Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

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


    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


    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]


    Last edited: Jan 22, 2012
  4. Jan 23, 2012 #3
    Thanks Micro, this answers my questions perfectly.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

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