▶️ Answer/Explanation
1. Separate variables and integrate: \[ \int y^{-2}dy = \int 2dx \] \[ -y^{-1} = 2x + C \]
2. Use initial condition \( y(1) = -1 \): \[ -(-1)^{-1} = 2(1) + C \] \[ C = -1 \]
3. Solve for \( y \) at \( x = 2 \): \[ -\frac{1}{y} = 2(2) – 1 \] \[ y = -\frac{1}{3} \]
✅ Correct answer: B) \(-\frac{1}{3}\)
▶️ Answer/Explanation
Solution
The solution to \(\frac{dy}{dt} = ky\) is \( y(t) = y(0) e^{kt} \).
Given the population doubles every 10 years, \( y(10) = 2y(0) \).
Substitute: \( 2y(0) = y(0) e^{10k} \).
Divide by \( y(0) \) (assuming \( y(0) \neq 0 \)): \( 2 = e^{10k} \).
Take natural log: \( \ln(2) = 10k \).
Solve: \( k = \frac{\ln(2)}{10} \approx \frac{0.693}{10} = 0.069 \).
✅ Answer: A
▶️ Answer/Explanation
Solution
The solution to \(\frac{dy}{dt} = ky\) is \( f(t) = f(0) e^{kt} \).
From the table, assume \( f(0) = 4 \) and \( f(2) = 12 \).
Then \( f(t) = 4 e^{kt} \). At \( t = 2 \): \( 4 e^{2k} = 12 \).
Solve: \( e^{2k} = 3 \), so \( 2k = \ln 3 \), \( k = \frac{\ln 3}{2} \).
Thus, \( f(t) = 4 e^{\left(\frac{\ln 3}{2}\right) t} = 4 e^{\frac{t}{2} \ln 3} \).
✅ Answer: A
▶️ Answer/Explanation
Solution
Separate variables: \(\frac{dy}{y} = \sec^2 x \, dx\).
Integrate: \(\ln |y| = \tan x + C\).
Exponentiate: \( y = A e^{\tan x} \).
With \( y(0) = 5 \): \( 5 = A e^0 \), so \( A = 5 \).
Thus, \( y = 5 e^{\tan x} \).
✅ Answer: C