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First Order Differential Equation

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] xy^{'} - 2y = x^{5} [/tex]

    2. Relevant equations
    [tex]e^{\int P(x)dx}[/tex]


    3. The attempt at a solution

    Rearranging the into the form,

    [tex]y^{'} - P(x)y = Q(x)[/tex]

    So,

    [tex]y^{'} - \frac{2y}{x} = x^{4}[/tex]

    Multiplying both sides by [tex] e^{\int P(x)dx} [/tex] or [tex]e^{-2\int \frac{dx}{x}}[/tex],

    Since [tex]e^{-2ln|x|} = \frac{1}{x^{2}}[/tex]

    [tex]\frac{y^{'}}{x^{2}} - \frac{2y}{x^{3}} = x^{2} [/tex]

    Integrating both sides,

    [tex]\int\left( \frac{y^{'}}{x^{2}} - \frac{2y}{x^{3}}\right)dx =\int x^{2}dx [/tex]

    Integrating the LHS should result in my integrating factor times y, [tex]e^{-2ln|x|}(y)[/tex] or [tex]\frac{1}{x^{2}}(y)[/tex]

    So,

    [tex]\frac{1}{x^{2}}(y) = \int x^{2}dx[/tex]

    [tex]\frac{1}{x^{2}}(y) = \frac{1}{3} x^{3} + C[/tex]

    Finally,

    [tex]y = \frac{x^{5}}{3} + Cx^{2}[/tex]

    Is this correct? (I don't have any type of solutions to check, so I thought I'd post)
     
    Last edited: Jul 25, 2010
  2. jcsd
  3. Jul 25, 2010 #2
    Why you did is correct.

    This is a linear non-homogeneous ODE with non-constant coefficients, you solved it correctly with an integration factor.

    Good job ! :)

    P.S.: for future reference, you can check if your answer is correct by replacing y(x) and y'(x) with the answer and its derivative ;)
    verify it. it should end up as 0=0, basically..
     
  4. Jul 25, 2010 #3
    I'll give verifying it a shot!

    [tex]y = \frac{x^{5}}{3} + Cx^{2}[/tex]

    [tex] y^{'} = \frac{5x^{4}}{3} + 2Cx[/tex]

    Plugging into the original equation,

    [tex] \frac{5x^{5}}{3} + 2Cx^{2} - \frac{2x^{5}}{3} - 2Cx^{2} = x^{5}[/tex]

    [tex] x^{5} = x^{5} [/tex]

    Thanks!
     
  5. Jul 25, 2010 #4

    Char. Limit

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    Gold Member

    Just a little note on this part here (which is correct, I'm not saying it isn't)...

    Setting, say, u = 1/x^2, we can see that u' = -2/x^3...

    and so the LHS would equal u y' + y u'

    which obviously equals (u y)' or (y/x^2)'

    Just saying, although you probably already knew that. Just clarifying for anyone else who might read.
     
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