AP®MathsCollege BoardCalculus BCExam QuestionsUnit 7: Differential EquationsFirst-Order Differential Equations

First-Order Differential Equations (College Board AP® Calculus BC): Exam Questions

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First-Order Differential Equations

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1
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1 mark

A slope field for the differential equation fraction numerator d A over denominator d t end fraction equals fraction numerator 50 minus A over denominator 3 end fraction is shown below.

A slope field for the differential equation dA/dt = (50-A)/3, between 0 and 20 on the horizontal (t) axis, and between 0 and 70 on the vertical (A) axis

Sketch the solution curve through the point open parentheses 0 comma space 10 close parentheses.

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2
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1 mark

A portion of the slope field for the differential equation fraction numerator d y over denominator d x end fraction equals sin open parentheses pi over 3 x close parentheses square root of y plus 5 end root is given below.

A slope field for the differential equation dy/dx =sin((pi/3)x)sqrt(y-5), between -10/3 and 10/3 on the x-axis, and between -2 and 4 on the y-axis

Sketch the solution curve through the point open parentheses 1 comma space minus 1 close parentheses.

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3
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2 marks

Consider the differential equation fraction numerator d y over denominator d x end fraction equals fraction numerator open parentheses y minus 1 close parentheses squared over denominator x minus 2 end fraction.

On the axes provided, sketch a slope field for the given differential equation at the six points indicated.

Graph displaying points with coordinates (1, 1), (1, 2), (1, 3), (3, 1), (3, 2) and (3, 3). Both axes are labelled from 0 to 4.

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4
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2 marks

Consider the differential equation fraction numerator d y over denominator d x end fraction equals open parentheses y minus 3 close parentheses squared cos open parentheses pi x close parentheses.

On the axes provided, sketch a slope field for the given differential equation at the nine points indicated.

Graph displaying points with coordinates (-1, 1), (-1, 2), (-1, 3), (0, 1), (0, 2), (0, 3), (1, 1), (1, 2), and (1, 3). The x-axis is labelled from -1 to 1, and the y-axis is labelled from 0 to 3.

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5
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1 mark

Consider the differential equation fraction numerator d y over denominator d x end fraction equals fraction numerator x minus 3 over denominator y end fraction.

Describe all the points in the x y-plane, y not equal to 0, for which fraction numerator d y over denominator d x end fraction equals negative 1 half.

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6
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2 marks

Consider the differential equation fraction numerator d y over denominator d x end fraction equals fraction numerator 2 x y over denominator 5 end fraction. Let y equals f open parentheses x close parentheses be the particular solution to that differential equation with the initial condition f open parentheses 0 close parentheses equals 4. Use Euler's method with a step size of 0.1, starting at x equals 0, to approximate f open parentheses 0.2 close parentheses. Show the work that leads to your answer.

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7
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2 marks

A population is modeled by a function P that satisfies the logistic differential equation fraction numerator d P over denominator d t end fraction equals 6 P open parentheses 13 minus P close parentheses.

If P open parentheses 0 close parentheses equals 4, what is limit as t rightwards arrow infinity of P open parentheses t close parentheses?

If P open parentheses 0 close parentheses equals 15, what is limit as t rightwards arrow infinity of P open parentheses t close parentheses?

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8
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1 mark

Consider the differential equation \frac{\text{d} y}{\text{d} x} = \frac{1}{2} \text{sin} \; \left(\frac{\pi}{2} x\right) \sqrt{y + 7}. Let y = f(x) be the particular solution to the differential equation with the initial condition f(1) = 2. The function f is defined for all real numbers.

A portion of the slope field for the differential equation is given below. Sketch the solution curve through the point (1, 2).

Slope field on x–y axes from −3 to 3 and −2 to 4, with short line segments tilting; a single point is marked near (1.5, 2).

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9
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1 mark

A medication is administered to a patient. The amount, in milligrams, of the medication in the patient at time t hours is modeled by a function y = A(t) that satisfies the differential equation \dfrac{\text{d}y}{\text{d}t} = \dfrac{12 - y}{3}. At time t = 0 hours, there are 0 milligrams of the medication in the patient.

A portion of the slope field for the differential equation \dfrac{\text{d}y}{\text{d}t} = \dfrac{12-y}{3} is given below. Sketch the solution curve through the point (0, 0).

Slope field on a t–y graph, with short line segments slanting upward; above a dashed horizontal line slopes are shallower than those below it

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10
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1 mark

A bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed. The increasing function M models the temperature of the milk at time t, where M(t) is measured in degrees Celsius (°C) and t is the number of minutes since the bottle was placed in the pan. M satisfies the differential equation \displaystyle \frac{\text{d} M}{\text{d} t} = \frac{1}{4} (40 - M). At time t = 0, the temperature of the milk is 5°C. It can be shown that M(t) < 40 for all values of t.

A slope field for the differential equation \displaystyle \frac{\text{d} M}{\text{d} t} = \frac{1}{4} (40 - M) is shown. Sketch the solution curve through the point (0, 5).

Slope field for M(t) versus time t, showing solution curves bending towards the horizontal equilibrium line midway between M=5 and M=70 marks.

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11
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1 mark

The depth of seawater at a location can be modeled by the function H that satisfies the differential equation

\frac{\text{d}H}{\text{d}t} = \frac{1}{2}(H - 1)\cos\frac{t}{2}

where H(t) is measured in feet and t is measured in hours after noon (t = 0). It is known that H(0) = 4.

A portion of the slope field for the differential equation is provided below.

Slope field on t–y axes from t=0–5 and y≈0–11, with short line segments curving upward; a point is marked at (t=0, y=4).

Sketch the solution curve y = H(t) through the point (0, 4) on the slope field above.

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1
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4 marks

A cylindrical tank contains water and is being used to fill up a garden pond. The tank is standing upright on its circular base. The rate of change of the height h of the water in the tank with respect to time t is modeled by fraction numerator d h over denominator d t end fraction equals negative 1 third square root of h cubed end root, where h is measured in feet and t is measured in seconds.

At time t equals 0 seconds, the height of water in the tank is 4 feet. Use separation of variables to find an expression for h in terms of t.

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2
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1 mark

Consider the differential equation fraction numerator d y over denominator d x end fraction equals open parentheses y minus 2 close parentheses cubed sin open parentheses pi x close parentheses.

There is a horizontal line with equation y equals c that satisfies this differential equation. Find the value of c.

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3
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5 marks

Consider the differential equation fraction numerator d y over denominator d x end fraction equals fraction numerator x minus 6 over denominator y squared end fraction.

Find the particular solution y equals f open parentheses x close parentheses to the differential equation with the initial condition f open parentheses 0 close parentheses equals 3.

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4
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1 mark

Consider the differential equation fraction numerator d y over denominator d x end fraction equals x cubed open parentheses y plus 3 close parentheses.

Describe all the points in the x y-plane for which the slopes of tangent lines drawn on a slope field for that differential equation will be negative.

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5
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5 marks

Consider the differential equation fraction numerator d y over denominator d x end fraction equals fraction numerator y squared over denominator 2 x minus 3 end fraction.

Find the particular solution y equals f open parentheses x close parentheses to the differential equation with the initial condition f open parentheses 2 close parentheses equals 2.

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6
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2 marks

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0

pi over 4

pi over 2

fraction numerator 3 pi over denominator 4 end fraction

pi

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3

5

7

2

1

f is a twice-differentiable function for all values of x, with f open parentheses 0 close parentheses equals 0. The table above gives values of the derivative of f, f to the power of apostrophe, for selected values of x.

Use Euler's method, with two steps of equal size starting at x equals 0, to approximate f open parentheses pi close parentheses. Show the computations that lead to your answer.

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7a
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1 mark

Consider the logistic differential equation fraction numerator d y over denominator d t end fraction equals y open parentheses 4 minus y over 3 close parentheses for a population y. Let y equals f open parentheses t close parentheses be the particular solution to the differential equation for a given initial condition.

If f open parentheses 0 close parentheses equals 15, what is the range of f for t greater or equal than 0?

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7b
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1 mark

If, instead, f open parentheses 0 close parentheses equals 2, for what value of y is the population growing the fastest?

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8
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5 marks

Use separation of variables to find y = f(x), the particular solution to the differential equation \frac{\text{d} y}{\text{d} x} = \frac{1}{2} \text{sin}\left(\frac{\pi}{2} x\right) \sqrt{y + 7} with the initial condition f(1) = 2.

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9
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4 marks

Use separation of variables to find y = A(t), the particular solution to the differential equation \dfrac{\text{d}y}{\text{d}t} = \dfrac{12-y}{3} with initial condition A(0) = 0.

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10
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2 marks

Consider the differential equation

\frac{dy}{dx} = \frac{y^{2}}{x - 1}

On the axes provided, sketch a slope field for the given differential equation at the six points indicated.

Scatter plot on x–y axes showing points at (0,0), (0,1), (0,2), (2,0), (2,1) and (2,2); axes labelled x and y with ticks at 1, 2 and 3

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11
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3 marks

At time t = 0, a boiled potato is taken from a pot on a stove and left to cool in a kitchen. The internal temperature of the potato is 91\,^{\circ}\text{C} at time t = 0, and the internal temperature of the potato is greater than 27\,^{\circ}\text{C} for all times t > 0. The internal temperature of the potato at time t minutes is modeled by the function H satisfying the differential equation \dfrac{dH}{dt} = -\dfrac{1}{4}(H - 27), where H(t) is measured in degrees Celsius and H(0) = 91.

Write an equation for the line tangent to the graph of H at t = 0. Use this equation to approximate the internal temperature of the potato at time t = 3.

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12
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2 marks

Consider the differential equation \dfrac{dy}{dx} = \dfrac{1}{3}x(y-2)^2.

A slope field for the given differential equation is shown below. Sketch the solution curve that passes through the point (0, 2), and sketch the solution curve that passes through the point (1, 0).

Slope field on x–y axes with short dashed line segments curving outward from the y-axis, showing changing gradient above and below the origin.

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1
Sme Calculator
5 marks

Consider the differential equation fraction numerator d y over denominator d x end fraction equals fraction numerator 8 minus x cubed over denominator y end fraction.

Find the particular solution y equals f open parentheses x close parentheses to the differential equation such that the line y equals negative 4 is tangent to the graph of f.

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2
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5 marks

Scientists are studying a population of leopards in a nature reserve. The rate of change of the population is proportional to the difference between the current population and the maximum population that the reserve can support. If P open parentheses t close parentheses is the population at time t months after the start of the study, then

fraction numerator d P over denominator d t end fraction equals 1 over 20 open parentheses 90 minus P close parentheses

Use separation of variables to find the particular solution to the differential equation with initial condition P open parentheses 0 close parentheses equals 30. Give your answer as an expression for P in terms of t.

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3
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5 marks

Use separation of variables to find y equals f open parentheses x close parentheses, the particular solution to the differential equation fraction numerator d y over denominator d x end fraction equals cos open parentheses pi over 5 x close parentheses square root of y minus 2 end root with the initial condition f open parentheses 5 close parentheses equals 11.

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4
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2 marks

Consider the differential equation fraction numerator d y over denominator d x end fraction equals x minus 1 half y plus 1.

Find the values of the constants m and b, for which y equals m x plus b is a solution to the differential equation.

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5
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6 marks

Consider the differential equation fraction numerator d y over denominator d x end fraction equals open parentheses y plus 3 close parentheses cubed sin open parentheses pi x close parentheses.

Find the particular solution y equals f open parentheses x close parentheses to the differential equation with the initial condition f open parentheses 1 close parentheses equals 0

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6
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5 marks

Let y equals f open parentheses x close parentheses be the particular solution to the differential equation fraction numerator d y over denominator d x end fraction equals x y squared ln x with initial condition f open parentheses 1 close parentheses equals 2. Use Euler's method, with two equal steps starting at x equals 1, to approximate f open parentheses 3 over 2 close parentheses. Determine whether this approximation is an over- or underestimate, being sure to justify your answer.

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7a
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5 marks

A population is modeled by a function P that satisfies the logistic differential equation fraction numerator d P over denominator d t end fraction equals 4 P open parentheses 1 minus P over 16 close parentheses.

Use separation of variables to find P open parentheses t close parentheses if P open parentheses 0 close parentheses equals 5.

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7b
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2 marks

Determine the value of the time, t, at which the population, P, is changing the fastest.

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8
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4 marks

Use separation of variables to find an expression for M(t), the particular solution to the differential equation \displaystyle \frac{\text{d} M}{\text{d} t} = \frac{1}{4} (40 - M) with initial condition M(0) = 5.

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9
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5 marks

Use separation of variables to find y = H(t), the particular solution to the differential equation

\frac{\text{d}H}{\text{d}t} = \frac{1}{2}(H - 1)\text{cos}\frac{t}{2}

with initial condition H(0) = 4.

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10
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5 marks

For t < 10, an alternate model for the internal temperature of the potato at time t minutes is the function G satisfying the differential equation \dfrac{dG}{dt} = -(G - 27)^{2/3}, where G(t) is measured in degrees Celsius and G(0) = 91. Find an expression for G(t). Based on this model, what is the internal temperature of the potato at time t = 3?

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