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Diff EQ - Simple Linear ODE

  1. Sep 26, 2006 #1
    Hey,

    So I solved this one linear DE and the answer I got isn't the same as the one in the back of my textbook... And I'm not sure why. I thought I was doing this right. Could someone tell me what I'm doing wrong?

    Here's what I'm given:

    [tex]x^{2}y'+xy=1[/tex]
    [tex]\frac{dy}{dx}+\frac{x}{x^{2}}y=\frac{1}{x^{2}}[/tex]
    [tex]\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x^{2}}[/tex]
    [tex]integrating-factor-p=\frac{1}{x}[/tex]
    [tex]integrating-factor\rightarrow e^{\int p(x) dx}=e^{\int \frac{1}{x}dx}= e^{lnx}=x[/tex]
    [tex](xy)'=\frac{1}{x^{2}}[/tex]
    [tex]xy=\int x^{-2}dx=\frac{-1}{x}+C[/tex]
    [tex]y=\frac{-1}{x^{2}}+\frac{C}{x}[/tex]

    That's what I get.

    The answer in the textbook is:
    [tex]y=x^{-1}lnx+Cx^{-1}[/tex]

    Which is similar but not the same... What'd I do wrong?
     
  2. jcsd
  3. Sep 26, 2006 #2
    When you get an integrating factor, multiply it through both sides to obtain

    [tex] (xy)' = \frac{1}{x} [/tex]
    Then, Integrate to obtain
    [tex] \int d(xy) = \int \frac{1}{x} dx [/tex]
    And this leads you to
    [tex] y = \frac{ln(x)}{x} + Cx^{-1} [/tex]

    The step where you made the error was after the integrating factor, you did not also apply it to the right side of the equation, the [tex] x^{-2} [/tex].
     
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