# solving differential equation

How to solve the following differential equation $\frac{dx}{dt}=x^2-x^6$
I tried in this way
$\frac{dx}{dt}=x^2(1-x^4)$
Put $x^2=y$, we get $\frac{dy}{dt}=2y\sqrt{y}(1-y^2)$
Integrating by parts method is going multiple times.
Is there any method to solve this kind of differential equations. Please give the solution.

We get $$\int \frac {1}{(x^2)(1-x^2)(1+x^2)} dx=t+c$$ Now just let $x^2=k$ . Note it isnt a substitution so by parts we have $\int (\frac{a}{k}+\frac{b}{1+k}+\frac {c}{1-k} )dx=t+c$ now find $a,b,c$ and then we have known formulae for every separate integral.

• You need another factor of $x$ for $2xdx=dk$ – Michael Jul 26 '17 at 7:57
• Its not substitution its just a name for x so that we can use fractional decomposition. – Archis Welankar Jul 26 '17 at 8:11

Hint: A seperable ODE $\frac{dx}{dt} = \frac{F(x)}{G(t)}$ can always be solved by rewriting $\frac{dx}{F(x)}=\frac{dt}{G(t)}$. Integration will yield:

$$\int \frac{dx}{F(x)}=\int\frac{dt}{G(t)}.$$

For your problem $F(x)=x^2-x^6$ and $G(t)=1$ (a there is no dependence on $t$ on the right hand side). Using the previous result we get:

$$\int \frac{dx}{x^2-x^6}=\int dt \implies \int \frac{dx}{x^2-x^6}=t+c.$$

Your, task is now to integrate the left-hand side. You can do this by partial fractions as Archis Welankar already implied.

Note that there are some special points $x=0$, $x=\pm 1$ and $x=\pm i \in \mathbb{C}$. These are equilibrium points of the system that are associated with the initial conditions $x(t=0)=0,\pm 1, \pm i$, that means if you start at these values you will always stay at these values. The complex equilibrium point is not meaningful, if you are only considering $x \in \mathbb{R}$.