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Homework Help: Linear Equations

  1. Jan 2, 2010 #1
    Gday,
    Just trying to work out the following linear differential equation. I'm trying to work it out via the Integrating Factor method and I am getting stuck at the point where you multiply both sides by I(x).

    The equation is:

    xy' - 2y = x2

    Can anyone help me out with this?

    Regards,
    Matt

    PS. Also, how do I type equations, etc in the math font?
     
  2. jcsd
  3. Jan 2, 2010 #2
    What do you have for I(x) ?

    For writing latex, click the sigma symbol on the toolbar, but I found it easier to get started using something like this
    http://www.codecogs.com/components/equationeditor/equationeditor.php [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Jan 2, 2010 #3
    x(dy/dx)-2y=x^2
    divide by x...(dy/dx)-2(y/x)=x
    then its in the form:(dy/dx)+p(x)y=q(x)
    we multiply by a(x), such that a(x)=exp([tex]\int[/tex]p(x)dx)
    we obtain a(x)=x^2
    then the solution is given by the following formula:
    y(x)=a(x)-1{[tex]\int[/tex]a(x).q(x)dx +C}
    hope that its the correct solution
     
  5. Jan 2, 2010 #4
    That looks wrong to me. Surely [tex]a(x)=I(x)=x^{-2}[/tex], giving a solution of [tex]y(x)=x^{2}(\ln(x)+C)[/tex]
     
  6. Jan 2, 2010 #5
    Ah okay, thanks for your reply.
    One thing I am not understanding is the multiplication into the equation of the integrating factor. It has confused me on equations other than this one also.
    Here is another example of an equation that is confusing me:

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

    How does the breaking down of the left side work? Where did y/x come from?

    Matt
     
  7. Jan 2, 2010 #6
    The first thing you need to do is express the DE in "standard form" so that nothing is multiplying y'...so in your last example you need to multiply through by x, so it becomes y'+y/x = x. Not sure where your last equation comes from - maybe you made some typos ? Then the integrating factor will be the exponential of the integral of 1/x, which is just 1/x.
     
  8. Jan 2, 2010 #7
    If you go back to the basic definition:
    the standard form of the equation is :
    (dy/dx)+p(x)y=q(x)
    we multiply by a(x), and the form becomes: a(x).(dy/dx)+a(x).p(x).y=q(x).a(x) eq.1
    we can write it in the form: [tex]\stackrel{d[a(x).y}{dx}[/tex]=a(x).p(x)
    this is because of the rule: d(a(x).y)/dx=d(a(x))/dx + dy/dx eq.2
    and comparing eq.1 and eq.2
    we get d(a(x))/dx=a(x).p(x)
    try to follow this explanation and do the example by your self...gd luck
     
  9. Jan 2, 2010 #8
    This is the method I use:

    Take a generic linear 1st order ODE:
    y' + g(x)y = h(x)

    Then, the integrating factor I(x) = exp(int(g(x) dx))

    After multiplying through by I(x) we can write:
    (I(x)y)' = I(x)h(x)
    because the integrating factor always has the property that after multiplication the LHS becomes a perfect derivative (following the definition of the product rule of differentiation)

    So then we integrate both sides (remembering to include the constant of integration) and then divide through by I(x) to give the general solution in explicit form.
     
  10. Jan 2, 2010 #9
    Of course, my method is the same as hisham's but I just find it easier to follow.
     
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