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

I How do you solve this equation?

  1. Jun 18, 2017 #1
    How can we solve a differential equation of the form $$a(r)Y' = Y - Y^2$$ for ##Y \equiv \phi(r)##? An equation of this form appears in https://en.wikipedia.org/wiki/Deriv...field_equations_to_find_A.28r.29_and_B.28r.29. If there were not the ##Y^2##, that would be a easy to solve differential equation: just a first order linear differential equation. The problem is the ##Y^2## though.
     
  2. jcsd
  3. Jun 18, 2017 #2

    Ssnow

    User Avatar
    Gold Member

    You must use the separable variables method.
     
  4. Jun 18, 2017 #3

    Ssnow

    User Avatar
    Gold Member

    Note: In order to have a closed form for the solution ##Y## you must integrate ##\int \frac{dr}{a(r)}##, so it depends by the expression of ##a(r)## ...
    Ssnow
     
  5. Jun 18, 2017 #4

    fresh_42

    Staff: Mentor

  6. Jun 18, 2017 #5
    Just too easy! Thanks for the hint

    $$ a(r) \frac{dy}{dr} = y - y^2 \\
    \frac{dy}{y} + \frac{dy}{1-y} = \frac{dr}{a(r)} \\
    \text{ln}y + c_1 - \text{ln}(1-y) + c_2 = \int \frac{1}{a(r)}\ dr \\
    C \equiv c_1 + c_2 \\
    \text{ln}\bigg(\frac{y}{1-y}\bigg) + C = \int \frac{1}{a(r)} \ dr \\
    \frac{y}{1-y} = e^{\int \frac{1}{a(r)}\ dr - C} \\
    y\bigg(1- e^{\int \frac{1}{a(r)}\ dr - C}\bigg) = e^{\int \frac{1}{a(r)}\ dr - C} \\
    \Rightarrow y = \frac{e^{\int \frac{1}{a(r)}\ dr - C}}{\bigg(1- e^{\int \frac{1}{a(r)}\ dr - C}\bigg)}$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How do you solve this equation?
Loading...