Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    [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]


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

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


    [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]

  5. Jul 25, 2010 #4

    Char. Limit

    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook