Linear Programming

y = x² + 5, y = x² - 3, y = x² + C

Step 1: Solve dy/dx = 2x by integrating both sides. Step 2: ∫dy = ∫2x dx gives y = x² + C. Step 3: Check each option by differentiating: d/dx(x² + 5) = 2x ✓, d/dx(x² - 3) = 2x ✓, d/dx(2x + 1) = 2 ✗, d/dx(x² + C) = 2x ✓, d/dx(x³) = 3x² ✗

True

Step 1: A linear differential equation in y has the form dy/dx + P(x)y = Q(x). Step 2: Rewrite the given equation: dy/dx + 2xy = x. Step 3: Compare with standard form: P(x) = 2x and Q(x) = x. Step 4: Since y appears only to the first power and dy/dx is to the first power, this is indeed linear in y.

e^(3x)

Step 1: For linear equation dy/dx + P(x)y = Q(x), integrating factor = e^∫P(x)dx. Step 2: Here P(x) = 3 (constant). Step 3: ∫P(x)dx = ∫3 dx = 3x. Step 4: Therefore, integrating factor = e^(3x).

(ln 2)/5

Step 1: Solve dP/dt = kP to get P = P₀e^(kt). Step 2: If population doubles in 5 years: P(5) = 2P₀. Step 3: Substitute: 2P₀ = P₀e^(5k). Step 4: Simplify: 2 = e^(5k). Step 5: Take natural log: ln 2 = 5k. Step 6: Solve for k: k = (ln 2)/5.