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Solving by seperation of variables.

  1. Nov 20, 2004 #1
    I've got a few 1st order ODEs which I have problems solving. I am new to the subject and self-taught so I may have a little difficulty absorbing. The question is.......

    1. [tex] \frac {dy}{dx} = \frac {y^3}{x^2} [/tex]

    for 1. I put it in the form,

    [tex] x^2 dy = y^3 dx [/tex]

    [tex] \frac {dy}{y^3} = \frac {dx}{x^2} [/tex]

    But I get stuck when i integrate both sides. I have not integrated dy/ y^n functions before.
     
  2. jcsd
  3. Nov 20, 2004 #2

    arildno

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    For ALL numbers except r=-1, we have:
    [tex]\int{t}^{r}dt=\frac{1}{r+1}t^{r+1}+C[/tex]
     
  4. Nov 20, 2004 #3
    hmmm, I'm quite familiar with the formula. But, usually when we have a

    [tex] \frac {dy}{y} [/tex] function, we usually get [tex] ln y +c [/tex] for the integral. So, i was wondering if there is anything remotely close to ln which I can use. I can actually integrate [tex] x^3 [/tex] but I don't know what to do with the dy on top.

    thanks.
     
  5. Nov 21, 2004 #4
    When you have the dy on top you just need to rewrite the power of the y by putting a negative before the three. This is because dy/(x³) is just the same as (x^-3)dy.

    [tex]
    \frac{{dy}}{{dx}} = \frac{{y^3 }}{{x^2 }}
    [/tex]

    [tex]
    \int {\frac{{dy}}{{y^3 }}} = \int {\frac{{dx}}{{x^2 }}}
    [/tex]

    [tex]
    \int {y^{ - 3} } dy = \int {x^{ - 2} } dx
    [/tex]

    [tex]
    - \frac{1}{{2y^2 }} = - \frac{1}{x} + c
    [/tex]
     
    Last edited: Nov 21, 2004
  6. Nov 21, 2004 #5

    That's is true. When you have r=-1, then the answer would be [tex]ln x [\tex]
     
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