All alternative solution for an equation I'm looking for all alternative solutions of this
$$x'=x(x-1)(x+1)$$
But I absolutely don't know what I have to do!
Thanks!
 A: $$\frac{x'}{x(x-1)(x+1)}=1$$
$$\frac{x'}{2(x+1)}+\frac{x'}{2(x-1)}-\frac{x'}{x}=1$$
$$\frac{dx}{2(x+1)}+\frac{dx}{2(x-1)}-\frac{dx}{x}=dt$$
A: Look for separable differential equations.
Equation can be divided by $x(x-1)(x+1)$ and integrated (left side with respect to $x$ where $x'$ becomes $dx$ and right side with respect to $t$ - we integrate 1).
Left side can be integrated using partial fractions method!
A: Let we start by noticing that
$$\frac{1}{(z-1)z(z+1)} = \frac{1/2}{z-1}-\frac{1}{z}+\frac{1/2}{z+1}\tag{1}$$
by partial fraction decomposition/the residue theorem. It follows that the separable DE
$$ \frac{x'}{(x-1)x(x+1)} = 1 \tag{2}$$
leads to
$$ \frac{1}{2}\log(x-1)-\log(x)+\frac{1}{2}\log(x+1) = t+C \tag{3} $$
or to:
$$ \log\left(1-\frac{1}{x^2}\right) = 2t+D\tag{4} $$
from which:
$$ x(t)^2 = \color{red}{\frac{1}{1-K e^{2t}}}\tag{5} $$
where $K>0$ depends on the initial conditions. In particular, $x(0)$ tells us exactly where the associated solution has a blowup.
A: When $\text{n}$ is a constant:
$$x'(t)=x(t)\left(x(t)-\text{n}\right)\left(x(t)+\text{n}\right)\Longleftrightarrow\int\frac{x'(t)}{x(t)\left(x(t)-\text{n}\right)\left(x(t)+\text{n}\right)}\space\text{d}t=\int1\space\text{d}t$$
Now, use:


*

*Substitute $u=x(t)$ and $\text{d}u=x'(t)\space\text{d}t$:
$$\int\frac{x'(t)}{x(t)\left(x(t)-\text{n}\right)\left(x(t)+\text{n}\right)}\space\text{d}t=\int\frac{1}{u\left(u-\text{n}\right)\left(u+\text{n}\right)}\space\text{d}u=$$
$$\frac{1}{2\text{n}^2}\int\frac{1}{u-\text{n}}\space\text{d}u+\frac{1}{2\text{n}^2}\int\frac{1}{u+\text{n}}\space\text{d}u-\frac{1}{\text{n}^2}\int\frac{1}{u}\space\text{d}u=$$
$$\frac{\ln\left|u-\text{n}\right|}{2\text{n}^2}+\frac{\ln\left|u+\text{n}\right|}{2\text{n}^2}-\frac{\ln\left|u\right|}{\text{n}^2}+\text{C}=\frac{\ln\left|x(t)-\text{n}\right|}{2\text{n}^2}+\frac{\ln\left|x(t)+\text{n}\right|}{2\text{n}^2}-\frac{\ln\left|x(t)\right|}{\text{n}^2}+\text{C}$$

*$$\int1\space\text{d}t=t+\text{C}$$


So, we get:
$$\frac{\ln\left|x(t)-\text{n}\right|+\ln\left|x(t)+\text{n}\right|-2\ln\left|x(t)\right|}{2\text{n}^2}=t+\text{C}\Longleftrightarrow\ln\left|\frac{\text{n}^2}{x(t)^2}-1\right|=2\text{n}^2t+\text{C}$$
