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First order non-separable linear deq's using an integration factor?

  1. Jan 22, 2012 #1
    For example:
    [tex]\frac{dy}{dx} + y = e^{3x}[/tex]



    I understand that these differential equations are most easily solved by multiplying in a factor of integration, and then comparing the equation to the product rule to solve et al..

    For example:

    [tex]t\frac{dy}{dx} + 2t^{2}y = t^{2}[/tex]
    [tex]\frac{dy}{dx} + 2ty = t[/tex]

    Multiplying in an integration factor u(x), which in this case:
    [tex]u(x) = e^{\int{2t}dt} = e^{t^{2}}[/tex]

    [tex]e^{t^{2}}\frac{dy}{dx} + 2te^{t^{2}}y = te^{t^{2}}[/tex]

    Now I can compress the left side down using the product rule and all that.

    I don't understand how they are getting u(x) or why it's equal to e^{\int{2t}dt} ?
     
    Last edited: Jan 22, 2012
  2. jcsd
  3. Jan 22, 2012 #2

    micromass

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    First we multiply the equation with an unkown function u(x)

    [tex]u(x)\frac{dy}{dx}+u(x)y=u(x)e^{3x}[/tex]

    The goal is to recognize the left side as the derivative of a known function. Because then we can integrate both sides.

    Now, remember the product rule

    [tex]\frac{d}{dx}(u(x)y)=u(x)\frac{dy}{dt} + \frac{du(x)}{x}y[/tex]

    This looks a lot like the left side, doesn't it?? The only thing we want is

    [tex]\frac{du(x)}{dx}=u(x)[/tex]

    Such a differential equation is easily solved. Indeed, it is clear [itex]u(x)=e^x[/itex] satisfies this.

    So now we got

    [tex]\frac{d}{dx}{e^x y}=e^{4x}[/tex]

    Integrating both sides gives us

    [tex]e^x y = \frac{1}{4}e^{4x}+C[/tex]

    or

    [tex]y=Ce^{-x}+\frac{1}{4}e^{3x}[/tex]
     
    Last edited: Jan 22, 2012
  4. Jan 23, 2012 #3
    Thanks Micro, this answers my questions perfectly.
     
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