Sugar water tank problem (again)

Thank you for any help that you can give me. I am very bad at differential equations and this one is really stumping me.

Homework Statement

A tank contains 2740 L of pure water. A solution that contains 0.05 kg of sugar per liter enters a tank at the rate 3 L/min The solution is mixed and drains from the tank at the same rate.

Homework Equations

I guess I need to solve for ds/dt, which I think would be something along the lines of
ds/dt=.15-3s/2740, or the rate at which the sugar enters minus the rate at which it leaves.

The Attempt at a Solution

So far I've found the equation above and I think that the solution is something like e^[(-3*t)/2740], but my online homework thing keeps saying that it is wrong. Also, that solutino doesn't work for s(0)=0 and subtracting one from the above equation does not work. Does anybody have any suggestions? I think I might be solving the differential equation wrong, but I'm sure that my integrating factor is correct. Thank you very much.

lanedance
Homework Helper
the rate of change will be the difference between the sugar in (conc of 0.05kg/l at 3L/M per min) & the sugar out (conc of tank)

I am guessing you have used s for the sugar in the tank?

one would expect it would start at zero and increase assymotically to the concentration of the incoming stream

inlet stream (conc c_0=0.05kg/L) sugar in
(ds/dt)_in = c_0*3L/min = 3*(0.05)
(ds/dt)_out = c_tank*3L/min = (s/2740)*3

so the differential equation looks correct
ds/dt=3*(0.05-s/2740) with initial condition s(0) = 0

(2740/3)*ds/dt = 2740*0.05-s

maybe show how you solve this & we can have a look at it

hint you could try the subtitution & variable change
u = s-2740*0.05

the rate of change will be the difference between the sugar in (conc of 0.05kg/l at 3L/M per min) & the sugar out (conc of tank)

I am guessing you have used s for the sugar in the tank?

one would expect it would start at zero and increase assymotically to the concentration of the incoming stream

inlet stream (conc c_0=0.05kg/L) sugar in
(ds/dt)_in = c_0*3L/min = 3*(0.05)
(ds/dt)_out = c_tank*3L/min = (s/2740)*3

so the differential equation looks correct
ds/dt=3*(0.05-s/2740) with initial condition s(0) = 0

(2740/3)*ds/dt = 2740*0.05-s

maybe show how you solve this & we can have a look at it

hint you could try the subtitution & variable change
u = s-2740*0.05

Yeah, I chose s for the amount of sugar in the tank.
Anyway, here is how I would try to solve this (probably wrong)
(2740/3)*ds/dt=2740*0.05-s
ds/dt=.15-3s/2740
ds/dt+3s/2740=.15 I(t)=e^(3t/2740)
ds/dt*e^(3t/2740)+(3s/2740)e^(3t/2740)=.15*e^(3t/2740)
integrate both sides:
e^(3t/2740)s=.15e^(3t/2740)*(2740/3)
simplify:
e^(3t/2740)s=137(e^3t/2740)
And this is where I get stuck. Because then I get that s(t)=137, which obviously is wrong. Sorry about the slow post.

lanedance
Homework Helper
i haven't used integrating factors for ages, but how about the substitution?

so if you have
(2740/3)*ds/dt = 2740*0.05-s

let u = s-2740*0.05
then
du/dt = ds/dt

do the equation becomes
(2740/3)*du/dt = -u

should be able to separate, integrate & substitute back in for s

Last edited:
i haven't used integrating factors for ages, but how about the substitution?

so if you have
(2740/3)*ds/dt = 2740*0.05-s

let u = s-2740*0.05
then
du/dt = dst/dt

do the equation becomes
(2740/3)*du/dt = -u

should be able to separate, integrate & substitute back in for s

I ended up with -e^(3t/2740)+137=s but my homework thing still said that the answer was wrong.

My steps were:
3dt/2740=du/-u integrate both sides:
3t/2740=ln|-u|
e^(3t/2740)=137-s
-e^(3t/2740)+137=s
The problem is that this just doesn't fit the initial condition of 0 kg sugar. And when I changed the constant so that it was (+/-)e^3t/2740) (either way) -1, it still said that I was wrong, even though the initial condition was met.

lanedance
Homework Helper
I ended up with -e^(3t/2740)+137=s but my homework thing still said that the answer was wrong.

My steps were:
3dt/2740=du/-u integrate both sides:
3t/2740=ln|-u|
e^(3t/2740)=137-s
-e^(3t/2740)+137=s
The problem is that this just doesn't fit the initial condition of 0 kg sugar. And when I changed the constant so that it was (+/-)e^3t/2740) (either way) -1, it still said that I was wrong, even though the initial condition was met.

you're right that its wrong, that answer makes no sense, s will tend toward negative infinity and the initial condition is not met -> s(0) =-e^(0) + 137, which is not 0

i think the problem is with the negative, move it to the t side of the equation before integrating, then keep a constant when you integrate as well, then choose it to satisfy your initial condition