Ordinary Differential Equations — Practice Quiz

sin²x

Step 1: The equation is dy/dx + Py = Q where P = 2cot x. Step 2: Compute ∫P dx = ∫2cot x dx = 2log|sin x| = log(sin²x). Step 3: I.F. = e^(∫P dx) = e^(log sin²x) = sin²x. Step 4: The integrating factor is sin²x. With this I.F., the solution becomes y·sin²x = ∫3x²csc²x·sin²x dx = ∫3x² dx = x³ + C. Common mistake: Students confuse ∫cot x dx = log|sin x| with log|cos x| (which is ∫(-tan x)dx). Also, multiplying by 2 inside the log gives sin²x, not 2sin x.

50(log 3/log 2) years

Step 1: Population model: dx/dt = kx, giving x = x₀e^(kt). Step 2: Population doubles in 50 years: 2x₀ = x₀e^(50k), so e^(50k) = 2, giving k = (1/50)log 2. Step 3: For tripling: 3x₀ = x₀e^(kt₁), so e^(kt₁) = 3. Step 4: kt₁ = log 3, therefore t₁ = (log 3)/k = (log 3)/[(1/50)log 2] = 50(log 3/log 2). Final: The population triples in 50(log 3/log 2) years ≈ 79.25 years. Common mistake: Students use t₁ = 150 years (triple the 50 years), which is wrong because exponential growth is not linear.

tan(x - y + 1) = x + C

Step 1: Let z = x - y + 1. Then dz/dx = 1 - dy/dx, so dy/dx = 1 - dz/dx. Step 2: Substituting into the DE: 1 - dz/dx = sin²z. Therefore dz/dx = 1 - sin²z = cos²z. Step 3: Separate variables: dz/cos²z = dx, i.e., sec²z dz = dx. Step 4: Integrate both sides: tan z = x + C. Step 5: Substitute back z = x - y + 1: tan(x - y + 1) = x + C. Common mistake: Students make sign errors when computing dz/dx and forget that dy/dx = 1 - dz/dx (not dz/dx - 1).

2

Step 1: Identify the highest order derivative: it is d³y/dx³ (order = 3). Step 2: The term (d³y/dx³)^(2/3) has a fractional power. We must eliminate this fractional power. Step 3: Isolate the fractional term: (d³y/dx³)^(2/3) = 3(d²y/dx²) - 5(dy/dx) - 4. Cube both sides: (d³y/dx³)² = [3(d²y/dx²) - 5(dy/dx) - 4]³. Step 4: Now the highest order derivative d³y/dx³ appears with power 2. Therefore degree = 2. Common mistake: Taking degree = 2/3 directly from the original equation without cubing to remove the fractional power.