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Stuck on one of the substitution method steps

  1. Jan 27, 2013 #1
    I put it in std form, did the homogeneous test. it passed with degree 2. I substituted y=ux and dy=udx+xdu and now I'm stuck. it needs to be simplified somehow but I don't know if ux is one var or if it's u*x. Same goes for udx and xdu. Is it really u*dx+x*du? even assuming that is correct, it doesn't get any simpler. What does the next step look like for this problem?

    52ck6vx.jpg
     
  2. jcsd
  3. Jan 27, 2013 #2

    lurflurf

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    That is good so far, the substitution has transformed the equation into separable form, divide to complete separation.

    $$\frac{\mathrm{f}(u)\mathrm{g}(x) \mathrm{dx}+\mathrm{p}(u) \mathrm{q}(x)\mathrm{du}}{\mathrm{f}(u) \mathrm{q}(x)}=\frac{\mathrm{g}(x)}{\mathrm{q}(x)}\mathrm{dx}+\frac{\mathrm{p}(u) }{\mathrm{f}(u)}\mathrm{du}$$


    $$\frac{((ux)^2-2x^2)\mathrm{dx}+2x(ux)(x \mathrm{du}+u\mathrm{dx})}{x^3(3u^2-2)}=\frac{\mathrm{dx}}{x}+\frac{2u\mathrm{du}}{3u^2-2}=0$$
     
  4. Jan 27, 2013 #3
    Where did you get the denominator x3(3u2-2) from?

    What does f(u), g(x), p(u), and q(x) each equal?
     
  5. Jan 27, 2013 #4

    lurflurf

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    In this example
    [tex]\mathrm{f}(u)=3u^2-2 \\
    \mathrm{g}(x)=x^2 \\
    \mathrm{p}(u)=2u \\
    \mathrm{q}(x)=x^3 [/tex]

    This is just separation of variables
    $$((ux)^2-2x^2)\mathrm{dx}+2x(ux)(x \mathrm{du}+u\mathrm{dx})=(3u^2-2)(x^2)\mathrm{dx}+(2u)(x^3)\mathrm{du}$$
    So we divide by the the factors that do not match their differentials.
     
  6. Jan 27, 2013 #5
    $$\frac{\mathrm{f}(u)\mathrm{g}(x) \mathrm{dx}+\mathrm{p}(u) \mathrm{q}(x)\mathrm{du}}{\mathrm{f}(u) \mathrm{q}(x)}=\frac{\mathrm{g}(x)}{\mathrm{q}(x)}\mathrm{dx}+\frac{\mathrm{p}(u) }{\mathrm{f}(u)}\mathrm{du}$$

    Is the above eq some sort of shortcut that is explained somewhere. I've never seen it in all of the lessons I've read.

    I understand how I got to here:
    $$((ux)^2-2x^2)\mathrm{dx}+2x(ux)(x \mathrm{du}+u\mathrm{dx})$$

    and how you got from here:
    $$(3u^2-2)(x^2)\mathrm{dx}+(2u)(x^3)\mathrm{du}$$ to the next part but not the step(s) in between. I tried working it out several times but could never separate the variables completely.
     
    Last edited: Jan 27, 2013
  7. Jan 28, 2013 #6

    lurflurf

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    $$\frac{\mathrm{f}(u)\mathrm{g}(x) \mathrm{dx}+\mathrm{p}(u) \mathrm{q}(x)\mathrm{du}}{\mathrm{f}(u) \mathrm{q}(x)}=\frac{\mathrm{g}(x)}{\mathrm{q}(x)}\mathrm{dx}+\frac{\mathrm{p}(u) }{\mathrm{f}(u)}\mathrm{du}$$
    That is just the separation of variables equation.
    Write the equation in the form of each differential multiplied by a function of each variable, divide by each non-matching function to give each differential multiplied by a matching function.


    [tex]((ux)^2-2x^2) \, \mathrm{dx}+2x(ux)(x \, \mathrm{du}+u \, \mathrm{dx})=
    x^2(u^2 -2) \, \mathrm{dx}+2x^2 u(x \, \mathrm{du}+u \, \mathrm{dx})\\
    =x^2(u^2 -2) \, \mathrm{dx}+(2x^2 u)x \, \mathrm{du}+(2x^2 u)u \,\mathrm{dx}\\
    =x^2(u^2 -2+2u^2) \, \mathrm{dx}+2x^3 u \, \mathrm{du} \\
    =(3u^2-2)(x^2) \, \mathrm{dx}+(2u)(x^3) \, \mathrm{du}[/tex]
     
  8. Jan 28, 2013 #7
    thanks. I got it now.
     
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