# Math Help - Trivial Problem...

1. ## Trivial Problem...

Apologies in advance, I'm rusty and having some trouble and really want to make more of an effort to keep up with my current year.

Have a simple differential:

dx/dt = x^3

Initial condition x(0) = x0

Apparently a simple integration by parts yields:

x(t) = x0 / (1-2x0^2 t)^1/2

I know i'm falling at something trivial and just not recognising the form. Could someone explain this expression? (how has it retained t terms?)
Thank you for bearing with me. Once I get the ball rolling, maybe I can come up with better questions

2. I'm not sure I would say that you use integration by parts. The differential equation you have there yields to separation of variables thus:

$\displaystyle \frac{dx}{dt}=x^{3},$ subject to $x(0)=x_{0}.$ Thus,

$\displaystyle \frac{dx}{x^{3}}=dt,$ and hence

$\displaystyle \int\frac{dx}{x^{3}}=\int dt.$ Can you continue from here?

3. Originally Posted by Ackbeet
I'm not sure I would say that you use integration by parts. The differential equation you have there yields to separation of variables thus:

$\displaystyle \frac{dx}{dt}=x^{3},$ subject to $x(0)=x_{0}.$ Thus,

$\displaystyle \frac{dx}{x^{3}}=dt,$ and hence

$\displaystyle \int\frac{dx}{x^{3}}=\int dt.$ Can you continue from here?
All that is correct if $x_{0} \ne 0$. In the case $x_{0} = 0$ the solution to the DE $x^{'} = x^{3}$ is $x=0$...

Kind regards

$\chi$ $\sigma$

4. Originally Posted by Ackbeet
I'm not sure I would say that you use integration by parts. The differential equation you have there yields to separation of variables thus:

$\displaystyle \frac{dx}{dt}=x^{3},$ subject to $x(0)=x_{0}.$ Thus,

$\displaystyle \frac{dx}{x^{3}}=dt,$ and hence

$\displaystyle \int\frac{dx}{x^{3}}=\int dt.$ Can you continue from here?
erm....

t = -1/2x^2 + c
1/x^2 = 2c -2t
x^2 = 1/ (C-2t)
x = 1/ sqt(C-2t)

x(0) = xsubscript0
which gives: C = 1/ (xsub0 ^2)

substitute:
x = 1/ sqt(1/xsub0^2 - 2t)

I think the last step i got confused mostly:

multiply by xsub0^2(rooted) on RHS top and bottom

x = x0 / sqt(1-2x0^2 x t)

5. So is it all clear now?