Let's solve **part C** step by step: Given: - After 1 week: \( y(1) = 0.90 \) - After 2 weeks: \( y(2) = 0.85 \) - Initial: \( y(0) = 1 \) - The formula: \( y(t) = a + (1-a)e^{-kt} \) We need to find **how much is remembered after 14 weeks** (\( t = 14 \)), i.e., \( y(14) \). --- ### **Step 1: Set up the equations** From the formula: \[ y(1) = a + (1-a)e^{-k \cdot 1} = 0.90 \] \[ y(2) = a + (1-a)e^{-k \cdot 2} = 0.85 \] Let \( x = e^{-k} \). Then \( e^{-2k} = (e^{-k})^2 = x^2 \). So: \[ 0.90 = a + (1-a)x \] \[ 0.85 = a + (1-a)x^2 \] --- ### **Step 2: Solve for \( a \) and \( x \)** Subtract the first equation from the second: \[ 0.85 - 0.90 = [a + (1-a)x^2] - [a + (1-a)x] \] \[ -0.05 = (1-a)(x^2 - x) \] So, \[ (1-a)(x^2 - x) = -0.05 \] \[ (1-a)(x^2 - x) = -0.05 \tag{1} \] Now, from the first equation: \[ 0.90 = a + (1-a)x \] \[ 0.90 - a = (1-a)x \] \[ x = \frac{0.90 - a}{1-a} \] Plug this into equation (1): \[ (1-a)\left[\left(\frac{0.90 - a}{1-a}\right)^2 - \frac{0.90 - a}{1-a}\right] = -0.05 \] \[ \left(\frac{0.90 - a)^2}{1-a} - (0.90 - a)\right) = -0.05 \] \[ \frac{(0.90-a)^2 - (0.90-a)(1-a)}{1-a} = -0.05 \] \[ \frac{(0.90-a)[0.90-a-(1-a)]}{1-a} = -0.05 \] \[ \frac{(0.90-a)[0.90-a-1+a]}{1-a} = -0.05 \] \[ \frac{(0.90-a)[-0.10]}{1-a} = -0.05 \] \[ -0.10(0.90-a) = -0.05(1-a) \] \[ -0.10 \times 0.90 + 0.10a = -0.05 + 0.05a \] \[ -0.09 + 0.10a = -0.05 + 0.05a \] \[ 0.10a - 0.05a = -0.05 + 0.09 \] \[ 0.05a = 0.04 \] \[ a = 0.8 \] --- ### **Step 3: Find \( k \) (or \( x \))** Recall: \[ x = \frac{0.90 - a}{1-a} = \frac{0.90 - 0.8}{1-0.8} = \frac{0.10}{0.2} = 0.5 \] But \( x = e^{-k} \implies e^{-k} = 0.5 \implies k = \ln 2 \approx 0.6931 \) --- ### **Step 4: Find \( y(14) \)** \[ y(t) = a + (1-a)e^{-kt} \] \[ y(14) = 0.8 + (1-0.8)e^{-k \cdot 14} \] \[ = 0.8 + 0.2 \cdot e^{-0.6931 \times 14} \] Calculate \( -0.6931 \times 14 = -9.7034 \): \[ e^{-9.7034} \approx 6.09 \times 10^{-5} \] So: \[ y(14) = 0.8 + 0.2 \times 6.09 \times 10^{-5} \] \[ = 0.8 + 0.00001218 \] \[ \approx 0.800012 \] **As a percent:** \[ 0.800012 \times 100 \approx 80.0\% \] --- ### **Final Answer:** **Percent remembered after 14 weeks:** \[ \boxed{80} \]

Let's proceed step-by-step including parts A and B, as requested. --- ### **Part A: Set up the differential equation and initial condition** **Given information:** - \( y(t) \): the fraction of material remembered after \( t \) weeks. - The model (Ebbinghaus) suggests: *"The rate at which a student forgets material is proportional to the difference between what is currently remembered and some positive constant \( a \)."* **Step 1: Express the rate of forgetting** The rate of change \( \frac{dy}{dt} \) is proportional to \( y(t) - a \), and since forgetting reduces \( y \), the proportionality constant is negative: \[ \frac{dy}{dt} = -k (y - a) \] where: - \( k > 0 \) is a proportionality constant, - \( a \) is a positive constant less than \( y(t) \) for all \( t \). **Step 2: Initial condition** At \( t = 0 \), the student just finished the course, so: \[ \boxed{ y(0) = 1 } \] --- ### **Part B: Solve the differential equation** The differential equation: \[ \frac{dy}{dt} = -k (y - a) \] is a first-order linear ODE. **Step 1: Rewrite** \[ \frac{dy}{dt} + k y = k a \] **Step 2: Find the integrating factor** \[ \mu(t) = e^{k t} \] **Step 3: Multiply through by integrating factor** \[ e^{k t} \frac{dy}{dt} + k e^{k t} y = k a e^{k t} \] which simplifies to: \[ \frac{d}{dt}\left( e^{k t} y \right) = k a e^{k t} \] **Step 4: Integrate both sides** \[ e^{k t} y = \int k a e^{k t} dt + C \] \[ = a e^{k t} + C \] **Step 5: Solve for \( y(t) \)** \[ y(t) = a + C e^{-k t} \] **Step 6: Use initial condition** At \( t = 0 \), \( y(0) = 1 \): \[ 1 = a + C e^{0} = a + C \] \[ C = 1 - a \] **Final solution:** \[ \boxed{ \boxed{ y(t) = a + (1 - a) e^{-k t} } } \] --- ### **Summary:** - **Part A:** Differential equation: \[ \frac{dy}{dt} = -k (y - a), \quad y(0) = 1 \] - **Part B:** Solution: \[ \boxed{ y(t) = a + (1 - a) e^{-k t} } \] --- ### **Part C (additional): Practical interpretation and calculation** Previously, we used the given data points: - \( y(1) = 0.90 \), - \( y(2) = 0.85 \), to determine \( a \) and \( k \), leading to the approximate solution: \[ y(t) \approx 0.8 + 0.2 e^{-k t} \] with \( k \approx 0.6931 \). **Estimate memory after 14 weeks:** \[ y(14) = 0.8 + 0.2 e^{-0.6931 \times 14} \approx 0.8 + 0.2 \times 6.09 \times 10^{-5} \approx 0.800012 \] which is roughly **80%**. --- **Final note:** - **Constants in the model:** - \( a \): the asymptotic level of recall; the amount of material that remains after a long time. - \( k \): the rate at which forgetting occurs. - **Interpretation:** - The model predicts that after about 14 weeks, the student will remember approximately **80%** of the original material.