Differential Equations Quiz: Modeling With Linear Odes

What this quiz covers

This quiz focuses on Modeling With Linear Odes, 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.

Question 1

1. $7.5^{\circ}$C
2. $15^{\circ}$C (correct answer)
3. $25^{\circ}$C
4. $30^{\circ}$C

Explanation: The model is $T' = -k(T-T_a)$. The solution is $T(t) = T_a + (T_0 - T_a)e^{-kt}$. Phase 1 ($0 \le t \le 10$): $T_a = 20$, $T_0 = 100$. So $T(t) = 20 + (100-20)e^{-kt} = 20 + 80e^{-kt}$. We are given $T(10) = 60$. So $60 = 20 + 80e^{-10k} \implies 40 = 80e^{-10k} \implies e^{-10k} = 1/2 \implies -10k = \ln(1/2) = -\ln(2) \implies k = \ln(2)/10$. Phase 2 ($t > 10$): The object is moved to a new environment with $T_a = 0$. The initial time for this phase is $t=10$, and the initial temperature is $T(10)=60$. Let $s = t-10$ be the time elapsed since the move. The model is $T(s) = 0 + (60 - 0)e^{-ks} = 60e^{-ks}$. We need to find the temperature 20 minutes after the move, which is at $s=20$. $T(20) = 60e^{-k(20)} = 60e^{-(\ln(2)/10)(20)} = 60e^{-2\ln(2)} = 60e^{\ln(2^{-2})} = 60(1/4) = 15^{\circ}$C.

Question 2

1. $62,500 + 412,500e^{-0.4}$
2. $187,500 + 287,500e^{0.4}$
3. $187,500 + 287,500e^{-0.4}$ (correct answer)
4. $-187,500 + 712,500e^{-0.4}$

Explanation: The differential equation modeling the principal $P(t)$ is $\frac{dP}{dt} = rP - W(t)$, where $r=0.04$. So, $\frac{dP}{dt} = 0.04P - (5000 + 100t)$. The standard linear form is $\frac{dP}{dt} - 0.04P = -5000 - 100t$. The integrating factor is $I(t) = e^{\int -0.04 dt} = e^{-0.04t}$. Multiplying by $I(t)$ gives $\frac{d}{dt}(Pe^{-0.04t}) = (-5000 - 100t)e^{-0.04t}$. Integrating both sides (using integration by parts for the $t e^{-0.04t}$ term) gives: $Pe^{-0.04t} = \int (-5000 - 100t)e^{-0.04t} dt = 125000e^{-0.04t} + 2500te^{-0.04t} + 62500e^{-0.04t} + C$. $Pe^{-0.04t} = (187500 + 2500t)e^{-0.04t} + C$. $P(t) = 187500 + 2500t + Ce^{0.04t}$. We are given $P(10) = 500,000$. $500,000 = 187500 + 2500(10) + Ce^{0.04(10)} = 187500 + 25000 + Ce^{0.4} = 212,500 + Ce^{0.4}$. $Ce^{0.4} = 500,000 - 212,500 = 287,500 \implies C = 287,500e^{-0.4}$. The initial investment is $P(0)$. $P(0) = 187500 + 2500(0) + Ce^{0} = 187500 + C$. Substituting the value of $C$: $P(0) = 187,500 + 287,500e^{-0.4}$.

Question 3

1. $0$ kg/m$^3$
2. $0.0083$ kg/m$^3$
3. $0.01$ kg/m$^3$ (correct answer)
4. $0.05$ kg/m$^3$

Explanation: Let $A(t)$ be the amount of pollutant and $V(t)$ be the volume of water at time $t$. Let $c(t) = A(t)/V(t)$ be the concentration. The volume changes according to $\frac{dV}{dt} = 500 - 600 = -100$, so $V(t) = 10^7 - 100t$. The lake is empty when $t = 10^5$ days. The amount of pollutant changes according to $\frac{dA}{dt} = (500)(0.01) - 600 \frac{A(t)}{V(t)} = 5 - 600c(t)$. Since $A(t) = c(t)V(t)$, we have $\frac{dA}{dt} = \frac{d(cV)}{dt} = c'V + cV' = c'(10^7-100t) + c(-100)$. Equating the two expressions for $\frac{dA}{dt}$: $c'(10^7-100t) - 100c = 5 - 600c$. $c'(10^7-100t) + 500c = 5$. This is a linear ODE for $c(t)$. In standard form: $c' + \frac{500}{10^7-100t} c = \frac{5}{10^7-100t}$. Let $T = 10^5$. $c' + \frac{5}{T-t} c = \frac{0.05}{T-t}$. The integrating factor is $I(t) = e^{\int \frac{5}{T-t} dt} = e^{-5\ln(T-t)} = (T-t)^{-5}$. $\frac{d}{dt}(c(T-t)^{-5}) = \frac{0.05}{T-t}(T-t)^{-5} = 0.05(T-t)^{-6}$. Integrating gives $c(T-t)^{-5} = 0.05 \frac{(T-t)^{-5}}{-5}(-1) + K = 0.01(T-t)^{-5} + K$. $c(t) = 0.01 + K(T-t)^5$. Since $c(0)=0$, $0 = 0.01 + KT^5$, so $K = -0.01/T^5$. $c(t) = 0.01 - 0.01(\frac{T-t}{T})^5$. As the volume approaches zero, $t \to T$. The limiting concentration is $\lim_{t\to T} c(t) = 0.01 - 0 = 0.01$ kg/m$^3$.

Question 4

1. $y(t)$ approaches $0$
2. $y(t)$ approaches a finite positive constant
3. $y(t)$ approaches $+\infty$
4. $y(t)$ approaches $-\infty$ (correct answer)

Explanation: The differential equation is $\frac{dy}{dt} - \frac{2}{t+1}y = -(t+1)^3$. This is a first-order linear ODE. The integrating factor is $I(t) = e^{\int -\frac{2}{t+1} dt} = e^{-2\ln(t+1)} = (t+1)^{-2}$. Multiplying the equation by $I(t)$ gives $\frac{d}{dt}(y(t+1)^{-2}) = -(t+1)^3(t+1)^{-2} = -(t+1)$. Integrate both sides with respect to $t$: $y(t+1)^{-2} = \int -(t+1) dt = -\frac{1}{2}(t+1)^2 + C$. To find $C$, we use the initial condition $y(0)=y_0$: $y_0(1)^{-2} = -\frac{1}{2}(1)^2 + C \implies y_0 = -1/2 + C \implies C = y_0 + 1/2$. The solution is $y(t+1)^{-2} = -\frac{1}{2}(t+1)^2 + y_0 + 1/2$. $y(t) = (t+1)^2 \left( -\frac{1}{2}(t+1)^2 + y_0 + 1/2 \right)$. To determine the long-term behavior, we examine the limit as $t \to \infty$. We look at the dominant terms: $y(t) \approx t^2 \left( -\frac{1}{2}t^2 \right) = -\frac{1}{2}t^4$. As $t \to \infty$, the term $-\frac{1}{2}t^4$ goes to $-\infty$. Therefore, $y(t)$ approaches $-\infty$.

Question 5

1. $1$ mg/hour
2. $2$ mg/hour (correct answer)
3. $10$ mg/hour
4. $20$ mg/hour

Explanation: Let $A(t)$ be the amount of the drug in the body at time $t$. The model is given by the differential equation $\frac{dA}{dt} = (\text{rate in}) - (\text{rate out})$. Rate in is the constant administration rate $R$. Rate out is proportional to the amount present, $kA(t)$. So, $\frac{dA}{dt} = R - kA(t)$. With $k=0.1$, the equation is $\frac{dA}{dt} = R - 0.1A$, or $\frac{dA}{dt} + 0.1A = R$. This is a linear first-order ODE. Assuming $A(0)=0$, the solution is $A(t) = \frac{R}{0.1}(1 - e^{-0.1t}) = 10R(1 - e^{-0.1t})$. The amount of drug in the body will approach a saturation level, or steady-state amount, as $t \to \infty$. $A_{ss} = \lim_{t\to\infty} A(t) = \lim_{t\to\infty} 10R(1 - e^{-0.1t}) = 10R(1-0) = 10R$. Alternatively, the steady-state amount is where $\frac{dA}{dt}=0$, so $R - 0.1A_{ss} = 0$, which gives $A_{ss} = R/0.1 = 10R$. We require that the amount of the drug never exceeds the safe limit of 20 mg. Since the amount $A(t)$ is an increasing function approaching its limit $A_{ss}$, we must ensure that this limit does not exceed 20 mg. So, $A_{ss} \le 20 \implies 10R \le 20 \implies R \le 2$ mg/hour. The maximum rate $R$ is therefore $2$ mg/hour.

Question 6

1. $\frac{dD}{dt} = -kD$
2. $\frac{dD}{dt} = -k \left( \frac{1}{m_1c_1} - \frac{1}{m_2c_2} \right) D$
3. $\frac{dD}{dt} = -k \left( \frac{1}{m_1c_1} + \frac{1}{m_2c_2} \right) D$ (correct answer)
4. $\frac{dD}{dt} = k(200 - D)$

Explanation: The rate of change of heat for an object is $dQ/dt = mc \frac{dT}{dt}$. For the aluminum bar (object 1), the heat is flowing in: $\frac{dQ_1}{dt} = k(T_2 - T_1) = kD$. Thus, $m_1c_1 \frac{dT_1}{dt} = kD$, which means $\frac{dT_1}{dt} = \frac{k}{m_1c_1}D$. For the copper block (object 2), heat is flowing out: $\frac{dQ_2}{dt} = -k(T_2 - T_1) = -kD$. Thus, $m_2c_2 \frac{dT_2}{dt} = -kD$, which means $\frac{dT_2}{dt} = -\frac{k}{m_2c_2}D$. The rate of change of the difference $D(t) = T_2(t) - T_1(t)$ is $\frac{dD}{dt} = \frac{dT_2}{dt} - \frac{dT_1}{dt}$. Substituting the expressions for $\frac{dT_1}{dt}$ and $\frac{dT_2}{dt}$: $\frac{dD}{dt} = \left( -\frac{k}{m_2c_2}D \right) - \left( \frac{k}{m_1c_1}D \right) = -k \left( \frac{1}{m_2c_2} + \frac{1}{m_1c_1} \right)D$. This is a linear first-order differential equation for the temperature difference $D(t)$.

Question 7

1. $2$
2. $1/2$
3. $\ln(2)$ (correct answer)
4. $\ln(3/2)$

Explanation: The differential equation $\frac{dA}{dt} + kA = R$ is a linear first-order ODE. Its general solution is $A(t) = Ce^{-kt} + \frac{R}{k}$. We are given three data points:

1. $A(0) = 10 \implies 10 = C + R/k$
2. $A(1) = 8 \implies 8 = Ce^{-k} + R/k$
3. $A(2) = 7 \implies 7 = Ce^{-2k} + R/k$ Let $X = R/k$. From (1), $C = 10 - X$. Substitute this into (2) and (3): 2') $8 = (10-X)e^{-k} + X$ 3') $7 = (10-X)e^{-2k} + X$ From (2'): $8 - X = (10-X)e^{-k} \implies e^{-k} = \frac{8-X}{10-X}$. From (3'): $7 - X = (10-X)e^{-2k} = (10-X)(e^{-k})^2$. Substitute the expression for $e^{-k}$ into this equation: $7 - X = (10-X) \left( \frac{8-X}{10-X} \right)^2 = \frac{(8-X)^2}{10-X}$. $(7-X)(10-X) = (8-X)^2$ $70 - 17X + X^2 = 64 - 16X + X^2$ $70 - 17X = 64 - 16X \implies 6 = X$. So, $R/k = 6$. Now find $k$ using $e^{-k} = \frac{8-X}{10-X} = \frac{8-6}{10-6} = \frac{2}{4} = \frac{1}{2}$. $e^{-k} = 1/2 \implies -k = \ln(1/2) = -\ln(2) \implies k = \ln(2)$.

Question 8

1. $50 + 2470e^{-1}$ (correct answer)
2. $2520e^{-1}$
3. $5050 - 2530e^{-1}$
4. $50t - 2450 + 2470e^{-t/50}$

Explanation: Let $A(t)$ be the amount of salt in kg at time $t$. The rate of change of salt is given by the differential equation $\frac{dA}{dt} = (\text{rate in}) - (\text{rate out})$. The rate in is $r_{in} \times c_{in}(t) = 10 \text{ L/min} \times 0.1(1+t) \text{ kg/L} = 1+t$ kg/min. The rate out is $r_{out} \times c_{out}(t) = 10 \text{ L/min} \times \frac{A(t)}{500} \text{ kg/L} = \frac{A}{50}$ kg/min. The resulting ODE is $\frac{dA}{dt} = 1+t - \frac{A}{50}$, which can be written in standard linear form as $\frac{dA}{dt} + \frac{1}{50}A = 1+t$. The initial condition is $A(0)=20$. The integrating factor is $I(t) = e^{\int (1/50) dt} = e^{t/50}$. Multiplying the ODE by $I(t)$ gives $\frac{d}{dt}(A e^{t/50}) = (1+t)e^{t/50}$. Integrating both sides (using integration by parts for $\int te^{t/50} dt$) yields $A e^{t/50} = 50e^{t/50} + 50te^{t/50} - 2500e^{t/50} + C = 50te^{t/50} - 2450e^{t/50} + C$. So, $A(t) = 50t - 2450 + Ce^{-t/50}$. Using $A(0)=20$, we find $20 = -2450 + C$, so $C = 2470$. The solution is $A(t) = 50t - 2450 + 2470e^{-t/50}$. At $t=50$, $A(50) = 50(50) - 2450 + 2470e^{-50/50} = 2500 - 2450 + 2470e^{-1} = 50 + 2470e^{-1}$.

Question 9

1. $0$ (correct answer)
2. $|A-B|$
3. $|A-B|e^{-kt}$
4. $\infty$

Explanation: Let $y_1(t)$ and $y_2(t)$ be two solutions to the linear ODE $y' + ky = q(t)$. The general solution is of the form $y(t) = y_c(t) + y_p(t)$, where $y_c(t) = Ce^{-kt}$ is the solution to the homogeneous equation and $y_p(t)$ is a particular solution that does not depend on the initial conditions. For $y_1(t)$, the initial condition $y_1(0) = A$ determines its constant $C_1$. $A = C_1e^0 + y_p(0) \implies C_1 = A - y_p(0)$. So, $y_1(t) = (A-y_p(0))e^{-kt} + y_p(t)$. For $y_2(t)$, the initial condition $y_2(0) = B$ determines its constant $C_2$. $B = C_2e^0 + y_p(0) \implies C_2 = B - y_p(0)$. So, $y_2(t) = (B-y_p(0))e^{-kt} + y_p(t)$. Now consider the difference $y_1(t) - y_2(t)$: $y_1(t) - y_2(t) = [(A-y_p(0))e^{-kt} + y_p(t)] - [(B-y_p(0))e^{-kt} + y_p(t)]$ $y_1(t) - y_2(t) = (A-y_p(0) - B + y_p(0))e^{-kt} = (A-B)e^{-kt}$. We are asked for the limit of the absolute difference as $t \to \infty$: $\lim_{t\to\infty} |y_1(t) - y_2(t)| = \lim_{t\to\infty} |(A-B)e^{-kt}| = |A-B| \lim_{t\to\infty} e^{-kt}$. Since we are given that $k>0$, $\lim_{t\to\infty} e^{-kt} = 0$. Therefore, $\lim_{t\to\infty} |y_1(t) - y_2(t)| = |A-B| \times 0 = 0$. This indicates that for stable linear systems (where $k>0$), the effect of the initial conditions vanishes over time, and all solutions converge.

Question 10

1. $0.2\ln(10)$
2. $0.4\ln(10)$ (correct answer)
3. $2\ln(10)$
4. $10\ln(10)$

Explanation: The differential equation for an RL circuit is $L\frac{dI}{dt} + RI = V(t)$. Substituting the given values, we get $2\frac{dI}{dt} + 10I = 50\cos(5t)$, which simplifies to $\frac{dI}{dt} + 5I = 25\cos(5t)$. The transient current $I_{tr}(t)$ is the solution to the homogeneous equation $\frac{dI}{dt} + 5I = 0$, which is $I_{tr}(t) = Ce^{-5t}$. To find $C$, we need the full solution. The steady-state solution is of the form $I_{ss}(t) = A\cos(5t) + B\sin(5t)$. Substituting this into the ODE gives $A=2.5$ and $B=2.5$. So, $I(t) = Ce^{-5t} + 2.5\cos(5t) + 2.5\sin(5t)$. Using the initial condition $I(0)=0$: $0 = C e^0 + 2.5\cos(0) + 2.5\sin(0) \implies 0 = C + 2.5 \implies C = -2.5$. So the transient current is $I_{tr}(t) = -2.5e^{-5t}$. The initial magnitude of the transient current is $|I_{tr}(0)| = |-2.5e^0| = 2.5$. We want to find $t$ such that $|I_{tr}(t)| = 0.01 \times |I_{tr}(0)|$. $|-2.5e^{-5t}| = 0.01 \times 2.5 \implies 2.5e^{-5t} = 0.025 \implies e^{-5t} = 0.01$. Taking the natural logarithm of both sides: $-5t = \ln(0.01) = \ln(10^{-2}) = -2\ln(10)$. Therefore, $t = \frac{-2\ln(10)}{-5} = \frac{2}{5}\ln(10) = 0.4\ln(10)$.

Question 11

1. $\frac{dA_2}{dt} = 2.5 - \frac{A_2}{40}$
2. $\frac{dA_2}{dt} = 2.5(1-e^{-t/20}) - \frac{A_2}{40}$ (correct answer)
3. $\frac{dA_2}{dt} = 0.25(1-e^{-t/20}) - \frac{A_2}{40}$
4. $\frac{dA_2}{dt} = 50(1-e^{-t/20}) - \frac{A_2}{200}$

Explanation: First, we model the amount of salt $A_1(t)$ in Tank 1. The volume is constant at 100 L. $\frac{dA_1}{dt} = (\text{rate in}) - (\text{rate out}) = (5 \text{ L/min})(0.5 \text{ kg/L}) - (5 \text{ L/min})\frac{A_1(t)}{100 \text{ L}} = 2.5 - \frac{A_1}{20}$. This is a linear ODE for $A_1$. With $A_1(0)=0$, the solution is $A_1(t) = 50(1-e^{-t/20})$. The concentration of salt flowing from Tank 1 into Tank 2 is $c_1(t) = \frac{A_1(t)}{100} = \frac{50(1-e^{-t/20})}{100} = 0.5(1-e^{-t/20})$. Now, we model the amount of salt $A_2(t)$ in Tank 2. The volume is constant at 200 L. $\frac{dA_2}{dt} = (\text{rate in}) - (\text{rate out})$. The rate in is the rate of flow from Tank 1 times the concentration from Tank 1: $(5 \text{ L/min}) \times c_1(t) = 5 \times 0.5(1-e^{-t/20}) = 2.5(1-e^{-t/20})$. The rate out is $(5 \text{ L/min}) \times \frac{A_2(t)}{200 \text{ L}} = \frac{A_2}{40}$. Therefore, the differential equation for $A_2(t)$ is $\frac{dA_2}{dt} = 2.5(1-e^{-t/20}) - \frac{A_2}{40}$.

Question 12

1. $\frac{100}{3} \approx 33.3$
2. $\frac{100k}{2k-k} = 100$
3. $\frac{100}{e} \approx 36.8$
4. $\frac{100}{2} = 50$ (correct answer)

Explanation: When you encounter a coupled system of differential equations describing chemical kinetics, you need to solve for each concentration as a function of time, then find when the intermediate product reaches its maximum. First, solve the equation for A: $\frac{d[A]}{dt} = -k[A]$ gives $[A](t) = 100e^{-kt}$. Next, substitute this into the equation for B: $\frac{d[B]}{dt} = k(100e^{-kt}) - 2k[B] = 100ke^{-kt} - 2k[B]$. This is a first-order linear ODE. Using an integrating factor of $e^{2kt}$, you can solve to get $[B](t) = 100(e^{-kt} - e^{-2kt})$. To find the maximum concentration of B, take the derivative and set it equal to zero: $\frac{d[B]}{dt} = 100(-ke^{-kt} + 2ke^{-2kt}) = 0$. This gives $e^{-kt} = \frac{1}{2}$, so $t = \frac{\ln(2)}{k}$. Substituting back: $[B]_{max} = 100(\frac{1}{2} - \frac{1}{4}) = 100 \cdot \frac{1}{4} = 50$. Answer A ($\frac{100}{3}$) incorrectly assumes equal rate constants. Answer B shows dimensional confusion—you can't have $\frac{k}{k}$ terms in a concentration. Answer C ($\frac{100}{e}$) results from incorrectly setting the maximum at $t = \frac{1}{k}$ instead of properly solving the optimization. Remember: for intermediate species in consecutive reactions, always solve the full system, then optimize. The maximum occurs when the formation rate equals the consumption rate.

Question 13

1. $18000
2. $20000 (correct answer)
3. $24000
4. $12000

Explanation: The equilibrium occurs when dB/dt = 0, so 0.06B - 1200 = 0, giving B = 20000. If the initial balance equals $20000, the account remains constant. If initial balance >$20000, it grows; if < $20000, it decreases toward zero. Therefore, the minimum initial balance to prevent going to zero is exactly$20000. Choice A is below the equilibrium and would lead to depletion. Choice C is above the minimum required. Choice D is the break-even withdrawal amount divided by interest rate incorrectly.

Question 14

1. $\frac{10}{\sqrt{5}}$ coulombs (correct answer)
2. $\frac{5}{\sqrt{5}}$ coulombs
3. $\frac{10}{3}$ coulombs
4. $\frac{20}{\sqrt{5}}$ coulombs

Explanation: Substituting values: 2(dq/dt) + 2q = 10sin(t). For steady state with input Asin(ωt), the response is of the form q = B sin(t + φ). The amplitude is B = A/√((ω²RC - 1/C)² + (ωR)²). Here A = 10, ω = 1, R = 2, C = 0.5, so B = 10/√((1·2·0.5 - 1/0.5)² + (1·2)²) = 10/√((-1)² + 4) = 10/√5. Choice B omits the factor of 2 in the denominator calculation. Choice C uses only the resistive term. Choice D incorrectly doubles the numerator.

Question 15

1. 20 minutes from the start time
2. 30 minutes from the start time (correct answer)
3. 25 minutes from the start time
4. 15 minutes from the start time

Explanation: The solution is T(t) = 20 + 60e^(-kt). Using T(10) = 60: 60 = 20 + 60e^(-10k), so e^(-10k) = 2/3. For T = 30: 30 = 20 + 60e^(-kt), so e^(-kt) = 1/6. Since e^(-10k) = 2/3, we have e^(-20k) = (2/3)² = 4/9 and e^(-30k) = (2/3)³ = 8/27 ≈ 0.296. But 1/6 ≈ 0.167, which is between e^(-30k) and e^(-40k). Solving: -kt = ln(1/6) and -10k = ln(2/3), so t = 10ln(1/6)/ln(2/3) = 30 minutes. Choice A uses an incorrect exponential relationship. Choice C assumes linear cooling. Choice D uses the wrong initial condition.

Question 16

1. $100(1 - e^{-5}) \approx 99.3\%$
2. $100(1 - e^{-1}) \approx 63.2\%$
3. $100(1 - e^{-2}) \approx 86.5\%$ (correct answer)
4. $100(1 - e^{-10}) \approx 100\%$

Explanation: When you encounter a differential equation describing motion with resistance proportional to velocity, you're dealing with a first-order linear ODE that models exponential approach to terminal velocity. The key insight is recognizing this as a standard form that leads to exponential behavior. Starting with $m\frac{dv}{dt} = mg - kv$, first find the terminal velocity where acceleration equals zero: $mg - kv_t = 0$, so $v_t = \frac{mg}{k} = \frac{70 \times 9.8}{14} = 49$ m/s. Rearranging the equation: $\frac{dv}{dt} = g - \frac{k}{m}v = 9.8 - 0.2v$. This separable equation yields the solution $v(t) = v_t(1 - e^{-kt/m})$ with initial condition $v(0) = 0$. At $t = 10$ seconds: $v(10) = 49(1 - e^{-14 \times 10/70}) = 49(1 - e^{-2})$ The percentage of terminal velocity is $\frac{v(10)}{v_t} \times 100\% = 100(1 - e^{-2}) \approx 86.5\%$, which is answer C. Answer A uses the exponent $-5$, which would correspond to $t = 25$ seconds. Answer B uses $-1$, corresponding to $t = 5$ seconds. Answer D uses $-10$, which incorrectly uses $kt$ instead of $kt/m$. Study tip: Always identify the time constant $\tau = m/k$ in resistance problems. The exponent in the solution is always $-t/\tau$, and after time $\tau$, you've reached about 63% of terminal velocity.