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First order differentials equations

  1. Dec 3, 2012 #1
    Using z = y/x to transform the given homogeneous differential equation into a differential equation in z and x. By first solving the transformed equation, find the general solution of the original equation, giving y in terms of x.

    [itex] z = \frac{y}{x} \rightarrow y = xz \rightarrow \frac{dy}{dx} = x\frac{dz}{dx} + z [/itex]
    [itex] x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4y^3}{3xy^2} [/itex]
    [itex] x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4z^3x^3}{3xz^2x^2} [/itex]
    [itex] \dfrac{d}{dx}(xz) = \dfrac{1 + 4z^3}{3z^2} [/itex]
    [itex] xz = \displaystyle\int (\dfrac{1 + 4z^3}{3z^2} ) [/itex]
    [itex] xz = \frac{-1}{3z} + \frac{2}{3}z^2 + c [/itex]
    subbing z = y/x
    [itex] x = \frac{-x^2}{3y^2} + \frac{2y}{3x} + \frac{cx}{y} [/itex]

    How do I rearrange from y from here? I just can't seem to do it :\
     
    Last edited: Dec 3, 2012
  2. jcsd
  3. Dec 3, 2012 #2

    Mark44

    Staff: Mentor

    In the next step, instead of writing the left side as d/dx(xz), subtract z from both sides. Then you will have ## x \frac{dz}{dx}## on one side and whatever you get on the right side. At this point you can separate the equation so that one side has all the stuff involving x and dx, and the other has all the stuff involving z and dz.
    The step below is flaky. What is the variable of integration - dx or dz?
     
  4. Dec 4, 2012 #3
    I seem to get the right answer by separating the variables as you mentioned.

    The variable of integration is dz (wont let me edit now...), is there an error in my original method?
     
  5. Dec 4, 2012 #4

    Mark44

    Staff: Mentor

    It doesn't seem right time.

    Starting from this step:
    $$ d/dx(xz) = \frac{1 + 4z^3}{3z^2}$$

    What you're saying is that the derivative of xz with respect to x is <stuff on right side>, so xz must be the antiderivative of <stuff on right side> with respect to x. You can't just stick a dz in there out of nowhere.
     
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