Derivative of function = square of the function?

[mnb96]

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.

 

[Mark44]

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.

 

[mnb96]

that's true! Thanks for the hint.
Using the separation of variables I got f(x)=-\frac{1}{x}, as one possible solution.

 

[JJacquelin]

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)

 

[mnb96]

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.

 

[JG89]

f = 0 is also a solution.

 

[JJacquelin]

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.