Differential Equations and Integral Flashcards

If dy/dx = xy and y(0) = 2, what is y(2)?

Answer: 2e²

Separating gives dy/y = x dx, so ln|y| = x²/2 + C. Since y(0) = 2, y = 2e^(x²/2), and y(2) = 2e².

A population satisfies dP/dt = 0.2P(1 − P/500). For what value of P is the population growing most rapidly?

Answer: 250

The growth rate 0.2P(1 − P/500) is a downward-opening parabola in P. Its maximum is halfway between the zeros at P = 0 and P = 500, which is P = 250.

Euler's method with two steps of size 0.5 is used to approximate y(1) for the differential equation dy/dx = x + y with y(0) = 1. What is the approximation?

Answer: 2.5

At (0, 1) the slope is 1, so y(0.5) ≈ 1 + 0.5(1) = 1.5. At (0.5, 1.5) the slope is 2, so y(1) ≈ 1.5 + 0.5(2) = 2.5.

What is the area of the region enclosed by the graphs of y = x² and y = 2x?

Answer: 4/3

The curves meet where x² = 2x, at x = 0 and x = 2, and 2x is the upper curve on that interval. The area is ∫₀² (2x − x²) dx = 4 − 8/3 = 4/3.

The region bounded by y = x², the x-axis, and the line x = 2 is revolved about the y-axis. What is the volume of the solid generated?

Answer: 8π

Using washers in y, the outer radius is 2 and the inner radius is √y, for 0 ≤ y ≤ 4. The volume is π∫₀⁴ (4 − y) dy = π(16 − 8) = 8π.

What is the length of the curve y = x^(3/2) from x = 0 to x = 4?

Answer: (8/27)(10√10 − 1)

Here y′ = (3/2)x^(1/2), so the integrand is √(1 + 9x/4). Integrating gives (8/27)(1 + 9x/4)^(3/2), and evaluating from 0 to 4 gives (8/27)(10√10 − 1).

Consider dy/dx = y² with y(0) = 1. On which interval is the particular solution defined, given that the interval must contain x = 0?

Answer: x < 1

Separating gives −1/y = x + C, and y(0) = 1 gives C = −1. So y = 1/(1 − x), which has a vertical asymptote at x = 1, and the solution through x = 0 lives on x < 1.

A population satisfies dP/dt = 0.03P − 0.0006P² with P(0) = 10. What is lim (t→∞) P(t)?

Answer: 50

Factor the right side as 0.03P(1 − P/50), which is a logistic equation with carrying capacity 50. Since P(0) = 10 > 0, the population approaches 50.

What is the area of the region enclosed by the curves x = y² and x = y + 2?

Answer: 9/2

The curves intersect where y² = y + 2, at y = −1 and y = 2, and x = y + 2 is the right-hand curve. The area is ∫₋₁² (y + 2 − y²) dy = 9/2.

Consider dy/dx = y(2 − y) with y(0) = 1. What is lim (x→∞) y?

Answer: 2

The equilibria are y = 0 and y = 2. For 0 < y < 2 the slope y(2 − y) is positive, so a solution starting at y = 1 increases toward y = 2.