Question

a tank initially contains 200 gal of brine in which 70 lb of salt are dissolved. a brine containing 4 lb/gal of salt runs into the tank at the rate of 3 gal/min. the mixture is kept uniform by stirring and flows out of the tank at the rate of 2 gal/min. let y represent the amount of salt at time t. complete parts a through e
d write down and solve the initial value problem describing the mixing process
\\(\frac{dy}{dt} = 4(200 + t) - \frac{29200000}{(200 + t)^2}\\), \\(y(0) = 70\\).
what is the solution to the initial value problem?
y =
e find the concentration of salt in the tank 24 min after the process starts
lb/gal
(type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Identify the differential equation form

The given differential equation is \(\frac{dy}{dt}+P(t)y = Q(t)\). First, rewrite \(\frac{dy}{dt}=4(200 + t)-\frac{29200000}{(200 + t)^{2}}\) in the standard linear form.

The volume of the tank at time \(t\) is \(V(t)=200+(3 - 2)t=200 + t\). The differential equation for the amount of salt \(y(t)\) in a mixing problem is \(\frac{dy}{dt}=\text{Rate in}-\text{Rate out}\). Here, \(\text{Rate in}=4\times3 = 12\) lb/min (since the concentration of the incoming brine is \(4\) lb/gal and the inflow rate is \(3\) gal/min), and \(\text{Rate out}=\frac{y}{200 + t}\times2\) lb/min.

The standard form of a first - order linear differential equation is \(\frac{dy}{dt}+\frac{2}{200 + t}y=12\). Here \(P(t)=\frac{2}{200 + t}\) and \(Q(t) = 12\).

Step2: Find the integrating factor

The integrating factor \(\mu(t)=e^{\int P(t)dt}\).

\(\int\frac{2}{200 + t}dt=2\ln(200 + t)=\ln((200 + t)^{2})\), so \(\mu(t)=(200 + t)^{2}\).

Step3: Multiply the differential equation by the integrating factor

Multiply \(\frac{dy}{dt}+\frac{2}{200 + t}y = 12\) by \((200 + t)^{2}\).

We get \((200 + t)^{2}\frac{dy}{dt}+2(200 + t)y=12(200 + t)^{2}\).

The left - hand side is the derivative of \((200 + t)^{2}y\) using the product rule \((uv)^\prime=u^\prime v+uv^\prime\) (where \(u=(200 + t)^{2}\), \(u^\prime = 2(200 + t)\) and \(v = y\)).

Step4: Integrate both sides

\(\int d((200 + t)^{2}y)=\int12(200 + t)^{2}dt\)

\((200 + t)^{2}y=12\times\frac{(200 + t)^{3}}{3}+C\)

\((200 + t)^{2}y = 4(200 + t)^{3}+C\)

\(y = 4(200 + t)+\frac{C}{(200 + t)^{2}}\)

Step5: Use the initial condition \(y(0)=70\)

Substitute \(t = 0\) and \(y=70\) into \(y = 4(200 + t)+\frac{C}{(200 + t)^{2}}\)

\(70=4\times200+\frac{C}{200^{2}}\)

\(70 = 800+\frac{C}{40000}\)

\(\frac{C}{40000}=70 - 800=-730\)

\(C=-29200000\)

So \(y = 4(200 + t)-\frac{29200000}{(200 + t)^{2}}\)

Step6: Find the concentration at \(t = 24\)

First, find \(y(24)\):

\(y(24)=4(200 + 24)-\frac{29200000}{(200 + 24)^{2}}\)

\(y(24)=4\times224-\frac{29200000}{224^{2}}\)

\(y(24)=896-\frac{29200000}{50176}\approx896 - 582.0\)

\(y(24)=314\)

The volume of the tank at \(t = 24\) is \(V(24)=200+24 = 224\) gal.

The concentration \(C=\frac{y(24)}{V(24)}=\frac{314}{224}\approx1.4\) lb/gal

Answer:

For part d: \(y = 4(200 + t)-\frac{29200000}{(200 + t)^{2}}\)
For part e: \(1.4\) lb/gal