Problem with an IVP

1. Sep 26, 2004

Tryingtolearn

I'm trying to solve an IVP problem but have not had a lot of Diff Eq to really understand this.

The IVP is

dx/dt = -(1-(1/x))*(1/(x^.5)) , x>1
x(0) = 2

I guess some of the things I'm looking for with this is:
Any unique solutions
equilibrium points
Initial velocity
distance increasing or decreasing

Any additional help on how to solve this would be very much appreciated.

Thanks in advance.

2. Sep 26, 2004

HallsofIvy

Staff Emeritus
We can write dx/dt= -(1-(1/x))*(1/x^.5)) as -(x- 1)/x^1.5)

There is not t on the right so we can write this in differential form as
- x^1.5/(x-1)dx= dt.

Now integrate both sides to find the solutions.

The rest are even easier. An "equilibrium point" is a point where x doesn't change: the derivative is 0: dx/dt = -(1-(1/x))*(1/(x^.5)) = 0. Solve that equation for x.

distance (x) is increasing where the derivative is positive and decreasing where it is negative. Since you have a formula for dx/dt, that should be as easy as determining where it was 0.

3. Sep 26, 2004

Tryingtolearn

I'm sorry, I guess I don't see how -(1-(1/x))*(1/x^.5)) is the same as -(x- 1)/x^1.5). I've been straining with with algebra but I just don't see that.

If it is -(x- 1)/x^1.5 then I'm at a loss for how to integrate this.

Thanks.

4. Sep 27, 2004

HallsofIvy

Staff Emeritus
Subtract the fractions: 1- 1/x= x/x- 1/x= (x-1)/x. Now multiply by (1/x.5): (x-1)/(x*x.5)= (x-1)/(x1.5). All that's missing now is the negative sign.

-(x-1)/x1.5= -(x/x1.5-1/x1.5)= -x1-1.5+x-1.5= -x-.5- x-1.5. Can you integrate that?

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