Solving an IVP Problem: dx/dt, x(0) & Equilibrium Points

[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.

 

[HallsofIvy]

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.

 

[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.

 

[HallsofIvy]

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)/(x^1.5). All that's missing now is the negative sign.

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