1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You need another factor of $x$ for $2xdx=dk$ $\endgroup$ – Michael Jul 26 '17 at 7:57
  • 2
    $\begingroup$ Its not substitution its just a name for x so that we can use fractional decomposition. $\endgroup$ – Archis Welankar Jul 26 '17 at 8:11
0
$\begingroup$

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 $x \in \mathbb{R}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.