Modeling with Linear ODEs - Differential Equations

1. An object at temperature 200°C is placed in an environment where the ambient temperature increases linearly according to $T_a(t) = 20 + 2t$ degrees Celsius. If Newton's law of cooling applies with cooling constant $k = 0.1$ min$^{-1}$, which differential equation correctly models the object's temperature $T(t)$?
   Answer: $\frac{dT}{dt} = -0.1(T - 20 - 2t)$ When you encounter Newton's law of cooling with a time-varying ambient temperature, you need to carefully identify what represents the temperature difference that drives the cooling process. Newton's law of cooling states that the rate of temperature change is proportional to the difference between the object's temperature and the ambient temperature: $\frac{dT}{dt} = -k(T - T_a)$. Here, the ambient temperature varies as $T_a(t) = 20 + 2t$, so you substitute this directly into the formula. The correct equation becomes $\frac{dT}{dt} = -0.1(T - (20 + 2t)) = -0.1(T - 20 - 2t)$, which is choice C. The negative sign indicates that when the object is warmer than its surroundings, it cools down. Choice A incorrectly expands the Newton's law formula by distributing the negative sign, giving $-0.1T + 0.1(20 + 2t)$. While mathematically equivalent to C, this form obscures the physical meaning of the temperature difference. Choice B uses $-0.1(T - 20) - 2t$ but fails to account for the time-varying ambient temperature properly. The $-2t$ term should be $+0.1(2t)$ when the formula is expanded correctly. Choice D treats the changing ambient temperature as if only the constant rate of change (2°C/min) matters, writing $-0.1(T - 20) + 2$. This ignores that the actual ambient temperature at time $t$ includes the $2t$ term. Always write Newton's law as $\frac{dT}{dt} = -k(T - T_a)$ first, then substitute the given ambient temperature function directly into $T_a$.
2. A chemical reaction follows first-order kinetics where substance A converts to substance B according to $\frac{d[A]}{dt} = -k[A]$, and B simultaneously decomposes according to $\frac{d[B]}{dt} = k[A] - 2k[B]$. If initially $[A]_0 = 100$ and $[B]_0 = 0$, what is the maximum concentration of B that will be achieved?
   Answer: $\frac{100}{2} = 50$ 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.
3. A savings account earns interest at a rate proportional to the current balance, but has a constant withdrawal rate of $1200 per year. If the balance $B(t)$ satisfies $\frac{dB}{dt} = 0.06B - 1200$ and the initial balance is $15000, what is the minimum initial balance needed to prevent the account from eventually going to zero?
   Answer: $20000 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.
4. Consider the differential equation $\frac{dy}{dt} + ky = q(t)$ with $k>0$. Let $y_1(t)$ be the solution with initial condition $y(0)=A$ and $y_2(t)$ be the solution with initial condition $y(0)=B$, where $A \ne B$. What is the long-term behavior of the absolute difference between these two solutions, $\lim_{t\to\infty} |y_1(t) - y_2(t)|$?
   Answer: $0$ 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.
5. A Newton's law of cooling problem states that a hot object cools according to $\frac{dT}{dt} = -k(T - T_a)$, where $T_a$ is the ambient temperature and $k > 0$. If an object at initial temperature 80°C is placed in a room at 20°C, and after 10 minutes the temperature is 60°C, how long will it take for the object to reach 30°C?
   Answer: 30 minutes from the start time 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.
6. A 500-liter tank initially contains 20 kg of salt dissolved in water. Brine is pumped into the tank at a rate of 10 L/min, and the concentration of salt in the incoming brine at time $t$ is given by $c(t) = 0.1(1+t)$ kg/L. The well-mixed solution is pumped out at the same rate. Which of the following expressions represents the amount of salt $A(t)$ in the tank at $t=50$ minutes?
   Answer: $50 + 2470e^{-1}$ 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 $t=50$, $A(50) = 50(50) - 2450 + 2470e^{-50/50} = 2500 - 2450 + 2470e^{-1} = 50 + 2470e^{-1}$.
7. An RL circuit has an inductor of $2$ H, a resistor of $10$ $\Omega$, and is connected to a voltage source $V(t) = 50\cos(5t)$ V. If the initial current is $I(0)=0$ A, the solution for the current $I(t)$ is the sum of a transient current $I_{tr}(t)$ and a steady-state current $I_{ss}(t)$. At what time $t>0$ does the magnitude of the transient current decay to $1\%$ of its initial magnitude?
   Answer: $0.4\ln(10)$ 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, $t = \frac{-2\ln(10)}{-5} = \frac{2}{5}\ln(10) = 0.4\ln(10)$.
8. A parachutist falls under gravity with air resistance proportional to velocity. The equation of motion is $m\frac{dv}{dt} = mg - kv$, where $m = 70$ kg, $g = 9.8$ m/s², and $k = 14$ kg/s. If the parachutist is dropped from rest, what percentage of terminal velocity is reached after 10 seconds?
   Answer: $100(1 - e^{-2}) \approx 86.5\%$ 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.
9. A resistor-capacitor (RC) circuit with resistance $R = 2$ ohms and capacitance $C = 0.5$ farads is connected to a voltage source $V(t) = 10\sin(t)$ volts. The charge $q(t)$ on the capacitor satisfies $R\frac{dq}{dt} + \frac{q}{C} = V(t)$. What is the steady-state charge amplitude?
   Answer: $\frac{10}{\sqrt{5}}$ coulombs 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.
10. An insulated aluminum bar of mass $m_1$ and specific heat $c_1$ is at an initial temperature $T_1(0)=100$ K. It is brought into thermal contact with a copper block of mass $m_2$ and specific heat $c_2$ at an initial temperature $T_2(0)=300$ K. The rate of heat transfer from the copper to the aluminum is proportional to the temperature difference, $k(T_2 - T_1)$, where $k>0$. Let $D(t) = T_2(t) - T_1(t)$ be the temperature difference. Which differential equation models $D(t)$?
    Answer: $\frac{dD}{dt} = -k \left( \frac{1}{m_1c_1} + \frac{1}{m_2c_2} \right) D$ 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 $D(t)$.
11. An investment account earns interest at a rate of 4% per year, compounded continuously. Money is withdrawn continuously from the account at a rate of $W(t) = 5000 + 100t$ dollars per year, where $t$ is the number of years. To ensure the account has a balance of $500,000 at the end of 10 years, what must the initial investment $P(0)$ be?
    Answer: $187,500 + 287,500e^{-0.4}$ 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 $C$: $P(0) = 187,500 + 287,500e^{-0.4}$.
12. Two tanks are connected in series. Tank 1 initially holds 100 L of pure water. Brine with a salt concentration of 0.5 kg/L is pumped into Tank 1 at a rate of 5 L/min. The well-mixed solution from Tank 1 is pumped into Tank 2 at 5 L/min. Tank 2 initially holds 200 L of pure water, and its well-mixed solution is pumped out at 5 L/min. Let $A_2(t)$ be the amount of salt (in kg) in Tank 2 at time $t$. Which of the following is the correct differential equation for $A_2(t)$?
    Answer: $\frac{dA_2}{dt} = 2.5(1-e^{-t/20}) - \frac{A_2}{40}$ 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 $\frac{dA_2}{dt} = 2.5(1-e^{-t/20}) - \frac{A_2}{40}$.
13. The amount $A(t)$ of a substance in a chemical reaction follows the model $\frac{dA}{dt} = -kA + R$, where $R$ is a constant production rate and $k$ is a positive decay constant. Experimental data shows that $A(0)=10$ grams, $A(1)=8$ grams, and $A(2)=7$ grams. What is the value of the decay constant $k$?
    Answer: $\ln(2)$ 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 $e^{-k} = \frac{8-X}{10-X} = \frac{8-6}{10-6} = \frac{2}{4} = \frac{1}{2}$. $e^{-k} = 1/2 \implies -k = \ln(1/2) = -\ln(2) \implies k = \ln(2)$.
14. A tank initially contains 100 gallons of pure water. A brine solution containing 2 pounds of salt per gallon flows into the tank at a rate of 3 gallons per minute. The well-mixed solution flows out at a rate of 4 gallons per minute. If $S(t)$ represents the amount of salt in the tank at time $t$ minutes, which differential equation correctly models this situation?
    Answer: $\frac{dS}{dt} = 6 - \frac{4S}{100-t}$ The rate of salt entering is (2 lb/gal)(3 gal/min) = 6 lb/min. The volume at time t is 100 + 3t - 4t = 100 - t gallons. The concentration is S/(100-t) lb/gal, so salt leaves at rate (4 gal/min)(S/(100-t) lb/gal) = 4S/(100-t) lb/min. Thus dS/dt = 6 - 4S/(100-t). Choice B uses the wrong outflow rate coefficient. Choice C has the wrong sign in the volume expression. Choice D uses the concentration instead of the rate of salt input.
15. A drug is administered intravenously at a constant rate of $R$ mg/hour. The body eliminates the drug at a rate proportional to the amount present, with a proportionality constant $k=0.1$ hour$^{-1}$. The maximum safe amount of the drug in the body is 20 mg. What is the maximum rate $R$ at which the drug can be administered so that the amount of the drug never exceeds this safe limit?
    Answer: $2$ mg/hour 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.
16. A system is described by the initial value problem $\frac{dy}{dt} = \frac{2y}{t+1} - (t+1)^3$ with an initial condition $y(0) = y_0$, where $y_0$ is a positive constant. What is the long-term behavior of the solution $y(t)$ as $t \to \infty$?
    Answer: $y(t)$ approaches $-\infty$ 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 $-\infty$.
17. An object at $100^{\circ}$C is placed in a large room with an ambient temperature of $20^{\circ}$C. After 10 minutes, the object's temperature is $60^{\circ}$C. The object is then immediately moved to a freezer with an ambient temperature of $0^{\circ}$C. Assuming Newton's Law of Cooling, what is the temperature of the object 20 minutes after it was moved to the freezer?
    Answer: $15^{\circ}$C 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.
18. A lake initially contains $10^7$ m$^3$ of pure water. Water containing a pollutant at a concentration of $0.01$ kg/m$^3$ flows into the lake at a rate of $500$ m$^3$/day. Due to consumption and evaporation, water flows out of the lake at a higher rate of $600$ m$^3$/day. The lake is well-mixed. What is the limiting concentration of the pollutant in the lake as the volume of the lake approaches zero?
    Answer: $0.01$ kg/m$^3$ 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$^3$.

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