Review: The Integrating Factor Method
To solve a first-order linear equation in standard form: \(y' + P(x)y = Q(x)\)
- Find the integrating factor: \(\rho = e^{\int P dx}\)
- Multiply the entire equation by \(\rho\), resulting in: \((\rho y)' = \rho Q\)
- Integrate both sides and solve for \(y\):
$$y = \frac{\int \rho Q dx + C}{\rho}$$
Example 1: Finding a General Solution
Find a general solution for: \((x^2 + 1)y' + 3xy = 6x\)
Step 1: Standard Form
Divide by \((x^2 + 1)\): $$y' + \frac{3x}{x^2 + 1}y = \frac{6x}{x^2 + 1}$$ Here, \(P = \frac{3x}{x^2 + 1}\) and \(Q = \frac{6x}{x^2 + 1}\).
Step 2: Integrating Factor
$$\rho = e^{\int \frac{3x}{x^2+1}dx} = e^{\frac{3}{2}\ln(x^2+1)} = (x^2+1)^{3/2}$$ (Note: We pick \(C=0\) for the integrating factor and omit absolute values as \(x^2+1 > 0\)).
Step 3: Solve
$$\int (\rho y)' dx = \int \rho Q dx \implies (x^2+1)^{3/2} y = \int (x^2+1)^{3/2} \frac{6x}{x^2+1} dx + C$$ $$(x^2+1)^{3/2} y = \int 6x(x^2+1)^{1/2} dx + C$$ $$(x^2+1)^{3/2} y = 2(x^2+1)^{3/2} + C$$ General Solution: \(y = 2 + C(x^2+1)^{-3/2}\)
Application: Mixture Problems
Modeling the amount of salt \(A(t)\) in a tank over time.
Scenario A: Constant Volume (\(r_{in} = r_{out}\))
A tank contains 400 gallons of water with 40 lbs of salt initially. Pure water with \(1/2\) lb/gal salt enters at 4 gal/min, and the well-mixed solution exits at 4 gal/min.
- \(r_{in} = 4\), \(c_{in} = 1/2\), \(r_{out} = 4\), \(A(0) = 40\), \(V_0 = 400\)
- ODE: \(A' = r_{in}c_{in} - r_{out}\frac{A(t)}{V(t)} \implies A' = 2 - \frac{4A}{400}\)
- Standard Form: \(A' + \frac{1}{100}A = 2\)
- Integrating Factor: \(\rho = e^{t/100}\)
- Particular Solution: \(A(t) = 200 - 160e^{-t/100}\)
Scenario B: Changing Volume (\(r_{out} > r_{in}\))
Suppose \(r_{out} = 6\) gal/min while \(r_{in}\) remains 4 gal/min. The tank will eventually empty.
Volume Function: \(V(t) = 400 + (4-6)t = 400 - 2t\)
The tank empties at \(t = 200\) mins. Range: \(0 \le t < 200\).
ODE: \(A' + \frac{6}{400-2t}A = 2 \implies A' + \frac{3}{200-t}A = 2\)
Integrating Factor:
\(\rho = e^{\int \frac{3}{200-t} dt} = e^{-3\ln(200-t)} = (200-t)^{-3}\)
Solution Process:
$$(200-t)^{-3}A = \int 2(200-t)^{-3} dt + C$$ $$(200-t)^{-3}A = (200-t)^{-2} + C$$ General Solution: \(A(t) = (200-t) + C(200-t)^3\)
Particular Solution:
Using \(A(0) = 40\): \(40 = 200 + C(200)^3 \implies C = \frac{-160}{200^3} = \frac{-1}{50,000}\)
\(A(t) = (200-t) - \frac{(200-t)^3}{50,000}\)
Finding Maximum Salt Amount
To find the maximum amount of salt before the tank empties, set \(A' = 0\):
$$A' = -1 + \frac{3}{50,000}(200-t)^2 = 0$$ $$(200-t)^2 = \frac{50,000}{3} \implies t = 200 - \sqrt{\frac{50,000}{3}} \approx 70.9 \text{ mins}$$
At \(t \approx 70.9\), the maximum salt amount is \(A \approx 86.07 \text{ lbs}\).