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Derivative of function = square of the function?

  1. Jan 27, 2012 #1
    I was wondering whether it is possible or not to find a function f:ℝ→ℝ, such that its first derivative is equal to its square: [tex]f'(x)=f(x)^2[/tex]

    It is known that if we replace the exponent 2 with 1, and require that [itex]f'(x)=f(x)[/itex], then a solution would be [itex]f(x)=e^x[/itex], but when we require the derivative to be equal to the function squared, the solution (if it exists at all) is less obvious.
  2. jcsd
  3. Jan 28, 2012 #2


    Staff: Mentor

    Write the equation as dy/dx = y2.
    This equation is separable and is actually pretty simple.
  4. Jan 28, 2012 #3
    that's true! Thanks for the hint.
    Using the separation of variables I got [itex]f(x)=-\frac{1}{x}[/itex], as one possible solution.
  5. Jan 28, 2012 #4
    And with the integration constant c :
    y = -1/(x+c)
  6. Jan 28, 2012 #5
    What if we consider instead a function of two variables [itex]f(x,y)[/itex] and we want to find the family of functions that satisfy the following:

    [tex] \frac{\partial f}{\partial x}=f(x,y)^2[/tex]

    Relying on my intuition I would say the solution is: [itex]f(x,y)=-\frac{1}{x+C(y)}[/itex] but I'd like to know how to arrive to that result.
    Sorry if the question is very easy. I have not much experience with differential equations.
    Last edited: Jan 28, 2012
  7. Jan 29, 2012 #6
    f = 0 is also a solution.
  8. Jan 29, 2012 #7
    OK. f=0 is a solution, included in the set of solutions f=-1/(x+c) in the particular case c= infinity.
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