Question

\[\frac {\text d^2 x}{\text dt^2}=-\omega_0^2x+\beta x^2\]

Answer 1

\[\omega_0=2\pi\]

Answer 2

i want to let \[ p=\frac{dx}{dt} \]
then u have
\[ \frac{d^2x}{dt^2}=\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=p \frac{dp}{dx} \]
equation becomes to

\( p \ dp=(-Ax+Bx^2) \ dx \) and

\[ \ \frac{p^2}{2} =-\frac{A}{2}x^2+\frac{B}{3}x^2 +C\]

Answer 3

and we arrive to

\[ p^2=-Ax^2+\frac{2}{3}B x^3+C\]it gives

\[(\frac{dx}{dt}) ^2=-Ax^2+\frac{2}{3}B x^3+C\]
or
\[\frac{dx}{\sqrt{-Ax^2+\frac{2}{3}B x^3+C}}=dt\]

Answer 4

oh oh thats very bad for integration

Answer 5

\[\frac {\text d^2 x}{\text dt^2}=-\omega_0^2x+\beta x^2\]
\[ \int \text dt^2=\int \frac {\text d^2 x}{-\omega_0^2x+\beta x^2}\]
\[ \text dt=[\ln (\frac { x}{-\omega_0^2+\beta x})+C \ ] \ dx\]
\[\int \text dt=\int [\ln (\frac { x}{-\omega_0^2+\beta x})+C \ ] \ dx\]
\[ \text t= x \ln (\frac { x}{-\omega_0^2+\beta x})+\frac{\omega_0^2}{\beta}\ln (-\omega_0^2+\beta x)+Cx+D \]

Answer 6

i made a bunch of mistakes in integrating

Answer 7

@eliassaab
are we allowed to doing something like my second work?

Answer 8

No

Answer 9

so thats wrong....and no need to correct my integration
so what we gonna do !!!!

Answer 10

perhaps an analytic solution is impossible,

Answer 11

yes; maybe

Answer 12

seems that works
http://www.wolframalpha.com/input/?i=y%27%27+%3D+ax+%2B+bx^2

Answer 13

i think u made a little mistake santosh

Answer 14

ah @experimentX
x=y
t=x

http://www.wolframalpha.com/input/?i=x%27%27+%3D+ax+%2B+bx%5E2

Answer 15

ah ... my bad!!

Answer 16

what a neat solution...lol

Answer 17

that wasn't linear!!

Answer 18

i guess the first method.

Answer 19

it seems like motion equation but its extremely complicated

Answer 20

acceleration depending on position?