: y = x^(2)e^(-3x) + Ce^(-3x)

    Not to sure if this is right, this is the first time I've studied diff eqns..

    Thanks for any help.
     
  2. jcsd
  3. Jan 17, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Yes that should be correct.
     
  4. Jan 17, 2009 #3
    There is one sure-fire way to check that your expression for y is a solution to the DE: plug it in!
     
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