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Solving a first order linear differential equation by variation of parameters

  1. Jul 19, 2009 #1
    1. The problem statement, all variables and given/known data

    I have to solve the following differential equation by the "variation of parameters" method.


    2. Relevant equations

    [tex]\frac{dy}{dx}x +2y = 3x[/tex]



    3. The attempt at a solution

    The associated homogeneous equation of the initial equation is:

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

    So

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

    [tex]ln(y) = -2ln(x)[/tex]

    [tex]ln(y) = ln(x^{-2})[/tex]

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

    Unfortunately, this doesn't satisfy the homogeneous equation.
     
  2. jcsd
  3. Jul 19, 2009 #2

    djeitnstine

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    Gold Member

    Where did the '3' go to? When you divide by x it is [tex]\frac{dy}{dx}=3-2\frac{y}{x}[/tex] not [tex]-2\frac{y}{x}[/tex]

    Then you do the substitution v=y/x, then it comes easy.
     
  4. Jul 19, 2009 #3

    djeitnstine

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    Gold Member

    Double post
     
  5. Jul 19, 2009 #4

    rock.freak667

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    You can use djeitnstine guide OR divide by x and then use an integrating factor. But you said you had a certain method to use.

    Since that is the case then you should know that y1=x-2 is a solution of the homogenous equation. So just find the Wronskian of y1, which should be easy and just put it into the formula.
     
  6. Jul 19, 2009 #5
    I'm sorry I didn't clarify in my earlier post - I'm supposed to take an equation of the form [tex] y' + p(x)y = q(x)[/tex], set the right hand side of my equation to zero, solve that equation by using separation of variables, and then take the homogeneous equation result as yh*u and substitute that back into the original differential equation, and get another separable equation. So I do the following with the answer to the homogeneous equation above and the original equation:

    [tex]\frac{dy}{dx}(x^{-2}u) + 2x^{-1}x^{-2}u = 3[/tex]

    =

    [tex]-2x^{-1}u + x^{-2}\frac{du}{dx} + 2x^{-3}u = 3[/tex]

    The way it's supposed to work is that the first term and the third term cancel, leaving me with a linear differential equation in du/dx to solve. However, in this case it doesn't seem to be working out that way...
     
  7. Jul 20, 2009 #6

    HallsofIvy

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    -2-1= -3, not -1.


    The derivative of x-2 is -2x-3, not -2x-1.
     
  8. Jul 20, 2009 #7
    :facepalm: Yeah, that's the problem. Getting my differentiation and integration confused again.:cry:
     
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