Question

Would anyone be able to assist in solving this DE:

\[x'' + \frac{ 2U }{ bm } (e ^{-2(x-x _{0})/b} - e ^{(x-x _{0})/b}) = 0\]

Answer 1

?

Answer 2

Sorry, equation dropped off.

\[x'' + \frac{ 2U }{ bm } (e ^{-2(x-x _{0})/b} - e ^{(x-x _{0})/b}) = 0\]

Answer 3

what is U ?, just some constant?

Answer 4

correct, some constant.

Answer 5

so it might make it eaisier if you call change 2U/bm = k

and change it back when you have solve the equation,

Answer 6

so that

\[x'' + k(e ^{-2(x-x _{0})/b} - e ^{(x-x _{0})/b}) = 0\]

now can you see a common factor in the brackets

Answer 7

good point, thanks Uncle

Answer 8

I get something that is long and messy as my solution. Can I ask what you get, if you have a moment?

Answer 9

where did the equation come from , it looks like physics?

Answer 10

correct, original equation I have is:

\[\frac{ 2U }{ b } (e ^{-2(x-x _{0})/b} - e ^{(x-x _{0})/b}) = F(x)\]

\[f(x) = mx''\]

Answer 11

sorry, meant to be \[F(x) = mx''\]

Answer 12

did you find any common factors in the large brackets?

Answer 13

you meant the (x - x0)/b?

Answer 14

yers

Answer 15

@dumbcow

Answer 16

Sorry, I may be sleep deprived but I dont see how to proceed. Even if you sub out the the common factor?

Answer 17

ok this is a 2nd order non-linear DE which i am not used to and are pretty difficult...i worked it out best i could and i believe there is no closed form "x(t) = ?" solution
according to wolfram solution uses hypergeometric function.

anyway like @UnkleRhaukus said you can simplify the equation by transforming variables
let
\[k = \frac{2U}{m}\]
\[w = \frac{x-x_0}{b} \rightarrow dw = \frac{dx}{b}\]
now DE looks like this
\[w'' +\frac{k}{b^{2}}(e^{-2w}-e^{w}) = 0\]
now there is a trick to change this to a first order DE.....ref link below

http://math.stackexchange.com/questions/42223/solving-the-nonlinear-second-order-ode-u-u2-9-0
this transforms DE to this:
\[\frac{1}{2}(w')^{2} = \frac{k}{b^{2}} \int\limits e^{w} -e^{-2w} dw\]
after integrating
\[\frac{1}{2}(w')^{2} = \frac{k}{b^{2}}(e^{w}+\frac{1}{2}e^{-2w}+c_1)\]
solving
\[\large w' = \frac{\sqrt{2ke^{w}+ke^{-2w}+c_1}}{b}\]
separate variables
\[\large \frac{dw}{\sqrt{2ke^{w}+ke^{-2w}+c_1}} = \frac{dt}{b}\]
and that integral cant be done using elementary functions..
http://www.wolframalpha.com/input/?i=+dw%2Fsqrt%282e%5Ew%2Be%5E%28-2w%29%29+%3D+dt

Answer 18

dumbcow, thanks so much for that. I would not have got this...

Answer 19

hey, great work @dumbcow !

Answer 20

Yes, thanks both :)