# 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.

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?