Differential Equations Quiz: Common Setup Pitfalls

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Common Setup PitfallsQuestion 1 of 12

The population of fish, $P(t)$ , in a lake is modeled by a logistic equation. The carrying capacity of the lake is 10,000 fish, and the intrinsic annual growth rate is $r=0.2$ . The fish are harvested at a constant rate of 300 fish per year. The initial population is 2,000 fish. Which initial value problem describes the fish population?

dP/dt = 0.2P(1 - P/10000) + 300, \quad P(0)=2000

dP/dt = 0.2P(1 - P/2000) - 300, \quad P(0)=10000

dP/dt = 0.2(1 - P/10000) - 300, \quad P(0)=2000

dP/dt = 0.2P(1 - P/10000) - 300, \quad P(0)=2000

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Differential Equations Quiz

Differential Equations Quiz: Common Setup Pitfalls

Practice Common Setup Pitfalls in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Common Setup Pitfalls, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$dP/dt = 0.2P(1 - P/10000) + 300, \quad P(0)=2000$
$dP/dt = 0.2P(1 - P/2000) - 300, \quad P(0)=10000$
$dP/dt = 0.2(1 - P/10000) - 300, \quad P(0)=2000$
$dP/dt = 0.2P(1 - P/10000) - 300, \quad P(0)=2000$ (correct answer)

Explanation: The standard logistic model with intrinsic growth rate

r

and carrying capacity

M

dP/dt = rP(1 - P/M)

. Here,

r=0.2

and

M=10000

. Harvesting removes fish from the population, so the harvesting rate is subtracted from the growth rate. The harvesting is at a constant rate of 300 fish per year. Therefore, the differential equation is

dP/dt = 0.2P(1 - P/10000) - 300

. The initial population is given as 2,000, so

P(0)=2000

Question 2

A 2 kg mass is attached to a spring, stretching it to its equilibrium position. The system is governed by the equation $2y'' + \gamma y' + ky = F(t)$ , where $y(t)$ is the displacement from equilibrium in meters, and the downward direction is defined as positive. The mass is first pulled down 0.1 meters from its equilibrium point and then, at time $t=0$ , it is pushed upwards with a speed of 0.5 m/s. Which set of initial conditions correctly describes this situation?

$y(0) = 0.1, \quad y'(0) = 0.5$
$y(0) = -0.1, \quad y'(0) = -0.5$
$y(0) = 0.1, \quad y'(0) = -0.5$ (correct answer)
$y(0) = -0.1, \quad y'(0) = 0.5$

Explanation: The displacement

y(t)

is measured from the equilibrium position, with the downward direction as positive. 'Pulled down 0.1 meters' means the initial position is 0.1 m in the positive direction. Therefore,

y(0) = 0.1

. The initial velocity

y'(0)

is the speed and direction at

t=0

. 'Pushed upwards with a speed of 0.5 m/s' means the initial velocity is in the negative direction. Therefore,

y'(0) = -0.5

. The correct initial conditions are

y(0) = 0.1

and

y'(0) = -0.5

Question 3

A 500-gram object is dropped from rest. The force due to air resistance is proportional to the object's velocity, with a proportionality constant of 0.2 N·s/m. Let $v(t)$ be the velocity of the object in m/s at time $t$ in seconds, and assume the downward direction is positive. The acceleration due to gravity is $g = 9.8$ m/s $^2$ . Which initial value problem models the velocity of the object?

$500 dv/dt = 500(9.8) - 0.2v, \quad v(0)=0$
$0.5 dv/dt = 0.5(9.8) + 0.2v, \quad v(0)=0$
$0.5 dv/dt = -0.5(9.8) - 0.2v, \quad v(0)=0$
$0.5 dv/dt = 0.5(9.8) - 0.2v, \quad v(0)=0$ (correct answer)

Explanation: By Newton's second law,

m(dv/dt) = F_{\text{net}} = F_{\text{gravity}} + F_{\text{resistance}}

. A critical first step is to ensure consistent units. The mass is given as 500 grams, which must be converted to kilograms to be compatible with Newtons and m/s

^2

m = 0.5

kg. The force of gravity acts in the positive (downward) direction, so

F_{\text{gravity}} = mg = 0.5(9.8)

N. Air resistance opposes motion, so it acts in the negative (upward) direction:

F_{\text{resistance}} = -0.2v

. The object is 'dropped from rest', so the initial velocity is

v(0)=0

. Combining these gives the initial value problem:

0.5 dv/dt = 0.5(9.8) - 0.2v

, with

v(0)=0

Question 4

Consider an RLC circuit with a resistor $R = 10 \ \Omega$ , an inductor $L = 0.5$ H, a capacitor $C = 0.01$ F, and a constant voltage source $E = 12$ V. At time $t=0$ , the switch is closed. The initial charge on the capacitor is 2 C, and there is no initial current flowing through the circuit. Which of the following is the correct initial value problem for the charge on the capacitor, $q(t)$ ?

$0.5q'' + 10q' + 100q = 12, \quad q(0)=2, \ q'(0)=0$ (correct answer)
$0.5q'' + 10q' + 100q = 12, \quad q(0)=0, \ q'(0)=2$
$0.5q'' + 10q' + 0.01q = 12, \quad q(0)=2, \ q'(0)=0$
$0.5q'' + 10q' + 100q = 0, \quad q(0)=2, \ q'(0)=0$

Explanation: The differential equation for an RLC circuit is given by Kirchhoff's voltage law:

Lq'' + Rq' + \frac{1}{C}q = E(t)

. Substituting the given values:

L=0.5

R=10

C=0.01

(so

1/C = 100

), and

E(t)=12

. This gives the equation

0.5q'' + 10q' + 100q = 12

. The initial conditions are derived from the problem statement. 'The initial charge on the capacitor is 2 C' means

q(0)=2

. 'There is no initial current' means

I(0)=0

. Since current

I

is the rate of change of charge,

I(t) = q'(t)

, so

q'(0)=0

. This yields the IVP in option A.

Question 5

A population of animals, $P(t)$ , experiences a birth rate proportional to its current size, with a positive proportionality constant $\beta$ . The population also experiences a death rate due to predation that is proportional to its size, with a positive proportionality constant $\delta$ . Which differential equation correctly models the net rate of change of the population?

$dP/dt = (\beta + \delta)P$
$dP/dt = (\beta - \delta)P$ (correct answer)
$dP/dt = \beta P - \delta$
$dP/dt = (\delta - \beta)P$

Explanation: When modeling population dynamics, you need to carefully account for all factors affecting the rate of change and determine whether each factor increases or decreases the population. Let's break down this problem systematically. The population experiences two effects:

Birth rate: proportional to current population size with constant $\beta$ , contributing $+\beta P$ to the rate of change
Death rate from predation: proportional to current population size with constant $\delta$ , contributing $-\delta P$ to the rate of change

The net rate of change is the sum of these effects:

\frac{dP}{dt} = \beta P - \delta P = (\beta - \delta)P

This confirms that choice B is correct. Let's examine why the other options are wrong: Choice A gives

(\beta + \delta)P

, which incorrectly adds both the birth and death rates as positive contributions. This would mean predation somehow increases the population. Choice C gives

\beta P - \delta

, which correctly makes births proportional to population size but treats deaths as a constant rate regardless of population size. This contradicts the problem statement that deaths are also proportional to

P

. Choice D gives

(\delta - \beta)P

, which reverses the signs, making deaths contribute positively and births negatively to population growth. Key strategy: In population models, always identify whether each factor increases (+) or decreases (-) the population, then check if the rate depends on population size (proportional) or is constant. Birth rates increase population; death rates decrease it.

Question 6

A 200-gallon tank initially contains pure water. For the first 5 minutes, a faulty pump introduces pure water into the tank at 4 gal/min. After $t=5$ minutes, the pump is fixed and begins introducing a salt solution with a concentration of 3 pounds/gallon at the same rate of 4 gal/min. The mixture is kept uniform and is drained at 4 gal/min. Let $S(t)$ be the amount of salt (in pounds) in the tank. Which initial value problem models the amount of salt for $t \ge 5$ ?

$dS/dt = 12 - S/50, \quad S(0) = 0$
$dS/dt = 12 - S/50, \quad S(5) = 0$ (correct answer)
$dS/dt = 3 - S/200, \quad S(5) = 0$
$dS/dt = 12 - S/50, \quad S(5) = 60$

Explanation: When modeling mixing problems with differential equations, you need to set up the rate equation

\frac{dS}{dt} = \text{rate in} - \text{rate out}

and carefully identify the initial condition at the specific time period you're analyzing. For

t \geq 5

, salt enters at 4 gal/min with concentration 3 lb/gal, so the rate in is

4 \times 3 = 12

lb/min. Since the tank volume stays constant at 200 gallons (inflow equals outflow), the rate out is

\frac{S(t)}{200} \times 4 = \frac{S}{50}

lb/min. This gives us

\frac{dS}{dt} = 12 - \frac{S}{50}

. The key insight is determining the initial condition. At

t = 5

, the tank has contained only pure water for 5 minutes, so

S(5) = 0

pounds of salt. This is when we start tracking salt accumulation. Answer B correctly captures both the differential equation and initial condition:

\frac{dS}{dt} = 12 - \frac{S}{50}, S(5) = 0

. Answer A uses the wrong initial condition

S(0) = 0

, but we're modeling the system starting at

t = 5

, not

t = 0

. Answer C incorrectly calculates the rate out as

\frac{S}{200}

instead of

\frac{4S}{200} = \frac{S}{50}

, missing the outflow rate multiplication. Answer D uses

S(5) = 60

, but there's no salt in the tank at

t = 5

since only pure water entered initially. Study tip: In mixing problems, always multiply concentration by flow rate for both inflow and outflow terms, and set your initial condition at the time when the scenario you're modeling begins.

Question 7

A reversible chemical reaction is described by $A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B$ . The forward reaction ( $A \to B$ ) occurs at a rate proportional to the concentration of A, with rate constant $k_1$ . The reverse reaction ( $B \to A$ ) occurs at a rate proportional to the concentration of B, with rate constant $k_2$ . Let $x(t)$ be the concentration of substance B. If the initial concentration of A is $a_0$ and the initial concentration of B is 0, which differential equation describes the rate of change of $x(t)$ ?

$dx/dt = k_1(a_0 - x) + k_2 x$
$dx/dt = k_2 x - k_1(a_0 - x)$
$dx/dt = k_1(a_0 - x) - k_2 x$ (correct answer)
$dx/dt = k_1 a_0 - k_2 x$

Explanation: The rate of change of the concentration of B,

dx/dt

, is equal to the rate of its formation minus the rate of its consumption. B is formed by the forward reaction

A \to B

, so the rate of formation is

k_1[A]

. B is consumed by the reverse reaction

B \to A

, so the rate of consumption is

k_2[B]

. This gives

dx/dt = k_1[A] - k_2[B]

. We are given

[B] = x(t)

. By stoichiometry, the concentration of A at time

t

is its initial concentration minus the amount that has reacted to form B, so

[A] = a_0 - x(t)

. Substituting these into the rate equation gives

dx/dt = k_1(a_0 - x) - k_2 x

Question 8

A cold metal probe, initially at a temperature of $5^\circ\text{C}$ , is placed into a large water bath maintained at a constant temperature of $40^\circ\text{C}$ . According to Newton's law of cooling (or warming), the rate of change of the probe's temperature $T(t)$ is proportional to the difference between the ambient temperature of the bath and the probe's temperature. Which of the following initial value problems correctly models the temperature of the probe $T(t)$ in degrees Celsius, where $t$ is time in minutes and the proportionality constant $k$ is positive?

$dT/dt = k(T - 40)$ , $T(0) = 5$
$dT/dt = -k(40 - T)$ , $T(0) = 5$
$dT/dt = k(40 - T)$ , $T(0) = 40$
$dT/dt = k(40 - T)$ , $T(0) = 5$ (correct answer)

Explanation: The rate of change,

dT/dt

, is proportional to the difference between the ambient temperature (

40^\circ\text{C}

) and the object's temperature (

T

). So,

dT/dt = k(40 - T)

for some constant

k

. Since the probe is warming up (

T(0)=5

and the bath is at

40

), the temperature

T

must increase, meaning

dT/dt

must be positive when

T < 40

. The term

(40 - T)

is positive for

T < 40

, so the constant

k

must be positive. The initial condition is explicitly given as

T(0) = 5

. Thus, the correct model is

dT/dt = k(40 - T)

with

T(0) = 5

Question 9

A vertical cylindrical tank with a radius of 2 meters is initially full of water. Water drains from a hole in the bottom. By Torricelli's law, the rate of change of the volume of water, $V$ , is $dV/dt = -k\sqrt{h}$ , where $h$ is the height of the water and $k$ is a positive constant. Which differential equation correctly models the height $h(t)$ of the water in the tank?

$dh/dt = -k\sqrt{h}$
$dh/dt = -\frac{k}{4\pi}\sqrt{h}$ (correct answer)
$dh/dt = \frac{k}{4\pi}\sqrt{h}$
$dh/dt = -4\pi k\sqrt{h}$

Explanation: We are given the rate of change of volume,

dV/dt = -k\sqrt{h}

. We need to find the rate of change of height,

dh/dt

. For a cylinder, the volume is

V = A \cdot h

, where

A

is the constant cross-sectional area. Here, the radius is

r=2

m, so the area is

A = \pi r^2 = 4\pi

^2

. Thus,

V = 4\pi h

. We can relate the rates using the chain rule:

dV/dt = \frac{dV}{dh} \frac{dh}{dt}

. From

V = 4\pi h

, we have

dV/dh = 4\pi

. Substituting this into the chain rule expression gives

dV/dt = 4\pi \frac{dh}{dt}

. Now we equate the two expressions for

dV/dt

4\pi \frac{dh}{dt} = -k\sqrt{h}

. Solving for

dh/dt

yields

dh/dt = -\frac{k}{4\pi}\sqrt{h}

Question 10

A student takes out a $30,000 loan with an annual interest rate of 6%, compounded continuously. The student makes continuous payments on the loan at a constant rate of $K $dollars per year. Let$ P(t) $be the outstanding loan balance in dollars after$ t$ years. Which initial value problem correctly models the loan balance?

$dP/dt = 0.06P + K, \quad P(0)=30000$
$dP/dt = K - 0.06P, \quad P(0)=30000$
$dP/dt = 0.06P - K, \quad P(0)=30000$ (correct answer)
$dP/dt = 0.005P - K, \quad P(0)=30000$

Explanation: The rate of change of the loan balance,

dP/dt

, is the rate at which the balance grows due to interest minus the rate at which it is reduced by payments. The interest accrues at a rate of

rP

, where

r=0.06

, so this term is

0.06P

. The payments reduce the loan, so they are subtracted at a rate of

K

per year. This leads to the differential equation

dP/dt = 0.06P - K

. The initial loan amount is $30,000, so the initial condition is $P(0) = 30000$.

Question 11

A substance decays radioactively with a half-life of 3 hours. At the same time, the substance is being created and added to a container at a constant rate of 12 grams per hour. Let $A(t)$ be the amount of the substance in grams at time $t$ in hours. Which differential equation models this process?

$dA/dt = 12 - \frac{\ln(2)}{3}A$ (correct answer)
$dA/dt = 12 + \frac{\ln(2)}{3}A$
$dA/dt = 12 - 3A$
$dA/dt = -12 - \frac{\ln(2)}{3}A$

Explanation: The rate of change of the amount,

dA/dt

, is the rate of creation minus the rate of decay. The rate of creation is given as 12 g/hr. Radioactive decay occurs at a rate proportional to the amount present, so the rate of decay is

kA

for some decay constant

k>0

. Thus, the model is

dA/dt = 12 - kA

. The decay constant

k

is related to the half-life

T_{1/2}

by the formula

T_{1/2} = \ln(2)/k

. Given

T_{1/2} = 3

hours, we can solve for

k

3 = \ln(2)/k \implies k = \ln(2)/3

. Substituting this value of

k

into the model gives

dA/dt = 12 - \frac{\ln(2)}{3}A

Question 12

A 1000-liter tank initially contains 400 liters of a brine solution with 10 kg of dissolved salt. A brine solution with a concentration of 0.05 kg/L is pumped into the tank at a rate of 10 L/min. The well-mixed solution is pumped out at a slower rate of 5 L/min. Let $A(t)$ represent the amount of salt in the tank (in kg) at time $t$ (in minutes). Which differential equation correctly models this system?

$dA/dt = 0.5 - \frac{5A}{400}$
$dA/dt = 0.5 - \frac{5A}{1000}$
$dA/dt = 0.5 - \frac{5A}{400 + 5t}$ (correct answer)
$dA/dt = 0.5 + \frac{5A}{400 + 5t}$

Explanation: The rate of change of salt is

dA/dt = (\text{rate in}) - (\text{rate out})

. The rate of salt flowing in is (concentration in)

\times

(flow rate in) =

0.05 \text{ kg/L} \times 10 \text{ L/min} = 0.5

kg/min. The volume of the solution in the tank is not constant; it changes at a rate of

(10 - 5) = 5

L/min. The initial volume is 400 L, so

V(t) = 400 + 5t

. The concentration of salt in the tank at time

t

C(t) = A(t)/V(t) = A(t)/(400 + 5t)

. The rate of salt flowing out is (concentration out)

\times

(flow rate out) =

(A(t)/(400 + 5t)) \times 5 \text{ L/min}

. Therefore, the differential equation is

dA/dt = 0.5 - \frac{5A}{400+5t}