How to solve this integral?

Question

I want to express x as a function of t given the following:

dx/dt = sqrt(c - [3 k x^2]/4)

Can I integrate it w.r.t. t and if so how? And if not, how do I go about solving it?

Sorry, I put the wrong the power in my question. The equation should read:

dx/dt = sqrt(c-[3 k x^4]/4)

I've rearranged it to seperated the variables but the integral I've come up with looks pretty nasty and the wolfram integrator website is giving me answers involving elliptical integrals, which is a bit beyond my maths!

Answer 1

dx/dt
= sqrt(c - [3 k x^2]/4)
= 1/2 sqrt[4c - 3kx^2)]
= [sqrt(3k)/2] sqrt[4c/(3k) - x^2]

1/[sqrt(3k)/2] . 1/[4c/(3k) - x^2] dx = dt
1/[sqrt(3k)/2] is constant


Integrating both sides
1/[sqrt(3k)/2] . Sin^(-1)[x/sqrt(4c/3k)] = t + a
Now do the simplification and sail

Note:
Apply the form: 1/sqrt[a^2 - x^2] dx to 1/[4c/(3k) - x^2] dx
where a = sqrt[4c/(3k)]

Answer 2

the reason that Wolfram is giving the answer as an elliptic function is because that is the analytic answer. Solutions to differential equations are unique so there is only one. Elliptic functions are complicated, but if you draw the phase space of this function the behavior will become more apparent. If you don't want to deal with elliptic functions then I suggest using a numerical integrator in matlab or something. Most of the algebra for computing this integral in the other messages is wrong, again the answer is an elliptic function.

Answer 3

Multiple both sides by the dt term, so you have dx = sqrt(c - [3 k x^2]/4)dt. Then just integrate.
The integral of dx is just x.
the integral of sqrt(c - [3 k x^4]/4)dt is just t*sqrt(c - [3 k x^2]/4)dt+C. Remember, you are integrating with respect ot t, so you treat x like a constant. Now move the constant over to the x side and divide both sides by sqrt(c - [3 k x^4]/4).
So your fianl answer is (x+C)/(sqrt(c - [3 k x^4]/4))= t.

Answer 4

1) You need to put the x under the x and seperate out the t so you bring the right side over under the dx by cross multiplying and put the dt on the other side and you get:

dx * (c - [3*k*x^2]/4)^(-1/2) = dt

2) now you take the integral of the problem and you get:

[(c - (3*k*x^2)/4)^(1/2)] * [(6*k*x)/4] + C = t

i think thats right. its been a while since i've done integrations.

Answer 5

dx/dt = sqrt(c-[3 k x^4]/4)
rearrange
dt={1/sqrt(c-[3 k x^4]/4)}dx
integrate
t=int{1/sqrt(c-[3 k x^4]/4)}dx

use wolfram integrator to get
the integral of the x-term and
don't forget to add your other
constant A
the x^4 term makes this a very
difficult integration problem

good luck to you

ra

Answer 6

Yes you can.

sqrt(c - [3 k x^2]/4) = sqrt(c)sqrt(1-[sqrt(3k/c^2)x/2]^2)

This can be integrated into a sin-1 function.

Answer 7

No, only wrt x.