Lesson 6 (09/04/26)

Lesson 6 (09/04/26)

Linear first order equations: Examples, Applications

Warm Up:

1. A tank has \(4\text{ L}\) of water \(32\text{ gm}\) of salt is mixed What is the concentration of the salt solution

\[ \text{Concentration} = \frac{\text{Amount of Solute}}{\text{Volume of Liquid}} = \frac{32}{4}\text{ gm/L} = 8\text{ gm/L} \]

2. Suppose \(4\text{ L/min}\) of salt solution is being added, and there is a hole at bottom that leaks the solution at \(2\text{ L/min}\). What is the volume of solution after \(5\text{ mins}\).

\(\text{initial} = 4\text{ L}\)

\(\text{increasing } 2\text{ L/min}\)

\(5\text{ mins} \implies 4 + (2 \times 5) = \underline{\underline{14\text{ L}}}\)

Visual Description: Schematic diagram of a mixing tank. At the top left, an arrow points into the open container labeled 'Liquid going in'. Inside the container, liquid is represented by dashed horizontal lines. At the bottom center, an opening drains fluid downward, labeled 'Liquid going out'.

\(\text{initial volume} = V_0\)
\(\text{initial Amount of solute} = A_0\) \(\left.\vphantom{\begin{matrix} V_0 \\ A_0 \end{matrix}}\right\}\ \text{initial concentration} = \frac{A_0}{V_0}\)

pouring in solution of concentration \(c_1\) at a rate of \(r_1\)

\(\Rightarrow\) Concentration of solution
Amount of solute
Volume of solution
change with time

+ Have a leak \(\rightarrow\) leaks out solution at a rate of \(r_2\)

What is volume at any given time.

\[ \text{change in volume} = \text{Volume in} - \text{Volume out} \] \[ \Delta V = r_1 \Delta t - r_2 \Delta t \quad (\Delta t \text{ is small amount of time}) \] \[ \frac{dV}{dt} \approx \frac{\Delta V}{\Delta t} = \underbrace{r_1 - r_2}_{\text{constant}} \]

by integrating both sides.

\[ V(t) = (r_1 - r_2)t + V(0) \]

What is Amount of Solute at Any given time?

\[ \text{Change in Amount} = \text{Amount in} - \text{Amount out} \]

Pouring in solution of concentration \(c_1\) at rate \(r_1\)

In \(\Delta t\) time, \(\text{Volume in} = r_1 \Delta t\)

\[ \text{Concentration} = \frac{\text{Amount}}{\text{Vol.}} \]

\[ \text{Amount} = (\text{Concentration})(\text{volume}) \]

\[ \text{Amount in} = c_1 r_1 \Delta t \]

Similarly,

\[ \text{Amount out} = c_2 r_2 \Delta t \]

(\(c_2\) is not known.)

\[ \Delta A = \text{change in Amount} = \text{Amount in} - \text{Amount out} \]

\[ = c_1 r_1 \Delta t - c_2 r_2 \Delta t \]

we know \(c_1, r_1, r_2\), but \(c_2\) is concentration of liquid going out which changes with time

\[ \text{Concentration at Any given time} = \frac{\text{Amount at that time}}{\text{Volume at that time}} \]

\[ c_2 = \frac{A(t)}{V(t)} \]

\[ \Delta A = c_1 r_1 \Delta t - \frac{A(t)}{V(t)} r_2 \Delta t \]

\[ \frac{dA}{dt} = c_1 r_1 - \frac{A(t)}{V(t)} r_2 \Delta t \]

Equation to determine Amount at any given time

\[ \frac{dA}{dt} = c_1 r_1 - r_2 \frac{A(t)}{V(t)}, \quad A(0) = A_0 \]

\(\rightsquigarrow\) Linear equation

\[ A'(t) + \frac{r_2}{V(t)} A(t) = c_1 r_1 \]

A tank contains \(4\text{ L}\) of water, \(32\text{ gm}\) of chemical is mixed. A solution containing \(2\text{ gm/L}\) of chemical flows in at rate \(4\text{ L/min}\), well stirred mixture flows out at \(2\text{ L/min}\).

1. Determine the amount of chemical in tank at \(t = 10\text{ mins}\)

2. Find concentration of solution in tank at \(t = 10\text{ mins}\)

Visual Description: Schematic diagram of a mixing tank. At the top left, fluid flows into the tank labeled with inflow parameters c_1 and r_1. The tank contains liquid marked with dashed lines, labeled with initial contents 32 gm and volume 4 L. At the bottom, an opening indicates outflow with rate r_2.

Given initial values and rates:

\[ V_0 = 4\text{ L}, \quad A_0 = 32\text{ gm} \] \[ c_1 = 2\text{ gm/L}, \quad r_1 = 4\text{ L/min} \] \[ r_2 = 2\text{ L/min} \]

\[ \frac{dV}{dt} = r_1 - r_2, \quad V(0) = 4\text{ L} \] \[ \frac{dV}{dt} = 2, \quad V(0) = 4 \implies V(t) = 2t + 4\text{ L} \] \[ \frac{dA}{dt} = c_1 r_1 - c_2 r_2 = c_1 r_1 - \frac{A(t)}{V(t)} r_2, \quad A(0) = 32 \] \[ \frac{dA}{dt} = 8 - \frac{A(t)}{2t + 4} \times 2, \quad A(0) = 32 \] \[ \frac{dA}{dt} + \left( \frac{1}{t + 2} \right) A(t) = 8, \quad A(0) = 32 \]

Review: Solving 1st Order Linear ODEs

\[ y' + p(x)y = q(x) \]

1. \( \mu(x) = e^{\int p(x)\,dx} \rightarrow \mu'(x) = \mu(x)p(x) \)

2. Multiply \( \mu(x) \)

   \[ \mu(x)y' + \underbrace{\mu(x)p(x)}_{\mu'(x)}y = \mu(x)q(x) \]

3. Product Rules

   \[ (\mu(x)y(x))' = \mu(x)q(x) \]

4. Integrate

   \[ \mu(x)y(x) = \int \mu(x)q(x)\,dx + C \]

5. Divide with \( \mu(x) \) to find \( y(x) \).

going back to finding the Amount of the Solute:

\[ A' + \frac{1}{t+2}A = 8, \quad A(0) = 32 \]

1. Integrating factor:

   \[ \mu(x) = e^{\int \frac{1}{t+2} \, dt} = e^{\ln(t+2)} = t+2 \]

2. Multiply \( t+2 \) on both sides

   \[ (t+2) A' + A = 8(t+2) \]

3. product Rule:

   \[ ((t+2) A)' = 8(t+2) \]

4. integrate:

   \[ (t+2) A(t) = \int 8(t+2) \, dt = 8\left(\frac{t^2}{2} + 2t\right) + C \]

5. \[ A(t) = \frac{1}{(t+2)} \left[ 8\left(\frac{t^2}{2} + 2t\right) + C \right] \]

   plug in \( A(0) = 32 \) to find \( C \)

   \[ 32 = \frac{1}{2} [ 8(0) + C ] \implies C = 64 \]

   \[ A(t) = \frac{1}{t+2} \left[ 8\left(\frac{t^2}{2} + 2t\right) + 64 \right] \text{ gm} \]

\[ V(10) = 2(10) + 4 = 24\text{ L} \]

\[ A(10) = \frac{1}{12} \left[ 8[50+20] + 64 \right] = 52\text{ gm} \]

\[ \text{Concentration} = \frac{\text{Amount}}{\text{Volume}} \]

\[ c(t) = \frac{A(t)}{V(t)} = \frac{\frac{1}{t+2} \left[ 8\left(\frac{t^2}{2}+2t\right) + 64 \right]}{2t+4}\text{ gm/L}. \]

\[ c(10) = \frac{A(10)}{V(10)} = \frac{52}{24}\text{ gm/L}. \]

Q: Find integrating factor for the following:

\[ y' + p(x)y = q(x) \implies \mu(x) = e^{\int p(x)\,dx} \]

Example 1

\[ y' - \sin x \, y = \sin x \]

\[ \mu(x) = e^{\int -\sin x\,dx} = e^{\cos x} \]

Example 2

\[ y' + \frac{1}{x} y = 3x^2 \]

\[ \mu(x) = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x \]

Example 3

\[ xy' + (2x - 3)y = x^4 \]

\( e^{\int 2x - 3\,dx} = e^{x^2 - 3x} \)

\[ y' + \left[2 - \frac{3}{x}\right]y = x^3 \implies \mu(x) = e^{\int \left(2 - \frac{3}{x}\right) dx} = e^{2x - 3\ln x} = e^{2x - \ln x^3} = \frac{e^{2x}}{x^3} \]

\[ x y' + (2x - 3)y = x^4 \]

\[ y' + \left(2 - \frac{3}{x}\right)y = x^3 \]

Multiply with integrating factor

\[ \mu(x) = \frac{e^{2x}}{x^3} \]

\[ \frac{e^{2x}}{x^3} y' + \left(2 - \frac{3}{x}\right)\frac{e^{2x}}{x^3} y = \frac{e^{2x}}{x^3} \cdot x^3 \]

\[ \left(\frac{e^{2x}}{x^3} y\right)' = e^{2x} \]

integrate

\[ \frac{e^{2x}}{x^3} y = \frac{e^{2x}}{2} + C \]

divide \( \mu(x) \)

\[ y(x) = \frac{x^3}{e^{2x}} \left[\frac{e^{2x}}{2} + C\right] \]