# Derivative of function = square of the function?

1. Jan 27, 2012

### 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.

2. Jan 28, 2012

### Staff: Mentor

Write the equation as dy/dx = y2.
This equation is separable and is actually pretty simple.

3. Jan 28, 2012

### mnb96

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

4. Jan 28, 2012

### JJacquelin

And with the integration constant c :
y = -1/(x+c)

5. Jan 28, 2012

### 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.

Last edited: Jan 28, 2012
6. Jan 29, 2012

### JG89

f = 0 is also a solution.

7. Jan 29, 2012

### JJacquelin

OK. f=0 is a solution, included in the set of solutions f=-1/(x+c) in the particular case c= infinity.