Differential Equations Flashcards

What is the particular solution to dy/dx = 2xy with initial condition y(0) = 3?

Answer: y = 3e^(x²)

Separating gives dy/y = 2x dx, so ln|y| = x² + C and y = Ke^(x²). Using y(0) = 3 gives K = 3.

A slope field is drawn for dy/dx = x − y. At which of the following points is the slope segment horizontal?

Answer: (2, 2)

A horizontal segment has slope 0, so x − y = 0, which means x = y. Only (2, 2) satisfies this.

Which function is the solution to dy/dx = −x/y with initial condition y(0) = 2?

Answer: y = √(4 − x²)

Separating gives y dy = −x dx, so y²/2 = −x²/2 + C and y² + x² = 4. Since y(0) = 2 is positive, y = √(4 − x²).

If dy/dx = y² and y(0) = 1, what is y(1/2)?

Answer: 2

Separating gives dy/y² = dx, so −1/y = x + C. With y(0) = 1, C = −1, so y = 1/(1 − x) and y(1/2) = 2.

A function y = f(t) satisfies dy/dt = −3y with f(0) = 6. What is f(ln 2)?

Answer: 3/4

The solution is y = 6e^(−3t). At t = ln 2, e^(−3 ln 2) = 2^(−3) = 1/8, so f(ln 2) = 6/8 = 3/4.

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

Answer: 2

For 0 < y < 2 the slope y(2 − y) is positive, so y increases. It cannot cross the equilibrium solution y = 2, so it approaches 2.

What is the particular solution to dy/dx = e^(x − y) with y(0) = ln 2?

Answer: y = ln(e^x + 1)

Writing e^(x−y) = e^x/e^y gives e^y dy = e^x dx, so e^y = e^x + C. Using y(0) = ln 2 gives 2 = 1 + C, so C = 1 and y = ln(e^x + 1).

A slope field is drawn for dy/dx = sin x. Which statement about the field is true?

Answer: Slope segments are parallel along every vertical line.

The slope depends only on x, so at a fixed x-value every point has the same slope. Segments along a vertical line are therefore parallel.

What is the particular solution to dy/dx = x/y with initial condition y(0) = −2?

Answer: y = −√(x² + 4)

Separating gives y dy = x dx, so y² = x² + C. With y(0) = −2, C = 4, and the negative initial value requires y = −√(x² + 4).

Which function is the solution to dy/dx = y − 4 with y(0) = 1?

Answer: y = 4 − 3e^x

Separating gives dy/(y − 4) = dx, so y − 4 = Ce^x. Using y(0) = 1 gives C = −3, so y = 4 − 3e^x.