AP Calc AB Differential Equations — Questions and Answers
Question 1: What is the particular solution to dy/dx = 2xy with initial condition y(0) = 3?
- y = 3e^(x²) (Correct answer)
- y = 3 + x²
- y = e^(x²) + 2
- y = 3e^(2x)
Correct 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.
The equation is separable: divide by y and multiply by dx to get (1/y)dy = 2x dx. Integrating gives ln|y| = x² + C. Exponentiating gives y = Ke^(x²), where K = ±e^C. The initial condition y(0) = 3 gives 3 = Ke^0, so K = 3. The particular solution is y = 3e^(x²).
Question 2: A slope field is drawn for dy/dx = x − y. At which of the following points is the slope segment horizontal?
- (2, 0)
- (0, 2)
- (2, 2) (Correct answer)
- (1, −1)
Correct answer: (2, 2)
A horizontal segment has slope 0, so x − y = 0, which means x = y. Only (2, 2) satisfies this.
In a slope field, the segment at (x, y) has slope given by the right-hand side of the differential equation. Here the slope is x − y. At (2, 0) the slope is 2, at (0, 2) it is −2, and at (1, −1) it is 2. Only (2, 2) gives slope 0. All points on the line y = x have horizontal segments.
Question 3: Which function is the solution to dy/dx = −x/y with initial condition y(0) = 2?
- y = −√(4 − x²)
- y = √(4 − x²) (Correct answer)
- y = √(4 + x²)
- y = 2e^(−x²/2)
Correct 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).
Multiply to separate: y dy = −x dx. Integrating gives y²/2 = −x²/2 + C, or x² + y² = 2C. The condition y(0) = 2 gives 4 = 2C, so x² + y² = 4. Solving for y gives y = ±√(4 − x²). Because the solution passes through (0, 2) where y is positive, the upper semicircle y = √(4 − x²) is the particular solution.
Question 4: If dy/dx = y² and y(0) = 1, what is y(1/2)?
- 1/2
- 1
- 3/2
- 2 (Correct answer)
Correct 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.
Separate to get y^(−2) dy = dx. Integrating gives −1/y = x + C. Using y(0) = 1 gives −1 = C. So −1/y = x − 1, which gives y = 1/(1 − x). This solution is valid for x < 1. Evaluating at x = 1/2 gives y = 1/(1/2) = 2.
Question 5: A function y = f(t) satisfies dy/dt = −3y with f(0) = 6. What is f(ln 2)?
- 3/2
- 3/8
- 3/4 (Correct answer)
- 3
Correct 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.
The equation dy/dt = ky has general solution y = Ce^(kt). Here k = −3 and the initial value is 6, so y = 6e^(−3t). Evaluating at t = ln 2 gives 6e^(−3 ln 2) = 6e^(ln(2^(−3))) = 6·(1/8) = 3/4. Remember that e^(a ln b) = b^a.
Question 6: Consider dy/dx = y(2 − y) with initial condition y(0) = 1. What is lim as x → ∞ of y?
- 0
- 1
- does not exist (y is unbounded)
- 2 (Correct answer)
Correct 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.
The slope is zero when y = 0 or y = 2, giving horizontal equilibrium solutions. For 0 < y < 2, both y and 2 − y are positive, so dy/dx > 0 and the solution increases. Solutions to this kind of equation cannot cross an equilibrium line, so the solution through (0, 1) stays below y = 2 and increases toward it. The limit is therefore 2.
Question 7: What is the particular solution to dy/dx = e^(x − y) with y(0) = ln 2?
- y = x + ln 2
- y = ln(e^x + 1) (Correct answer)
- y = ln(e^x) + 2
- y = e^x + ln 2 − 1
Correct 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).
Since e^(x−y) = e^x · e^(−y), multiply both sides by e^y to get e^y dy = e^x dx. Integrating gives e^y = e^x + C. At x = 0, y = ln 2, so e^(ln 2) = 2 = 1 + C, giving C = 1. Taking the natural log gives y = ln(e^x + 1).
Question 8: A slope field is drawn for dy/dx = sin x. Which statement about the field is true?
- Slope segments are parallel along every vertical line. (Correct answer)
- Slope segments are parallel along every horizontal line.
- All slope segments have the same slope everywhere.
- Every slope segment is horizontal along the y-axis only.
Correct 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.
When dy/dx is a function of x alone, the slope at (x, y) does not change as y changes. Moving up or down a vertical line keeps x fixed, so all segments there are parallel. Moving horizontally changes x, so the slopes vary. Solutions to such an equation are vertical translates of one another, since they differ by a constant, as in y = −cos x + C.
Question 9: What is the particular solution to dy/dx = x/y with initial condition y(0) = −2?
- y = √(x² + 4)
- y = √(x² − 4)
- y = −√(x² − 4)
- y = −√(x² + 4) (Correct answer)
Correct 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).
Separate to get y dy = x dx and integrate: y²/2 = x²/2 + C, so y² = x² + K. The condition y(0) = −2 gives 4 = K. Then y² = x² + 4, so y = ±√(x² + 4). The solution passes through a point with negative y and y never equals 0 since x² + 4 > 0, so it stays negative. The particular solution is y = −√(x² + 4).
Question 10: Which function is the solution to dy/dx = y − 4 with y(0) = 1?
- y = 4 − 3e^x (Correct answer)
- y = 4 + 3e^x
- y = 1 + 4e^x
- y = −3e^x + 1
Correct 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.
Separate to get 1/(y − 4) dy = dx. Integrating gives ln|y − 4| = x + C, so y − 4 = Ke^x. At x = 0, y = 1 so 1 − 4 = K, giving K = −3. Thus y = 4 − 3e^x. As a check, dy/dx = −3e^x and y − 4 = −3e^x, so the equation is satisfied.
