Find analytically $x(t)$ $ \dot{x}= x-x^3$

Question

Find analytically $x(t)$

$$ \dot{x}= x-x^3$$

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Asked computer gave out

Answer 1

$$\frac{\mathrm{d}x}{\mathrm{d}t}=x-x^3$$ $$\frac{1}{x-x^3}\frac{\mathrm{d}x}{\mathrm{d}t}=1$$ Integrate both sides with respect to $t$: $$t=\int \frac{1}{x-x^3}\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}t$$ $$t=\int \frac{1}{x-x^3}\mathrm{d}x$$ Now we can use partial fraction decomposition: $$t=\int \frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}\mathrm{d}x$$ $$t=A\log(x)-B\log(1-x)+C\log(1+x)+\text{constant}$$ You can find $A$,$B$ and $C$ easily : $A=1$, $B=\frac{1}{2}$, $C=-\frac{1}{2}$, So: $$t=\log(x)-\frac{1}{2}\log(1-x)-\frac{1}{2}\log(1+x)+\text{constant}$$ $$t=\log\left(\frac{x}{\sqrt{1-x^2}}\right)+\text{constant}$$

Answer 2

you can write $$\frac{dx}{x-x^3}=dt$$

Answer 3

First of all notice that we have two trivial solutions $$ x=1 \text { and } x=0 $$ In order to solve this separable differential equation $$\dot{x}= x-x^3$$ we divide both side by $x-x^3$ to get $$ \frac {dx}{x-x^3} =dt $$

Upon integrating both side and using partial fraction we come up with $$ ln(|x|)-(1/2)ln(|1-x^2|)=t+c$$

$$ln(\frac {|x|}{\sqrt {1-x^2}})=t+c $$

$$\frac {x}{\sqrt {1-x^2}}= Ce^t$$

Where C is an arbitrary constant.

Answer 4

This is a Bernoulli-type equation. Set $u=x^{-2}$ to find $$ \dot u = -2x^{-3}\dot x = -2u+2 $$ to get a linear order one equation that can be solved via the usual means.