Help with decompression formulae

[Goose1232]

Hello, I've just joined and am doing a school project about decompression in scuba diving. I'm struggling at the moment about how the author of this paper used equations 2 and 3 to find equation 5.

Many thanks!!

 

[Deleted member 4993]

Hello, I've just joined and am doing a school project about decompression in scuba diving. I'm struggling at the moment about how the author of this paper used equations 2 and 3 to find equation 5.

Many thanks!!

They solved the Differential Equation (2) with initial conditions prescribed in (3)

Have you taken a math. course in solving DEs?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

 

[Goose1232]

Thanks so much! I've only really learned the very basics of solving DEs (I'm 17). I've given it my best attempt below, but I'm struggling ...


Once again, thanks in advance for any help you can give.

 

[HallsofIvy]

If I am reading this correctly, the differential equation is \(\displaystyle \frac{dA}{dt}= k_2(P- A)\) with initial condition \(\displaystyle A(0)= A_0\) (here, "A" is a function of t, and \(\displaystyle A_0\) is a number, the value of A(t) when t= 0). We can rewrite the equation as \(\displaystyle \frac{dA}{P- A}= k_2 dt\). Now integrate both sides.
To integrate \(\displaystyle \int\frac{dA}{P- A}\), let x= P- A. Then dx= -dA, dA= -dx so the integral becomes \(\displaystyle \int \frac{-dx}{x}= -ln(x)+ C= -ln(P- A)+ C\) where C is an undetermined constant.
The integral of \(\displaystyle \int k_2 dt\) is just \(\displaystyle k_2t+ C'\) where C' is also a constant, not necessarily the same as C.
So we have \(\displaystyle -ln(P- A)+ C= k_2t+ C'\). We can combine the two constants and call their difference, since it is yet another undetermined constant, C'': \(\displaystyle -ln(P- A)= k_2t+ C''\).
In order that \(\displaystyle A= A_0\) when t= 0 we must have \(\displaystyle -ln(P- A_0)= C''\) so we can write the solution as \(\displaystyle -ln(P- A)= k_2t- ln(P- A_0)\)
Multiply by -1 to get \(\displaystyle ln(P- A)= ln(P- A_0)- k_2t\).
Subtract \(\displaystyle ln(P- A_0)\) from both sides:
\(\displaystyle ln(P- A)- ln(P- A_0)= ln\left(\frac{P- A}{P- A_0}\right)= -k_2t\).
FINALLY, take the exponential of both sides:
\(\displaystyle \frac{P- A}{P- A_0}= e^{-k_2t}\).
You can, if you want, solve for A:
\(\displaystyle P- A= (P- A_0)e^{-k_2t}\)
\(\displaystyle -A= (P- A_0)e^{-k_2t}- P\)
\(\displaystyle A= P- (P- A_0)e^{-k_2t}\).

 

[Deleted member 4993]

If I am reading this correctly, the differential equation is \(\displaystyle \frac{dA}{dt}= k_2(P- A)\) with initial condition \(\displaystyle A(0)= A_0\) (here, "A" is a function of t, and \(\displaystyle A_0\) is a number, the value of A(t) when t= 0). We can rewrite the equation as \(\displaystyle \frac{dA}{P- A}= k_2 dt\). Now integrate both sides.
To integrate \(\displaystyle \int\frac{dA}{P- A}\), let x= P- A. Then dx= -dA, dA= -dx so the integral becomes \(\displaystyle \int \frac{-dx}{x}= -ln(x)+ C= -ln(P- A)+ C\) where C is an undetermined constant.
The integral of \(\displaystyle \int k_2 dt\) is just \(\displaystyle k_2t+ C'\) where C' is also a constant, not necessarily the same as C.
So we have \(\displaystyle -ln(P- A)+ C= k_2t+ C'\). We can combine the two constants and call their difference, since it is yet another undetermined constant, C'': \(\displaystyle -ln(P- A)= k_2t+ C''\).
In order that \(\displaystyle A= A_0\) when t= 0 we must have \(\displaystyle -ln(P- A_0)= C''\) so we can write the solution as \(\displaystyle -ln(P- A)= k_2t- ln(P- A_0)\)
Multiply by -1 to get \(\displaystyle ln(P- A)= ln(P- A_0)- k_2t\).
Subtract \(\displaystyle ln(P- A_0)\) from both sides:
\(\displaystyle ln(P- A)- ln(P- A_0)= ln\left(\frac{P- A}{P- A_0}\right)= -k_2t\).
FINALLY, take the exponential of both sides:
\(\displaystyle \frac{P- A}{P- A_0}= e^{-k_2t}\).
You can, if you want, solve for A:
\(\displaystyle P- A= (P- A_0)e^{-k_2t}\)
\(\displaystyle -A= (P- A_0)e^{-k_2t}- P\)
\(\displaystyle A= P- (P- A_0)e^{-k_2t}\).

Here however, P is also a function of 't' (look at eqn. 1 in the paper).

 

[HallsofIvy]

Ouch! Thanks.

 

[Goose1232]

Thanks very much!! This is very helpful.