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Math 308 Week in Review 2 (with Solutions)

Differential Equations · Spring 2026 • Sections 2.1, 2.3

Directions: Full worked solutions for Week in Review 2 (Differential Equations). Click Solution under any problem to reveal the worked answer. Use the Contents panel or the Prev / Up / Next buttons to move between sections, and Search to find terms on this page.

2.1 Linear Equations; Method of Integrating Factors

1.

Determine whether each of the following differential equations is linear. If it is linear, write it in standard form and identify $p(t)$ and $g(t)$.

(a) Linear. Standard form:

\[ y' + (\sin t) y = t^2 \]

\[ p(t) = \sin t, \quad g(t) = t^2 \]

(b) Nonlinear due to $y^2$ term.

(c) Linear. Rewrite for $t \neq 0$:

\[ y' - \dfrac{2}{t} y = \dfrac{1}{t} e^{t^2} \]

\[ p(t) = -\dfrac{2}{t}, \quad g(t) = \dfrac{1}{t} e^{t^2} \]

(d) Linear. Rewrite for $t \neq 0$:

\[ y' + \dfrac{2t+1}{t} y = \dfrac{\cos t}{t} \]

\[ p(t) = \dfrac{2t+1}{t}, \quad g(t) = \dfrac{\cos t}{t} \]

2.

Solve the differential equation

Standard form: $y' + 2y = 6e^{-t}$

Integrating factor: $\mu(t) = e^{\int 2\, dt} = e^{2t}$

\[ e^{2t}y' + 2e^{2t}y = 6e^{2t}e^{-t} = 6e^{t} \]
\[ \dfrac{d}{dt}\left(e^{2t}y\right) = 6e^{t} \]
\[ e^{2t}y = 6\int e^{t}\, dt = 6e^{t} + C \]
\[ y = 6e^{t}e^{-2t} + Ce^{-2t} = 6e^{-t} + Ce^{-2t} \]
\[ \boxed{y = 6e^{-t} + Ce^{-2t}} \]

3.

Solve the initial value problem

Standard form: $y' - \dfrac{2}{t}y = t^2\cos t$

Integrating factor: $\mu(t) = e^{\int -\dfrac{2}{t}\, dt} = e^{-2\ln t} = t^{-2}$

\[ t^{-2}y' - 2t^{-3}y = \cos t \]
\[ \dfrac{d}{dt}\left(t^{-2}y\right) = \cos t \]
\[ t^{-2}y = \int \cos t\, dt = \sin t + C \]
\[ y = t^2\sin t + Ct^2 \]

Apply $y(\pi) = 0$:

\[ 0 = \pi^2\sin\pi + C\pi^2 = 0 + C\pi^2 \Rightarrow C = 0 \]
\[ \boxed{y(t) = t^2\sin t} \]

4.

Find the general solution of

Standard form: $y' + (\cot x) y = \csc x$

Integrating factor: $\mu(x) = e^{\int \cot x\, dx} = e^{\ln|\sin x|} = \sin x$

\[ \sin x \cdot y' + y\cos x = 1 \]
\[ \dfrac{d}{dx}\left(\sin x \cdot y\right) = 1 \]
\[ \sin x \cdot y = \int 1\, dx = x + C \]
\[ y = \dfrac{x + C}{\sin x} \]
\[ \boxed{y = \dfrac{x + C}{\sin x}} \]

5.

Solve the initial value problem

Standard form for $t \neq 0$: $y' + \dfrac{t+1}{t}y = 1$

Integrating factor: $\mu(t) = e^{\int \dfrac{t+1}{t}\, dt} = e^{\int \left(1 + \dfrac{1}{t}\right) dt} = e^{t + \ln t} = te^{t}$

\[ te^{t}y' + (t+1)e^{t}y = te^{t} \]
\[ \dfrac{d}{dt}\left(te^{t}y\right) = te^{t} \]

Integrate using integration by parts: $\int te^{t}\, dt = te^{t} - e^{t} + C$

\[ te^{t}y = te^{t} - e^{t} + C \]
\[ y = 1 - \dfrac{1}{t} + \dfrac{C}{t}e^{-t} \]

Apply $y(1) = 2$:

\[ 2 = 1 - 1 + Ce^{-1} \Rightarrow C = 2e \]
\[ \boxed{y(t) = 1 - \dfrac{1}{t} + \dfrac{2}{t}e^{1-t}} \]

2.3 Modeling with First-Order Differential Equations

6.

A radioactive substance has an initial mass of 400 grams. After 8 years, 280 grams remain. Assuming the decay follows the usual exponential law,

(a) find the decay constant $k$,

(b) determine how much of the substance remains after 20 years,

(c) find the half-life of the substance.

Model: Exponential decay $m(t) = m_0 e^{-kt}$, $m_0 = 400\,\text{g}$.

(a) Find decay constant $k$:

\[ m(8) = 400e^{-8k} = 280 \]
\[ e^{-8k} = \dfrac{280}{400} = 0.7 \]
\[ -8k = \ln 0.7 \]
\[ k = -\dfrac{\ln 0.7}{8} \approx 0.0446\,\text{yr}^{-1} \]

(b) Mass after 20 years:

\[ m(20) = 400e^{-20k} = 400(0.7)^{20/8} = 400(0.7)^{2.5} \]
\[ m(20) \approx 188.8\,\text{g} \]

(c) Half-life $\tau$:

\[ \tau = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{0.0446} \approx 15.54\,\text{years} \]
\[ \boxed{k \approx 0.0446\,\text{yr}^{-1}} \]
\[ \boxed{m(20) \approx 188.8\,\text{g}} \]
\[ \boxed{\tau \approx 15.54\,\text{years}} \]

7.

A savings account starts with \$800 and earns interest at 5% per year compounded continuously. Deposits are made continuously at a rate of $60 per year.

(a) Set up the differential equation for the balance $B(t)$.

(b) Solve the differential equation.

(c) What is the balance after 12 years?

(a) Differential equation:

\[ B' = 0.05B + 60, \quad B(0) = 800 \]

(b) Solution:

Write in standard form: $B' - 0.05B = 60$

Integrating factor: $\mu(t) = e^{-0.05t}$

\[ (e^{-0.05t}B)' = 60e^{-0.05t} \]
\[ e^{-0.05t}B = \int 60e^{-0.05t}\, dt = -1200e^{-0.05t} + C \]
\[ B = -1200 + Ce^{0.05t} \]

Apply $B(0) = 800$: $800 = -1200 + C \Rightarrow C = 2000$

\[ B(t) = -1200 + 2000e^{0.05t} \]

(c) Balance after 12 years:

\[ B(12) = -1200 + 2000e^{0.05 \times 12} = -1200 + 2000e^{0.6} \]
\[ B(12) \approx 2444.20 \]
\[ \boxed{B' = 0.05B + 60,\ B(0)=800} \]
\[ \boxed{B(t) = -1200 + 2000e^{0.05t}} \]
\[ \boxed{B(12) \approx \$2444.20} \]

8.

A tank initially contains 60 liters of pure water. A solution containing 0.4 kg/L of salt enters the tank at 5 L/min. The well-stirred mixture leaves at the same rate.

(a) Set up the differential equation for the amount of salt $q(t)$ in the tank.

(b) Solve for $q(t)$.

(c) When does the concentration of salt in the tank reach 0.25 kg/L?

Constant volume: $V = 60\,\text{L}$ (inflow = outflow).

(a) Differential equation:

In rate: $0.4\,\text{kg/L} \times 5\,\text{L/min} = 2\,\text{kg/min}$

Out rate: $\dfrac{q}{60}\,\text{kg/L} \times 5\,\text{L/min} = \dfrac{q}{12}\,\text{kg/min}$

\[ q' = 2 - \dfrac{q}{12}, \quad q(0) = 0 \]

(b) Solution:

Standard form: $q' + \dfrac{1}{12}q = 2$

Integrating factor: $\mu(t) = e^{t/12}$

\[ (e^{t/12}q)' = 2e^{t/12} \]
\[ e^{t/12}q = 24e^{t/12} + C \]
\[ q = 24 + Ce^{-t/12} \]

Apply $q(0) = 0$: $0 = 24 + C \Rightarrow C = -24$

\[ q(t) = 24(1 - e^{-t/12}) \]

(c) Time when concentration = 0.25 kg/L:

Concentration $c = \dfrac{q}{60} = 0.25 \Rightarrow q = 15\,\text{kg}$

\[ 24(1 - e^{-t/12}) = 15 \]
\[ 1 - e^{-t/12} = \dfrac{15}{24} = \dfrac{5}{8} \]
\[ e^{-t/12} = \dfrac{3}{8} \]
\[ -\dfrac{t}{12} = \ln\left(\dfrac{3}{8}\right) \]
\[ t = -12\ln\left(\dfrac{3}{8}\right) \approx 11.77\,\text{min} \]
\[ \boxed{q' = 2 - \dfrac{q}{12},\ q(0)=0} \]
\[ \boxed{q(t) = 24(1 - e^{-t/12})} \]
\[ \boxed{t \approx 11.77\,\text{min}} \]

9.

A population of fish follows the logistic growth model with growth rate $r = 0.18$ per year and carrying capacity $K = 12$ million. However, a constant harvesting rate of $H$ million fish per year is applied.

(a) Write the differential equation for the population $P(t)$ in millions.

(b) Find the value of $H$ that keeps the population steady at 9 million.

(c) For the value of $H$ found in (b), describe the long-term behavior if the initial population is 6 million.

(a) Logistic growth with harvesting:

\[ P' = 0.18P\left(1 - \dfrac{P}{12}\right) - H \]

(b) Find $H$ for steady state at $P = 9$:

\[ 0 = 0.18 \times 9 \left(1 - \dfrac{9}{12}\right) - H \]
\[ 0 = 1.62 \times \dfrac{3}{12} - H = 0.405 - H \]
\[ H = 0.405\,\text{million/year} \]

(c) Long-term behavior with $H = 0.405$ and $P(0) = 6$:

Equilibria from $0.18P(1 - P/12) = 0.405$:

\[ P - \dfrac{P^2}{12} = \dfrac{0.405}{0.18} = 2.25 \]
\[ \dfrac{P^2}{12} - P + 2.25 = 0 \]
\[ P^2 - 12P + 27 = 0 \]
\[ (P-3)(P-9) = 0 \]

Equilibria at $P = 3$ and $P = 9$ million.

Analysis: $P' > 0$ for $3 < P < 9$, $P' < 0$ for $P < 3$ or $P > 9$.

Thus $P = 9$ is stable, $P = 3$ is unstable.

With $P(0) = 6$ (between 3 and 9), population increases toward 9 million.

\[ \boxed{P' = 0.18P\left(1 - \dfrac{P}{12}\right) - H} \]
\[ \boxed{H = 0.405\,\text{million/year}} \]
\[ \boxed{\text{Population increases to 9 million}} \]

10.

A pizza at room temperature (75°F) is placed in an oven whose temperature is rising at a constant rate of 20°F per minute. The oven starts at 75°F (same as the pizza) and rises until it reaches 375°F. Assume the temperature of the pizza $u(t)$ changes at a rate proportional to the difference between the oven temperature and the pizza temperature, with proportionality constant $k = 0.1$ per minute.

(a) Write a differential equation for the temperature of the pizza $u(t)$,

(b) solve for $u(t)$,

(c) determine the temperature of the pizza when the oven temperature first reaches 375°F.

Oven temperature: $T(t) = 75 + 20t$ for $0 \leq t \leq 15$ minutes.

(Oven reaches 375°F at $t = \dfrac{375-75}{20} = 15$ min)

(a) Differential equation (Newton's Law of Cooling/Heating):

\[ u' = 0.1(T(t) - u) = 0.1(75 + 20t - u) \]
\[ u' + 0.1u = 7.5 + 2t \]

(b) Solution:

Integrating factor: $\mu(t) = e^{0.1t}$

\[ (e^{0.1t}u)' = (7.5 + 2t)e^{0.1t} \]

Integrate using formula: $\int (a + bt)e^{ct}\, dt = e^{ct}\left(\dfrac{a}{c} + \dfrac{bt}{c} - \dfrac{b}{c^2}\right) + C$

With $a = 7.5$, $b = 2$, $c = 0.1$:

\[ \int (7.5 + 2t)e^{0.1t}\, dt = e^{0.1t}\left(\dfrac{7.5}{0.1} + \dfrac{2t}{0.1} - \dfrac{2}{0.01}\right) + C \]
\[ = e^{0.1t}(75 + 20t - 200) + C = e^{0.1t}(20t - 125) + C \]
\[ e^{0.1t}u = e^{0.1t}(20t - 125) + C \]
\[ u = 20t - 125 + Ce^{-0.1t} \]

Apply $u(0) = 75$: $75 = 0 - 125 + C \Rightarrow C = 200$

\[ u(t) = 20t - 125 + 200e^{-0.1t} \]

(c) Pizza temperature at $t = 15$ min:

\[ u(15) = 20(15) - 125 + 200e^{-0.1 \times 15} \]
\[ u(15) = 300 - 125 + 200e^{-1.5} \]
\[ u(15) \approx 175 + 200 \times 0.2231 \approx 219.62^\circ\text{F} \]
\[ \boxed{u' + 0.1u = 7.5 + 2t,\ u(0)=75} \]
\[ \boxed{u(t) = 20t - 125 + 200e^{-0.1t}} \]
\[ \boxed{u(15) \approx 219.62^\circ\text{F}} \]

11.

Brine containing 0.3 kg/L of salt enters a tank at 8 L/min. The tank initially holds 120 L of water with 4 kg of salt already dissolved. The well-mixed solution exits at 6 L/min.

(a) Find the differential equation for the mass of salt $m(t)$ in the tank.

(b) Solve for $m(t)$.

(c) Assume the tank has a capacity of 200L, what is the concentration of salt in the tank when it fills up?

Volume: $V(t) = 120 + (8-6)t = 120 + 2t$ L

(a) Differential equation:

In rate: $0.3\,\text{kg/L} \times 8\,\text{L/min} = 2.4\,\text{kg/min}$

Out rate: $\dfrac{m}{V} \times 6 = \dfrac{6m}{120 + 2t}\,\text{kg/min}$

\[ m' = 2.4 - \dfrac{6m}{120 + 2t}, \quad m(0) = 4 \]
\[ m' + \dfrac{3}{60 + t}m = 2.4 \]

(b) Solution:

Integrating factor: $\mu(t) = e^{\int \dfrac{3}{60+t}\, dt} = e^{3\ln(60+t)} = (60+t)^3$

\[ [(60+t)^3m]' = 2.4(60+t)^3 \]
\[ (60+t)^3m = 2.4\int (60+t)^3\, dt = 0.6(60+t)^4 + C \]
\[ m = 0.6(60+t) + C(60+t)^{-3} \]

Apply $m(0) = 4$:

\[ 4 = 0.6(60) + C(60)^{-3} = 36 + \dfrac{C}{216000} \]
\[ \dfrac{C}{216000} = -32 \Rightarrow C = -6.912 \times 10^6 \]
\[ m(t) = 0.6(60+t) - 6.912 \times 10^6(60+t)^{-3} \]

(c) Concentration when tank is full (200 L):

Volume: $120 + 2t = 200 \Rightarrow t = 40$ min

\[ m(40) = 0.6(60+40) - 6.912 \times 10^6(100)^{-3} \]
\[ m(40) = 60 - 6.912 \times 10^6 \times 10^{-6} = 60 - 6.912 = 53.088\,\text{kg} \]

Concentration: $c = \dfrac{m}{V} = \dfrac{53.088}{200} \approx 0.2654\,\text{kg/L}$

\[ \boxed{m' + \dfrac{3}{60+t}m = 2.4,\ m(0)=4} \]
\[ \boxed{m(t) = 0.6(60+t) - 6.912 \times 10^6(60+t)^{-3}} \]
\[ \boxed{c \approx 0.265\,\text{kg/L}} \]
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Math 308 Week in Review 2 (with Solutions) — Solutions · Spring 2026
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