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1.5 First-order Linear Eq. (continued)

\[ y' + py = Q \] \[ \text{integrating factor } I = e^{\int p dx} \] \[ \frac{d}{dx}(Iy) = IQ \quad \text{integrate to solve for } y. \]

Application: Mixing Problem

Example:

A tank initially contains 40 lbs of salt dissolved in 600 gals of water.

Water containing \( \frac{1}{2} \) lb of salt per gallon is poured in at the rate of 4 gal/min.

The well-mixed solution is let out of the tank at 4 gal/min.

Find an eq. describing the amount of salt in the tank.

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Let \( y(t) \) be the amount of salt in lb as function of time.

Problem tells us the rates of change:

\[ \frac{dy}{dt} = (\text{rate in}) - (\text{rate out}) \]

rate of change of salt

\[ = \underbrace{(\frac{1}{2} \text{ lb/gal})(4 \text{ gal/min})}_{\text{salt water going in}} - \underbrace{(\frac{y}{600} \frac{\text{lb}}{\text{gal}})(4 \text{ gal/min})}_{\text{salt coming out water}} \]

\[ \frac{dy}{dt} = 2 - \frac{y}{150} \] \[ y(0) = 40 \]

differential eq. for \( y \)

solve for \( y \)

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Solving the Linear Differential Equation

Rewrite the equation in standard form:

\[ y' + \underbrace{\frac{1}{150}}_{P} y = \underbrace{2}_{Q} \]

Integrating Factor

Calculate the integrating factor \( I \):

\[ I = e^{\int P dt} = e^{\int \frac{1}{150} dt} = e^{\frac{1}{150}t} \]

General Solution

Multiply the differential equation by the integrating factor:

\[ e^{\frac{1}{150}t} y' + \frac{1}{150} e^{\frac{1}{150}t} y = 2 e^{\frac{1}{150}t} \]

Apply the product rule in reverse:

\[ \frac{d}{dt} \left( e^{\frac{1}{150}t} y \right) = 2 e^{\frac{1}{150}t} \]

Integrate both sides:

\[ e^{\frac{1}{150}t} y = \int 2 e^{\frac{1}{150}t} dt = 300 e^{\frac{1}{150}t} + C \]

Solve for \( y \):

\[ y = 300 + C e^{-\frac{1}{150}t} \]

Initial Value Problem

Given the initial condition \( y(0) = 40 \):

\[ 40 = 300 + C \implies C = -260 \]

\[ y = 300 - 260 e^{-\frac{1}{150}t} \]

Asymptotic Behavior

As time approaches infinity:

\[ \lim_{t \to \infty} y = 300 \text{ lb} \]

\[ \frac{300 \text{ lb}}{600 \text{ gal}} = \frac{1}{2} \text{ lb/gal} \]

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Concentration Dynamics and Flow Rate Changes

Concentration of salt water in the tank approaches the concentration of the incoming flow.

After a long time, the tank is essentially a pipe because same flow rate in and out.

Figure: A coordinate graph showing y(t) increasing from 40 and leveling off at a horizontal asymptote of 300.

Changing Flow Rates

Now let's change the flow rate out to \( 6 \text{ gal/min} \).

In: same \( \to \frac{1}{2} \text{ lb/gal} \), \( 4 \text{ gal/min} \)
Out: \( \frac{\text{amount in tank}}{\text{volume in tank}} \text{ lb/gal} \), \( 6 \text{ gal/min} \)

Volume Change

Now volume changes:

Start with \( 600 \)
\( 4 \) in per min, \( 6 \) out per min

\[ V(t) = 600 - 2t \]

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Differential Equation Modeling: Rate In and Rate Out

\[ \frac{dy}{dt} = (\text{rate in}) - (\text{rate out}) \]

\[ = \left( \frac{1}{2} \right)(4) - \left( \frac{y}{600 - 2t} \right)(6) \]

Simplifying the expression:

\[ y' = 2 - \frac{3}{300 - t} y \]

Initial Value Problem

\[ y' + \frac{3}{300 - t} y = 2 \quad , \quad y(0) = 40 \]

Solve this for y

Calculating the Integrating Factor

\[ I = e^{\int p dt} = e^{\int \frac{3}{300 - t} dt} = e^{-3 \ln(300 - t)} = e^{\ln(300 - t)^{-3}} \]

\[ = (300 - t)^{-3} = \frac{1}{(300 - t)^3} \]

Multiplying the differential equation by the integrating factor:

\[ \frac{1}{(300 - t)^3} y' + \frac{3}{300 - t} \cdot \frac{1}{(300 - t)^3} y = \frac{2}{(300 - t)^3} \]

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Integration and General Solution

Applying the product rule in reverse:

\[ \frac{d}{dt} \left[ \frac{1}{(300 - t)^3} y \right] = \frac{2}{(300 - t)^3} \]

Integrating both sides:

\[ \frac{1}{(300 - t)^3} y = \frac{1}{(300 - t)^2} + C \]

Solving for \( y \):

\[ y = 300 - t + C(300 - t)^3 \]

Applying Initial Conditions

Given \( y(0) = 40 \):

\[ 40 = 300 + C(300)^3 \]

\[ C = -\frac{260}{27,000,000} = -\frac{13}{1,350,000} \]

Final Particular Solution

\[ y = 300 - t - \frac{13}{1,350,000} (300 - t)^3 \]

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Graphical Analysis of Function Maximum

The following graph illustrates a function with a local maximum. An annotation points to the peak of the curve, indicating the goal to find this maximum value.

Figure: A coordinate graph showing a blue parabolic curve with a peak near t=114 and y=124, crossing the t-axis at 300.

let's find this max

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Optimization and Differential Equation Verification

Finding the Maximum Value

Given the function for \( y \):

\[ y = 300 - t - \frac{13}{1350000} (300 - t)^3 \]

Find \( t \) at which \( y \) is maximized, and the maximum \( y \). First, we find the derivative \( y' \):

\[ y' = -1 + \frac{39}{1350000} (300 - t)^2 = 0 \]

Solve for \( t \):

\[ t \approx 114 \]

The maximum value of \( y \) occurs at \( t = 114 \):

\[ y_{\text{max}} \approx 124 \]

Verification with Differential Equation

Bring back the differential equation:

\[ y' = 2 - \frac{3}{300 - t} y \implies \text{max } y \text{ at } y' = 0 \]

Plug in \( y_{\text{max}} \) and \( t \approx 114 \) to verify \( y' = 0 \).

max reached when in matched by out

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Since \( t \approx 114 \) is when \( y \) is max

\( t < 114, \quad y' > 0 \rightarrow \text{saltier water in than out} \)
\( t > 114, \quad y' < 0 \rightarrow \text{saltier water out} \)

What if flow rate out is 2 gal/min

flow in as before

\( \text{volume} = 600 + 2t \)

\( \text{diff. eq:} \quad y' = 2 - \frac{y}{600 + 2t}(2) \quad \quad y(0) = 40 \)

\( \vdots \)

\( y = \frac{t^2 + 600t + 12000}{t + 300} \)

\( \lim_{t \to \infty} y = \infty \)

this is a tank w/ no hole or a small hole

as hole gets larger salt accumulation slows

when hole is at the size where flow in = flow out \( \rightarrow \) 1st example

when hole is larger \( \rightarrow \) 2nd example