1. ## First-Order Separable Equation

Hi Guys

been about 20years since i flexed my maths muscle and it apears to have weakened somewhat.

Im reading a research paper and going through the equations therein and have come across a part I cant solve.

I dont want to just use the papers equations without understanding how they were derived (not my style) so any help with this bottleneck would be really appreciated

Heres what the paper gives:

V = wA/2Nr

where V is a radial velocity hence can be rewrited as = -dr/dt (where r is radial position and t is time)

so we have -dr/dt =wA/2Nr (1)

We need to solve for r.

the paper gives the initial condition, r=ro at t=0 (where ro is a constant)

It then gives a solution for r as:- r= (ro^2- (wAt/N)^0.5

I cant see how they get that.....

can anyone shed any light on this?

I assumed I just needed to substitute ro into eq (1) and integrate with respect to t

giving r = wAt/2Nro?

but thats not what the paper is showing.

I can only assume im missing something/forgotten something. any help would be appreciated

2. ## re: First-Order Separable Equation

$\displaystyle \frac{dr}{dt} = \frac{wA}{2Nr}$

$\displaystyle 2Nr~dr = wA~dt$

$\displaystyle \int 2Nr~dr = \int wA~dt$

$\displaystyle Nr^2 = wAt+C$

Now sub $r(0)= r_0$ to find $C$

After you know that, the rest falls out as follows

$\displaystyle r^2 = \frac{wAt+C}{N}$

$\displaystyle r = \sqrt{\frac{wAt+C}{N}}$

3. ## re: First-Order Separable Equation

thank you so much!!! greatly appreciated

4. ## Re: First-Order Separable Equation

i didn't understand the solution

5. ## Re: First-Order Separable Equation

Originally Posted by faria2159
i didn't understand the solution
Unless you say what part you don't understand, this statement is not particularly useful.