How Do You Solve the Differential Equation x'=x(M-x) for x(t)?

[AronH]

I was making a problem about population grow, and I wasn't able to solve this:
x'=x*(M-x), for x(t).
Can anyone help me?
Thanks.

 

[ahrkron]

I have the impresion that there is no analytical solution to that one. IIRC, its solution has a chaotic behavior depending on M (though there may be another parameter).

 

[Hurkyl]

I think it's seperable.

 

[Tide]

Integrate directly to find
$$\frac{x}{x_0} \times \frac{M-x}{M-x_0} = e^{M t}$$
from which you can find x(t) by solving the quadratic equation.

 

[Galileo]

scribbly scribbly...

$$\frac{dx}{dt}=xM-x^2$$
$$\int \frac{dx}{xM-x^2}=\int dt$$

... doesn't work, never mind...

edit: wait,wait,wait... partial fractions:

$$\frac{1}{xM-x^2}=\frac{1}{Mx}+\frac{1}{M(M-x)}$$

$$\int \frac{dx}{Mx}+\int \frac{dx}{M(M-x)}=\frac{1}{M}(\ln(\frac{x}{x_0})-\ln(\frac{M-x}{M-x_0}))=\frac{1}{M}\ln\left(\frac{x(M-x_0)}{x_0(M-x)}\right)$$
So:
$$\frac{x}{x_0}\frac{M-x_0}{M-x}=e^{Mt}$$
Look solvable now...

 

[Tide]

Thanks - I flipped a sign!