Derivative of function = square of the function?

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Discussion Overview

The discussion revolves around the possibility of finding a function \( f: \mathbb{R} \to \mathbb{R} \) such that its first derivative equals its square, specifically \( f'(x) = f(x)^2 \). Participants explore both single-variable and multi-variable cases, examining the implications and potential solutions of the differential equation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant questions whether a function exists such that \( f'(x) = f(x)^2 \), noting that a simpler case with \( f'(x) = f(x) \) has a known solution, \( f(x) = e^x \).
  • Another participant suggests rewriting the equation as \( \frac{dy}{dx} = y^2 \) and mentions that it is separable and relatively simple to solve.
  • A participant claims to have found a solution using separation of variables, stating \( f(x) = -\frac{1}{x} \) as one possible solution.
  • Further elaboration includes the integration constant, leading to the general form \( f(x) = -\frac{1}{x+c} \).
  • Another participant introduces a multi-variable case, asking about the function \( f(x,y) \) such that \( \frac{\partial f}{\partial x} = f(x,y)^2 \), proposing a solution of the form \( f(x,y) = -\frac{1}{x+C(y)} \) but seeking clarification on how to derive it.
  • One participant notes that \( f = 0 \) is also a solution, which is acknowledged as part of the broader set of solutions including \( f = -\frac{1}{x+c} \) when \( c \) approaches infinity.

Areas of Agreement / Disagreement

Participants express varying viewpoints on the solutions to the differential equations, with some agreeing on specific solutions while others propose alternative forms. The discussion remains unresolved regarding the completeness of the solution set and the methods for deriving solutions in the multi-variable case.

Contextual Notes

Some participants note the separability of the equations and the role of integration constants, but there is no consensus on the full set of solutions or the methods for arriving at them, particularly in the multi-variable context.

mnb96
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Hi,
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: f'(x)=f(x)^2

It is known that if we replace the exponent 2 with 1, and require that f'(x)=f(x), then a solution would be f(x)=e^x, but when we require the derivative to be equal to the function squared, the solution (if it exists at all) is less obvious.
 
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mnb96 said:
Hi,
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: f'(x)=f(x)^2

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

Write the equation as dy/dx = y2.
This equation is separable and is actually pretty simple.
 
that's true! Thanks for the hint.
Using the separation of variables I got f(x)=-\frac{1}{x}, as one possible solution.
 
mnb96 said:
that's true! Thanks for the hint.
Using the separation of variables I got f(x)=-\frac{1}{x}, as one possible solution.

And with the integration constant c :
y = -1/(x+c)
 
What if we consider instead a function of two variables f(x,y) and we want to find the family of functions that satisfy the following:

\frac{\partial f}{\partial x}=f(x,y)^2

Relying on my intuition I would say the solution is: f(x,y)=-\frac{1}{x+C(y)} 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.
Thanks.
 
Last edited:
f = 0 is also a solution.
 
JG89 said:
f = 0 is also a solution.

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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