SOLUTION: The volume,x litres of water present in a solution during a chemical process varies with time t seconds and satisfies the relation dx/dt=-3x/(1+t)^2. initially at t=0, x=1000. Show

Algebra ->  Volume -> SOLUTION: The volume,x litres of water present in a solution during a chemical process varies with time t seconds and satisfies the relation dx/dt=-3x/(1+t)^2. initially at t=0, x=1000. Show      Log On


   

Question 1012678: The volume,x litres of water present in a solution during a chemical process varies with time t seconds and satisfies the relation dx/dt=-3x/(1+t)^2. initially at t=0, x=1000. Show that at time t the volume is given by x=1000exp[-3t/(1+t)].
Found 2 solutions by Fombitz, fractalier:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
dx%2Fdt=-3x%2F%281%2Bt%29%5E2
dx%2Fx=-3%281%2Bt%29%5E%28-2%29%2Adt
Use a substitution,
u=1%2Bt
du=dt
So,
dx%2Fx=-3u%5E%28-2%29du
Integrating both sides,
ln%28x%29=3u%5E%28-1%29%2BC
ln%28x%29=3%2F%281%2Bt%29%2BC
x=Ce%5E%283%2F%281%2Bt%29%29
When t=0,x=1000
1000=Ce%5E%283%2F%281%2B0%29%29
C=1000%2Fe%5E3
x=%281000%2Fe%5E3%29%2Ae%5E%283%2F%281%2Bt%29%29
Check your problem setup because our functions don't match.

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
From dx/dt=-3x/(1+t)^2, I'm thinking to solve this by separation of variables...thus, we can rearrange and get
(1/x) dx = -3 dt / (t+1)^2
Now integrate
ln x = 3/(t+1) + C
Now exponentiate
x(t) = Ce^(3/(t+1))
Now apply initial conditions...
x(0) = 1000 = Ce^3 so that
C = 1000/e^3 = 1000e^(-3)
and then
x(t) = 1000e^(-3)*e^(3/(t+1))
Now combine exponents and you get
x(t) = 1000e^(-3t/(1+t))