# An initial value problem, and rate of change problem

• Sep 29th 2008, 07:02 PM
darch_angel
An initial value problem, and rate of change problem
Hi, I was having a bit of trouble with the following problems, any advice would be greatly appreciated.

" dy/dx +2y=f(x) where f(x)= {x, 0<=x<3 and 0, x>=3} with the known value y(0)=0 "

and also

" A tank contains 40 gallons of pure water. Brine with 3 lb of salt per gallon flows in at the
rate of 2 gal/min. The thoroughly stirred mixture then flows out at the rate of 3 gal/min.
a. Find the amount of salt in the tank when there are 20 gallons of brine in the tank.
b. When is the amount of salt in the tank greatest? Provide an exact (symbolic) answer
as well as an approximation to one decimal place. Use Calculus techniques. "
• Sep 29th 2008, 07:17 PM
Jhevon
Quote:

Originally Posted by darch_angel
Hi, I was having a bit of trouble with the following problems, any advice would be greatly appreciated.

" dy/dx +2y=f(x) where f(x)= {x, 0<=x<3 and 0, x>=3} with the known value y(0)=0 "

solve this in general using the integration factor method. then specify your f(x) as given with the initial data.

Quote:

" A tank contains 40 gallons of pure water. Brine with 3 lb of salt per gallon flows in at the rate of 2 gal/min. The thoroughly stirred mixture then flows out at the rate of 3 gal/min.
Let $Q$ be the amount of brine in the tank at time $t$ (we should really write $Q(t)$, but that gets annoying)

clearly $Q(0) = 0$

now, $Q' = \text{ rate of brine in } - \text{ rate of brine out}$

$\Rightarrow Q' = [ \text{concentration in} \times \text{amount flowing in}]$ $- [ \text{concentration out} \times \text{amount flowing out}]$

$\Rightarrow Q' = \frac {3~\text{lb}}{\text{gal}} \cdot \frac {2~\text{gal}}{\text{min}} - \frac Q{40 - t} \cdot \frac {3~\text{gal}}{\text{min}}$

note, the concentration out is amount (Q) per volume. the volume is decreasing by 1 gallon every minute, so after t minutes, the volume decreases by t gallons. since we started with 40, the volume after time t is (40 - t)

thus we have

$Q' = 6 - \frac {3Q}{40 - t}$

this is a first order ODE, you can solve for $Q$

Quote:

a. Find the amount of salt in the tank when there are 20 gallons of brine in the tank.
i suppose to mean 20 gallons of the mixture.

that is just asking for $Q(20)$, since we have 20 gallons left after 20 mins. you found $Q$ above

Quote:

b. When is the amount of salt in the tank greatest? Provide an exact (symbolic) answer as well as an approximation to one decimal place. Use Calculus techniques. "
you want to maximize $Q$. that is, solve for $Q' = 0$ and take the local maximum value, this will give you the time
• Sep 29th 2008, 08:06 PM
darch_angel
if a differential equation is in the form $y' + p(t) y = g(t)$, then the integrating factor is always $e^{\int p(t)~dt}$, that is, $\text{exp} \left( \int p(t)~dt \right)$