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ODE with integrating factor

  1. Sep 12, 2013 #1
    Problem:
    [tex]xy'+2y=3x[/tex]
    Attempt:
    Divide by x...
    [tex]y'+\frac{2y}{x}=3[/tex]
    I think I find the integrating factor by doing:
    [tex]e^{\int \frac{2}{x}dx}[/tex]

    Not sure if that's right but if it is then the solution to the integral is just 2x.

    Any help is appreciated
     
  2. jcsd
  3. Sep 12, 2013 #2

    rock.freak667

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    Homework Helper

    You will need to add in the constant of integration to the right side before you do any simplification.
     
  4. Sep 12, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, the integrating factor is [itex]e^{\int (2/x)dx}[/itex] but then the integral is 2 ln(x) so the integrating factor is [itex]e^{2ln(x)}= e^{ln(x^2)}= x^2[/itex], not 2x.

    Multiplying the equation by that gives [itex]x^2y'+ 2xy= (x^2y)'= 3x^2[/itex]
     
    Last edited: Sep 12, 2013
  5. Sep 13, 2013 #4
    Yeah I figured it out ends up:
    [tex]x^{2}y'+2xy=3x^{2}[/tex]
    Take integral/derivative of integrating factor:
    [tex]\int \frac{d}{dx} x^{2}y=\int 3x^{2}dx[/tex]
    Simplifies to:
    [tex]x^{2}y=x^{3}+C \implies y=\frac{x^{3}+C}{x^{2}}[/tex]
    (can be simplified a little more too)

    Thanks for the help.
     
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