What this is for. Use this diagnostic to find which topics of Unit 7, Differential Equations, to revisit. There is one question per topic. These are original Marlbridge practice questions, not past exam questions. All contexts and data are invented. The questions are not calibrated, and your result is not a predicted score.
Rules. No calculator; about 30 minutes. Answer everything before opening any answer. k is a positive constant wherever it appears. The unit is shared by Calculus AB and Calculus BC. Questions 5 and 9 are BC only (Euler’s method and logistic models). If you take Calculus AB, skip them; the other seven questions are for both courses.
Question 1 (multiple choice · 7.1)
A cake is put into a freezer kept at −18 °C. In an invented model, the temperature T (°C) of the cake decreases at a rate proportional to the square of the difference between T and the freezer temperature. Time t is in minutes. Which differential equation fits this model?
- (A) dT/dt = −k(T − 18)²
- (B) dT/dt = k(T + 18)²
- (C) dT/dt = −k(T + 18)²
- (D) dT/dt = −k(T² + 18²)
Answer and explanation
Answer: (C). The difference between T and −18 is T − (−18) = T + 18. “The square of the difference” is (T + 18)². The temperature decreases and k > 0, so a minus sign is needed.
- (A) uses +18 °C as the freezer temperature. Watch the sign when the fixed value is negative.
- (B) has the right form but describes a cake that warms up.
- (D) squares each term separately. (T + 18)² is not T² + 18².
If you missed this: “Getting the sign right” in the Topic 7.1 study guide.
Question 2 (multiple choice · 7.2)
Which function is a solution of the differential equation x · dy/dx = 2y + x³ for x > 0?
- (A) y = x³ + 5x²
- (B) y = x³ + 5
- (C) y = x²
- (D) y = x³/3
Answer and explanation
Answer: (A). Substitute into both sides. For y = x³ + 5x², dy/dx = 3x² + 10x, so the left side is x(3x² + 10x) = 3x³ + 10x². The right side is 2(x³ + 5x²) + x³ = 3x³ + 10x². They match for every x > 0. In fact y = x³ + Cx² works for every constant C.
- (B) Adding a constant to x³ breaks the equation: the left side is 3x³ but the right side is 3x³ + 10.
- (C) is tempting because x · 2x = 2x² matches 2y, but the extra x³ on the right is left over.
- (D) gives x³ on the left and (5/3)x³ on the right.
If you missed this: “The verification method” in the Topic 7.2 study guide.
Question 3 (multiple choice · 7.3)
Which statement about the slope field for dy/dx = x² − y² is true?
- (A) The segments are horizontal at every point of the line y = x and of the line y = −x.
- (B) The segments are horizontal only at points of the line y = x.
- (C) Every segment in a vertical column has the same slope.
- (D) At every point of the y-axis other than the origin, the segments rise from left to right.
Answer and explanation
Answer: (A). The slope is 0 when x² = y², that is when y = x or y = −x. For example, the slope at (2, 2) and at (2, −2) is 4 − 4 = 0.
- (B) solves x² = y² as y = x only and forgets the negative root.
- (C) would be true only if the slope depended on x alone. Here it depends on y too: in the column x = 1 the slopes at y = 0, 1, 2 are 1, 0 and −3.
- (D) On the y-axis, x = 0, so the slope is −y², which is negative. The segments fall.
If you missed this: “Patterns that let you read a field quickly” in the Topic 7.3 study guide.
Question 4 (short answer · 7.4)
Consider the differential equation dy/dx = (y − 2)(y + 1). Let y = f(x) be the particular solution through the point (0, 1).
(a) Find all constant solutions of the differential equation. (b) Is f increasing or decreasing at x = 0? Is the graph of f concave up or concave down there? Justify both answers. (c) Describe the behaviour of f(x) as x → ∞. Explain your reasoning.
Worked answer
(a) A constant solution has dy/dx = 0 for every x. (y − 2)(y + 1) = 0 gives y = 2 and y = −1.
(b) At (0, 1), dy/dx = (1 − 2)(1 + 1) = −2 < 0, so f is decreasing. Differentiate implicitly: d²y/dx² = [(y + 1) + (y − 2)] · dy/dx = (2y − 1) · dy/dx. At (0, 1), d²y/dx² = (1)(−2) = −2 < 0, so the graph is concave down.
(c) For −1 < y < 2, one factor is negative and one is positive, so dy/dx < 0. The solution keeps falling, but it cannot cross the constant solution y = −1. So f(x) decreases towards −1: the limit of f(x) as x → ∞ is −1.
If you missed this: “Constant solutions” and Worked example 1 in the Topic 7.4 study guide.
Question 5 (multiple choice · 7.5) (BC only)
Let y = f(x) be the solution of dy/dx = x + 2y with f(1) = 0. Using Euler’s method with two steps of size 0.5, starting at x = 1, what is the approximation of f(2)?
- (A) 1
- (B) 1.25
- (C) 1.75
- (D) 4.5
Answer and explanation
Answer: (C). Step 1: the slope at (1, 0) is 1 + 0 = 1, so f(1.5) ≈ 0 + 0.5(1) = 0.5. Step 2: the slope at (1.5, 0.5) is 1.5 + 1 = 2.5, so f(2) ≈ 0.5 + 0.5(2.5) = 1.75.
- (A) is one step of size 1, the tangent line at x = 1.
- (B) works out the second slope at (1.5, 0), using the old y-value instead of the new one.
- (D) forgets to multiply each slope by the step size 0.5: 0 + 1 = 1, then 1 + 3.5 = 4.5.
If you missed this: “The procedure” in the Topic 7.5 study guide.
Question 6 (multiple choice · 7.6)
What is the general solution of dy/dx = eʸ cos x?
- (A) y = −ln(C − sin x)
- (B) y = −ln(sin x + C)
- (C) y = Ce^(sin x)
- (D) y = ln(sin x + C)
Answer and explanation
Answer: (A). Separate: e^(−y) dy = cos x dx. Antidifferentiate: −e^(−y) = sin x + C. So e^(−y) = C − sin x (the constant just changes sign), and −y = ln(C − sin x), giving y = −ln(C − sin x). Check: dy/dx = cos x/(C − sin x), and eʸ cos x = cos x/(C − sin x).
- (B) antidifferentiates e^(−y) as e^(−y), missing the factor −1 from the chain rule.
- (C) solves dy/dx = y cos x. Here the factor is eʸ, not y.
- (D) multiplies by eʸ instead of dividing by it, so the variables are not really separated.
If you missed this: “The method” and Worked example 3 in the Topic 7.6 study guide.
Question 7 (short answer · 7.7)
Let y = g(x) be the particular solution of dy/dx = 2x/cos y with g(0) = 0.
(a) Find g(x). (b) State the largest open interval containing x = 0 on which g is a solution. Explain. (c) Find g(1/√2) exactly.
Worked answer
(a) Separate: cos y dy = 2x dx. Antidifferentiate: sin y = x² + C. At (0, 0): sin 0 = 0 + C, so C = 0 and sin y = x². The solution passes through y = 0, so y stays between −π/2 and π/2, where y = arcsin(sin y). So g(x) = arcsin(x²).
(b) arcsin(x²) needs x² ≤ 1. At x = ±1, g = arcsin 1 = π/2 and cos(π/2) = 0, so the differential equation dy/dx = 2x/cos y is undefined there. The interval must be open and contain 0, so it is −1 < x < 1.
(c) g(1/√2) = arcsin(1/2) = π/6.
If you missed this: “Domain restrictions” and Worked example 3 in the Topic 7.7 study guide.
Question 8 (multiple choice · 7.8)
A quantity Q satisfies dQ/dt = kQ, where k is a constant (here k may be negative). Q(0) = 500 and Q(4) = 125. What is Q(6)?
- (A) 62.5
- (B) 15.625
- (C) 31.25
- (D) −62.5
Answer and explanation
Answer: (A). The solution is Q = 500e^(kt). From Q(4) = 125: e^(4k) = 1/4, so k = −(ln 2)/2 and Q halves every 2 time units. From t = 4 to t = 6 is one half-life, so Q(6) = 125/2 = 62.5. (Directly: Q(6) = 500e^(−3 ln 2) = 500/8.)
- (B) starts at Q(4) = 125 but then lets 6 time units pass instead of 2.
- (C) applies the factor 1/4 to the last 2 time units. The factor 1/4 belongs to 4 time units.
- (D) assumes Q falls by the same amount each unit (a straight line). An exponential model never changes sign.
If you missed this: Worked example 2 in the Topic 7.8 study guide.
Question 9 (multiple choice · 7.9) (BC only)
A population P of fish in a lake satisfies dP/dt = 0.3P − 0.0015P², where t is in years, and P(0) = 50. Which statement is true at the moment when P = 160?
- (A) P is increasing, and its graph is concave up.
- (B) P is increasing, and its graph is concave down.
- (C) P is decreasing, and its graph is concave down.
- (D) P is growing at its greatest rate.
Answer and explanation
Answer: (B). Factor: dP/dt = 0.0015P(200 − P), so the carrying capacity is 200. At P = 160, dP/dt = 48 − 38.4 = 9.6 > 0: increasing. The population grows fastest at half the carrying capacity, P = 100. It is past that point, so the rate is falling and the graph is concave down. Check: d²P/dt² = (0.3 − 0.003P) · dP/dt = (−0.18)(9.6) < 0.
- (A) would be true below P = 100, before the point of inflection.
- (C) Below the carrying capacity, P always increases.
- (D) The greatest rate happens at P = 100, not 160.
If you missed this: “Reading the equation without solving it” in the Topic 7.9 study guide.
Your next step
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 7.1 Modelling with differential equations | 1 | Guide 7.1 |
| 7.2 Verifying solutions | 2 | Guide 7.2 |
| 7.3 Sketching slope fields | 3 | Guide 7.3 |
| 7.4 Reasoning using slope fields | 4 | Guide 7.4 |
| 7.5 Euler’s method (BC only) | 5 | Guide 7.5 |
| 7.6 General solutions by separation | 6 | Guide 7.6 |
| 7.7 Particular solutions | 7 | Guide 7.7 |
| 7.8 Exponential models | 8 | Guide 7.8 |
| 7.9 Logistic models (BC only) | 9 | Guide 7.9 |
How to use your result
- Mark each topic secure, shaky (unsure or a slip) or gap (wrong).
- Fix gaps in Topics 7.2 and 7.6 first. Checking a solution by substitution (7.2) lets you catch your own mistakes everywhere else, and separation of variables (7.6) is the engine of Topics 7.7 and 7.8.
- Check your written answers to Questions 4 and 7 as well as the values. Did you give a reason (the sign of dy/dx, a constant solution that cannot be crossed, where the equation is undefined) and not just a conclusion?
- For a gap, read the guide, then do the topic’s practice set.
- Then try the Unit 7 mixed review, where each question combines topics.