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ODE help

  1. Aug 31, 2006 #1
    just started ordinary differential equation class, and i have to find the general soln of the eqn: y'=y(3-y)... just started so i only know of the method of seperation of varibles.

    Thought of multiplying the y through so it becomes:
    y'=3y-y^2 then do..

    dy/dx=3y-y^2 then... im stuck... I tried doing seperation of varibles by

    Dividing 3y-y^2 to the other side so you get:

    dy/3y-y^2=0? But that doesn't seem right.
    Last edited: Aug 31, 2006
  2. jcsd
  3. Aug 31, 2006 #2
    [tex] \frac{dy}{dx} = y(3-y) [/tex]

    [tex] \frac{dy}{y(3-y)} = dx [/tex]

    [tex] \int \frac{dy}{y(3-y)} = \int dx [/tex]

    Break up the left-hand side by the method of partial fractions.
  4. Aug 31, 2006 #3
    after partial fractions and integrating both sides:

    ln(y) - ln(y-3) = 3 x + C

    by logarithmic identities:

    ln(y/(y-3)) = 3 x + C

    Taking the exponential of both sides:

    y/(y-3) = C1 e^(3 x)

    Where C1 = e^C

    Solving for y:

    y = 3 C1 e^(3 x)/(C1 e^(3 x)-1)
  5. Aug 31, 2006 #4
    Well, geez... why don't ya just do the whole thing for him? Oh wait, you just did.

    Way to encourage the joy of discovery. :uhh:
  6. Aug 31, 2006 #5

    After my partial fraction decomposition I got:

  7. Aug 31, 2006 #6
    You have the same thing. Just change your - sign into an exponent and multiply both sides by 3.
  8. Aug 31, 2006 #7
    how do you change a minus sigh into a exponent, im lost when you said that.
  9. Aug 31, 2006 #8
    Use the property of logarithms: [tex] r log b = log b^{r} [/tex]

    where [tex] r = -1 [/tex]

    and [tex] b = \frac{y-3}{y} [/tex]
  10. Sep 1, 2006 #9


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    I'm afraid you are going to find differential equations extremely difficult if you cannot do basic algebra!
    Starting from dy/dx=3y-y2 and dividing both sides by 3y- y2, you do NOT get dy/(3y- y2)= 0 any more than dividing both sides of xy= 3 by 3 would give xy/3= 0.
  11. Sep 1, 2006 #10
    It was a brain fart ease up on me
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