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Linear diff eq - correctly done?

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve the linear differential equation:

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


    2. Relevant equations

    If you have a linear differential equation of the form:

    [tex] y'+P(x)y=Q(x) [/tex]

    then your integrating factor is:

    [tex] I(x)=e^{\int P(x) dx} [/tex]

    3. The attempt at a solution

    If we divide both sides by x then the equation is in standard form:

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

    where

    [tex] P(x)=-\frac{2}{x} [/tex]

    thus:

    [tex] I(x)=e^{\int -\frac{2}{x} dx} = e^{-2\int\frac{1}{x}dx}= e^{-2ln|x|}=x^{-2} [/tex]

    so we then multiple both sides of the diff eq by I(x):

    [tex] x^{-2}y'-2x^{-3}y=\frac{1}{x} [/tex]

    which is:

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

    if we the integrate both sides:

    [tex] x^{-2}y=ln|x|+c [/tex]

    therefore:

    [tex] y= x^{2}ln|x|+cx^{2} [/tex]
     
  2. jcsd
  3. Jun 28, 2011 #2

    Char. Limit

    User Avatar
    Gold Member

    Looks good to me.
     
  4. Jun 28, 2011 #3

    Mark44

    Staff: Mentor

    Your solution is easy enough to check. Substitute your solution for y and take its derivative. When you evaluate xy' - 2y, you should get x2.
     
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