Unit 7: Differential Equations

Interpreting Verbal Descriptions of Change as Separable Differential Equations

Translating Words into Mathematical Models

• Start by identifying the dependent variable (the quantity that changes) and the independent variable (often time). The dependent variable will be represented by a symbol such as \(y\) or \(P\), and the independent variable by \(t\) or \(x\). Verbal clues like "population," "temperature," or "amount of substance" tell you what to model as \(y\).
• Look for phrases describing the rate of change. Statements like “rate proportional to the current amount” translate to \(\frac{dy}{dt} = k y\). If the rate depends on the difference from some value, such as “approaches 20°C,” it becomes \(\frac{dT}{dt} = k (20 - T)\).
• When other quantities appear in the description, determine if they are constants or functions of the independent variable. If the rate depends on \(x\) directly, as in “changes at a rate equal to \(3x\) times the current value,” the equation becomes \(\frac{dy}{dx} = 3x\,y\), which is separable.
• Translate “gains” and “losses” into positive or negative terms. For example, “decreases at a rate proportional to the square of its size” becomes \(\frac{dy}{dt} = -k y^2\). The sign of the constant \(k\) matters for determining the growth or decay behavior.
• Common mistake: forgetting to match units or omitting constants. Always check that the units on both sides of the equation are consistent; this ensures your model is dimensionally correct and will integrate properly later.

Sketching Slope Fields and Families of Solution Curves

Constructing Slope Fields

• A slope field for \(\frac{dy}{dx} = f(x,y)\) is built by plotting short line segments at grid points \((x,y)\) with slope equal to \(f(x,y)\). This visualizes the local direction of all possible solutions without solving analytically.
• To create one, choose a grid of points (e.g., integer values for \(x\) and \(y\)), plug each point into \(f(x,y)\) to find the slope, and draw a small segment at that location. Keep segments short so the overall shape of solution curves emerges without clutter.
• Patterns in the slope field reveal qualitative behavior. If slopes are zero along a horizontal line \(y = c\), that line is an equilibrium solution. Positive slopes above and negative slopes below indicate stability, while the reverse indicates instability.
• Example: For \(\frac{dy}{dx} = x - y\), along the line \(y = x\) slopes are zero. Above this line, slopes are negative, and below, they are positive, showing that solutions tend to move toward the line as \(x\) increases.
• AP Tip: For quick sketching on exams, note symmetry and constant-slope lines to reduce computations. Avoid over-plotting; clear, representative slopes are more valuable than a dense, messy field.

Sketching Families of Solution Curves

• A family of solution curves represents all solutions to a differential equation, each determined by a different constant of integration \(C\). On a slope field, these appear as smooth curves tangent to the segments at every point.
• To sketch, start at different initial points and follow the slope segments in both directions, keeping the curve smooth and continuous. Do not cross other solution curves when the uniqueness theorem applies, as each initial point determines exactly one curve.
• Recognize that all curves in the family are related by the same general equation, such as \(y = Ce^{x}\) or \(y = \frac{1}{x + C}\). The constant \(C\) shifts or stretches the curve but preserves its general shape.
• Example: For \(\frac{dy}{dx} = y\), the general solution \(y = Ce^{x}\) produces exponential curves of varying steepness depending on \(C\). On the slope field, these curves all have slopes proportional to their height.
• Common mistake: mixing up different families of curves when multiple equilibria exist. Always verify the form of the general solution before deciding which curves belong to the same family.

Solving Separable Differential Equations to Find General and Particular Solutions

Separation and Integration

• Identify if the equation can be written as \(\frac{dy}{dx} = g(x)h(y)\). If so, rearrange to isolate variables: \(\frac{1}{h(y)}\,dy = g(x)\,dx\). This is the “separation” step and is valid as long as \(h(y) \neq 0\).
• Integrate both sides with respect to their respective variables: \(\int \frac{1}{h(y)}\,dy = \int g(x)\,dx\). Keep the constant of integration on one side only to avoid confusion; combining constants after both integrals is fine.
• Example: \(\frac{dy}{dx} = 2x y^3\). Separate: \(y^{-3}dy = 2x\,dx\). Integrate: \(-\frac{1}{2y^2} = x^2 + C\). Solve explicitly for \(y\) if possible: \(y = \pm \frac{1}{\sqrt{-2x^2 - 2C}}\), with domain restrictions.
• If given an initial condition \(y(x_0) = y_0\), substitute \(x_0\) and \(y_0\) into the general solution to find \(C\). This particular solution represents one curve from the family that matches the given condition.
• Common mistake: dropping the absolute value in logarithmic integrals too soon. For example, \(\int \frac{1}{y}dy = \ln|y| + C\) — the absolute value is needed until the initial condition fixes the sign of \(y\).

Checking and Interpreting Solutions

• Always check by differentiating your solution and substituting into the original differential equation. If the equality holds, your algebra and integration are likely correct.
• Interpret the constant \(C\) in context: in population growth problems, \(C\) can represent the initial scaled size; in cooling problems, it can relate to the starting temperature difference.
• Domain restrictions come from where the separated form was undefined (e.g., division by zero) or where radicals/logarithms require positivity. These restrictions determine where the solution is valid.
• Connection to Unit 6: The process of integrating a separated equation is exactly like finding an antiderivative, but now each side corresponds to a different variable. Many integration techniques from Unit 6 (substitution, properties of integrals) reappear here.
• AP Tip: If time is short and an explicit solution is messy, leave your answer in implicit form unless the prompt specifies “solve for \(y\).” This can still earn full credit on FRQs.

Exponential Growth and Decay Models

Deriving \(\frac{dy}{dt}=ky\) and Solving Step by Step

• Many real processes change at a rate proportional to the current amount, which reads as “rate equals constant times amount.” Translating this yields \(\frac{dy}{dt}=k y\), where \(k\) carries units of per time and its sign decides growth or decay. This sentence-to-equation move is the most important modeling step and must match the wording in the prompt.
• Separate variables to prepare for integration: \(\frac{1}{y}\,dy = k\,dt\). Integrate both sides to obtain \(\ln|y| = kt + C\), which captures the cumulative effect of a constant proportional rate over time. The absolute value appears because the antiderivative of \(1/y\) is \(\ln|y|\) and you will handle sign with the initial condition.
• Exponentiate to solve for \(y\): \(y = Ce^{kt}\) with \(C=\pm e^{C_0}\). Use the initial value \(y(0)=y_0\) to find \(C=y_0\), giving \(y(t)=y_0 e^{kt}\). This solution behaves smoothly, and the direction and speed of change follow from the sign and magnitude of \(k\).
• Mini-example: A culture satisfies \(\frac{dP}{dt}=0.12P\) with \(P(0)=500\). Then \(P(t)=500e^{0.12t}\) and the doubling time is \(T_d=\frac{\ln 2}{0.12}\). If the question gives a target population, solve \(500e^{0.12t}=P_{\text{target}}\) for \(t\) with natural logs, and include units in the final time.
• Common mistakes include mixing discrete and continuous models, forgetting that \(k\) has units of per time, and dropping the absolute value before using the initial condition. Always confirm that the interpreted \(k\) matches the context, for example half-life implies negative \(k\) and doubling time implies positive \(k\).

Parameter Identification, Half-life, Doubling Time, and Model Checking

• Given two measurements \((t_1,y_1)\) and \((t_2,y_2)\), divide the solution forms to eliminate \(C\): \(\frac{y_2}{y_1}=e^{k(t_2-t_1)}\). Solve for the proportional rate \(k=\frac{1}{t_2-t_1}\ln\!\left(\frac{y_2}{y_1}\right)\) and check that \(k\) carries units of per time, which ensures dimensional consistency. This formula is robust and avoids algebra with large exponentials.
• Half-life \(T_{1/2}\) is the time for a quantity to fall to half, so \(e^{kT_{1/2}}=\tfrac{1}{2}\). Solving gives \(T_{1/2}=\frac{\ln 2}{|k|}\) when \(k<0\), which matches physics and chemistry usage. Doubling time for growth is \(T_d=\frac{\ln 2}{k}\) when \(k>0\), and these serve as quick interpretation tools for graphs and tables.
• To convert a discrete percent rate \(r\) per period to a continuous rate, use \(k=\ln(1+r)\). This prevents mixing the discrete model \(y=P_0(1+r)^n\) with the continuous model \(y=P_0e^{kt}\), which can change answers significantly. Always decide which model the context implies before computing.
• Model checking involves comparing predicted values \(y_0e^{kt}\) to given data and inspecting residuals. If deviations grow systematically, a simple proportional model may not fit and suggests saturation or external inputs that point to logistic or piecewise models. On FRQs, a one sentence justification about residual direction earns interpretation credit.
• Mini-example: A substance decays from 80 to 50 in 3 hours. Compute \(k=\frac{1}{3}\ln(50/80)\approx -0.1633\ \text{hr}^{-1}\). Predict at 10 hours with \(y(10)=80e^{-0.1633\cdot 10}\) and round only at the end to maintain accuracy.

Logistic Growth Models

Deriving the Logistic Equation and Understanding Carrying Capacity

• When resources limit growth, verbal descriptions often say the rate is proportional to both the current population and the remaining capacity. Translate this to \(\frac{dP}{dt}=kP\left(M-P\right)\), where \(M\) is the carrying capacity and \(k>0\) sets the growth scale. This model reduces to exponential growth for small \(P\) but slows as \(P\) approaches \(M\).
• Key features follow directly from the right hand side. Rate is zero at \(P=0\) and \(P=M\), so those are equilibria. The rate is positive for \(0M\), which explains why solutions rise toward \(M\) from below and fall toward \(M\) from above.
• Solving by separation is possible but algebraic, and AP Calc AB usually emphasizes interpretation and parameter use rather than the explicit formula. The classic explicit solution is \(P(t)=\frac{M}{1+Ae^{-kMt}}\) where \(A=\frac{M-P(0)}{P(0)}\), and you can verify it by differentiation if asked. The constants have clear meanings that connect initial level and long term behavior.
• Mini-example: Suppose \(\frac{dP}{dt}=0.6P(500-P)\) with \(P(0)=50\). Since \(0
• Common pitfalls are treating logistic exactly like exponential at all scales, ignoring the stable equilibrium at \(M\), or failing to discuss units for \(k\) and \(M\). Always state that \(M\) has the same units as \(P\) and that \(k\) has per time per unit population when written in the \(kP(M-P)\) form.

Parameter Identification and Interpreting Data

• From a graph or table, identify the carrying capacity \(M\) as the horizontal level approached by solutions as \(t\) grows. Check that the slope decreases as the curve nears \(M\) because the factor \(M-P\) shrinks. This visual check distinguishes logistic from pure exponential growth.
• The maximum growth rate occurs at \(P=M/2\) where the parabola \(P(M-P)\) attains its peak. If data show the steepest slope near a value, that value is a good estimate for \(M/2\), which lets you infer \(M\). This is a common qualitative reasoning step on FRQs.
• To estimate \(k\), measure an early time interval where \(P\ll M\) so the equation behaves like \(\frac{dP}{dt}\approx kMP\). Then fit a short exponential segment to estimate \(kM\) and divide by \(M\). State that this is an approximation valid before crowding effects dominate.
• Mini-example: If data approach 1200 and the steepest observed slope is near \(P\approx 600\), infer \(M\approx 1200\). If over the time from \(t=0\) to \(t=5\) population roughly doubles while still small, estimate \(kM\approx \ln 2/5\) and compute \(k\approx \frac{\ln 2}{5M}\).
• Model validation includes comparing predicted slopes \(\frac{dP}{dt}\) to finite difference slopes from data. If the predicted peaks near \(M/2\) but the data show the peak elsewhere, consider that the assumed \(M\) or \(k\) may be off or that external factors are present.

Equilibrium Solutions and Stability Analysis

Finding Equilibria and Classifying Stability

• An equilibrium solution is any constant function \(y=c\) that satisfies \(f(x,c)=0\) for a first order equation \(\frac{dy}{dx}=f(x,y)\). To find these, fix \(x\) as a parameter and solve \(f(x,c)=0\) for constants \(c\), then confirm that substitution produces zero for all \(x\) in the domain. These horizontal solution lines organize the dynamics of the system.
• Stability describes how nearby solutions behave relative to an equilibrium as \(x\) increases. If nearby slopes point toward the equilibrium line, it is stable, if away then it is unstable, and if toward on one side but away on the other then it is semistable. You can decide this from the sign of \(f\) just above and below the equilibrium.
• Mini-example: For \(\frac{dy}{dx}=y(3-y)\), the equilibria are \(y=0\) and \(y=3\). For \(03\) the right hand side is negative so solutions decrease, which makes \(y=3\) stable and \(y=0\) unstable. A quick sign diagram in \(y\) communicates this logic clearly.
• Connection to slope fields is straightforward because equilibria appear as rows of zero slope. Families of solution curves bend toward stable lines and away from unstable lines, which helps you sketch credible qualitative solutions even without formulas. This mirrors first derivative tests for function behavior from earlier units.
• Common pitfalls include forgetting to check both sides of the equilibrium, assuming stability from a single sample slope, or ignoring domain limits where \(f\) is undefined. Always justify your classification with a two sided sign analysis or a brief reference to the slope field pattern.

Using Slope Fields and Sign Charts to Predict Long Term Behavior

• To predict the limit of \(y(x)\) as \(x\) grows, combine the slope field with knowledge of equilibria. If all nearby curves drift to a particular constant line, then solutions with initial values in that band will approach that equilibrium. This conclusion must match the sign pattern of \(f\) around that line.
• For autonomous equations \(\frac{dy}{dx}=g(y)\), stability reduces to a one dimensional sign chart in \(y\). Mark zeros of \(g\), test intervals for sign, and note the direction of motion in \(y\), then read long term limits by following arrows. This is fast and reliable under exam time pressure.
• Mini-example: With \(\frac{dy}{dx}=y(1-y)(2-y)\) the equilibria are \(0,1,2\). Testing intervals shows solutions between 0 and 1 move up, between 1 and 2 move down, and above 2 move up, which makes 0 and 2 unstable and 1 stable. A short paragraph explaining the sign chart earns clarity points.
• When parameters vary, describe how stability changes by tracking the factors that control the sign of \(f\). For instance, in a logistic model changing the carrying capacity moves the stable equilibrium and can change where the fastest growth occurs. This sensitivity analysis links calculus to modeling decisions.
• Common mistakes include concluding convergence only from decreasing slopes in \(x\) rather than from the sign of \(f\) relative to \(y\). Always ground claims about long term behavior in the direction of motion indicated by the differential equation rather than the visual steepness alone.

Applications of Differential Equations

Motion Problems

• Motion problems involving acceleration or velocity often lead to a differential equation \(\frac{dv}{dt}=a(t,v)\) or \(\frac{dx}{dt}=v(t,x)\). The variable \(t\) is time, \(x\) is position, and \(v\) is velocity, with acceleration given either as a function of time or depending on velocity/position. The task is to integrate step-by-step to recover velocity and then position, applying initial conditions at each stage.
• Example: A particle has acceleration \(a(t)=4t\) m/s\(^2\) and initial velocity \(v(0)=3\) m/s. Integrate: \(v(t)=2t^2+3\). If \(x(0)=5\) m, then \(x(t)=\frac{2}{3}t^3+3t+5\). This layered integration mirrors Unit 6’s net change applications but uses DE notation explicitly.
• Always keep track of units: acceleration (m/s\(^2\)) integrates to velocity (m/s), which integrates to position (m). This double integration approach works for constant or variable acceleration, and is a standard AP problem type.
• Connection: Motion DEs link back to Unit 6 because they require evaluating integrals of rate functions to determine accumulated change. In Unit 8, similar reasoning applies to physics-based work/energy problems.
• Common mistake: skipping the constant of integration at the velocity or position stage, which leads to wrong final answers when initial conditions are given. Solve for constants immediately after each integration step.

Mixing Problems

• Mixing problems model the change in concentration or amount of a solute in a tank as a DE: \(\frac{dQ}{dt} = \text{rate in} - \text{rate out}\). Here \(Q(t)\) is the amount of solute, and rates are computed as flow rate (volume/time) × concentration (amount/volume).
• Example: A tank contains 100 L of saltwater with 10 g of salt. Fresh water enters at 5 L/min, and the mixture leaves at the same rate. Since rate in is \(0\) g/min and rate out is \(\frac{Q(t)}{100} \cdot 5\), the DE is \(\frac{dQ}{dt}=-\frac{1}{20}Q\). Solve: \(Q(t)=10e^{-t/20}\) g.
• This is an exponential decay model with \(k = -\frac{1}{20}\). The solution approaches 0 as \(t \to \infty\), meaning the tank eventually contains pure water. The exponential form comes directly from separation of variables.
• Mixing DEs link to Unit 6 concepts because they integrate a net rate over time, and to Unit 8 because they often appear in environmental or population decay/growth contexts. Recognizing them quickly can save time on FRQs.
• Common mistake: forgetting that concentration changes as \(Q(t)\) changes — never treat it as constant unless the problem explicitly says so. Recalculate concentration each time it’s used in the DE.

Population Models

• Population models often start with exponential growth/decay or logistic growth DEs: \(\frac{dP}{dt} = kP\) or \(\frac{dP}{dt}=kP(M-P)\). Initial conditions provide the starting population \(P(0)\), which determines constants in the explicit solution.
• Example: A population satisfies \(\frac{dP}{dt} = 0.08P(1000-P)\) with \(P(0)=50\). The logistic model predicts an eventual population near 1000, with the fastest growth when \(P\approx 500\). This insight comes from analyzing equilibria and concavity of solutions.
• Real-world interpretation: The rate of change depends on both current size and environmental limits. Early growth mimics exponential behavior, mid-range growth accelerates, and late-stage growth slows near the carrying capacity.
• Link to Unit 6: Integrating \(\frac{dP}{dt}\) gives total population change over an interval. Link to Unit 8: Logistic and exponential forms can be used in optimization or average value contexts.
• Common mistake: applying exponential formulas when logistic is needed (e.g., when data levels off). Always check if growth rate decreases as the population approaches a maximum.

Euler’s Method

Step-by-Step Numerical Approximation Process

• Euler’s Method approximates solutions to \(\frac{dy}{dx}=f(x,y)\) starting from an initial point \((x_0,y_0)\). You move in small steps \(h\) along \(x\), estimating the slope at each step and using it to predict the next \(y\)-value: \(y_{\text{new}} = y_{\text{old}} + h \cdot f(x_{\text{old}},y_{\text{old}})\).
• Set up a table with columns for \(x_n\), \(y_n\), \(f(x_n,y_n)\), and the update formula. Start with \(n=0\) at \((x_0,y_0)\), compute the slope from the DE, multiply by \(h\), and add to the current \(y\) to get the next value. Repeat until reaching the desired \(x\).
• Example: Approximate \(y(0.4)\) for \(\frac{dy}{dx} = x + y\), \(y(0) = 1\), \(h=0.2\). Step 1: slope = \(0+1=1\), new \(y=1+0.2(1)=1.2\). Step 2: slope = \(0.2+1.2=1.4\), new \(y=1.2+0.2(1.4)=1.48\). So \(y(0.4) \approx 1.48\).
• Euler’s approximations improve as \(h\) decreases, but require more steps. In AP settings, you typically use the given \(h\) without refinement unless asked. Showing the table earns method points even if arithmetic errors occur later.
• Common mistake: forgetting to update \(x\) each time or using \(x_{\text{new}}\) in the slope instead of \(x_{\text{old}}\). Always evaluate slope at the current point before moving forward.

Error Prediction and Direction-of-Change Reasoning

• Error in Euler’s Method is influenced by the concavity of the true solution. If the solution curve is concave up, Euler’s Method underestimates; if concave down, it overestimates. This comes from the tangent line lying below or above the curve locally.
• On the AP exam, you may be asked to predict over/underestimation without computing. Use the sign of the second derivative \(\frac{d^2y}{dx^2}\) (found by differentiating \(f(x,y)\) with respect to \(x\) along the solution) to decide the direction.
• Example: For \(\frac{dy}{dx} = x+y\), \(y'' = 1 + \frac{dy}{dx} = 1 + (x+y)\), which is positive near the initial condition, meaning concave up and therefore an underestimation. You can state this with a brief justification for full credit.
• Reducing \(h\) decreases local error and improves approximation accuracy. Halving \(h\) often cuts error by a factor of about 4 for smooth solutions, but AP questions rarely require error magnitude calculations.
• Connection: Euler’s step-by-step accumulation is similar to Riemann sum reasoning from Unit 6 — both approximate an accumulated change using rate information over small intervals, but Euler applies it to functions defined implicitly by a DE.

Practice Problems

Problem 1 (Separable DE, Net Change, Long-Term Limit)

• A tank holds a constant volume of \(120\) L of brine. Inflow is \(6\) L/min with salt concentration \(2\) g/L, and the well-stirred mixture leaves at \(6\) L/min. Let \(Q(t)\) be the amount of salt (g) in the tank at time \(t\) minutes. Build the differential equation from a verbal rate statement and solve for \(Q(t)\) using separation/linear methods.
• (a) Write the rate-in/rate-out model \(\displaystyle \frac{dQ}{dt}=\text{(rate in)}-\text{(rate out)}\) and specify each term in units. Use the outflow concentration \(\frac{Q(t)}{120}\) g/L times the outflow \(6\) L/min to express the removal rate.
• (b) With initial salt \(Q(0)=30\) g, find the explicit solution \(Q(t)\). Show how the solution’s constant and the steady-state value are determined and interpret the meaning of the exponential term for transient behavior.
• (c) Compute the time when the tank first reaches \(60\) g of salt. Solve your equation algebraically for \(t\) and report the result to three significant figures with correct units. Briefly justify why the solution time is small/large based on the parameters.
• (d) Find the average amount of salt in the tank during the first \(10\) minutes: \(\displaystyle \frac{1}{10}\int_0^{10}Q(t)\,dt\). Evaluate the integral exactly (leaving \(e^{-0.5}\) if you prefer) and give a numerical approximation with units.

Solution to Problem 1

• (a) Rate in: \(6\ \text{L/min}\times 2\ \text{g/L}=12\ \text{g/min}\). Rate out: \(6\ \text{L/min}\times \frac{Q(t)}{120}\ \text{g/L}=\frac{Q(t)}{20}\ \text{g/min}\). Differential equation: \(\displaystyle \frac{dQ}{dt}=12-\frac{Q}{20}\) with consistent units of g/min; this linear DE encodes “in minus out.”
• (b) Solve \(\displaystyle \frac{dQ}{dt}+ \frac{1}{20}Q=12\). The steady-state (equilibrium) satisfies \(0=12-\frac{Q_\infty}{20}\Rightarrow Q_\infty=240\ \text{g}\). General solution: \(Q(t)=240+\left(Q_0-240\right)e^{-t/20}\), hence with \(Q_0=30\) g we get \(\boxed{Q(t)=240-210\,e^{-t/20}}\). The exponential term decays to \(0\), so \(Q(t)\to 240\) g as \(t\to\infty\).
• (c) Set \(60=240-210e^{-t/20}\Rightarrow e^{-t/20}=\frac{180}{210}=\frac{6}{7}\). Then \(-\frac{t}{20}=\ln\!\left(\frac{6}{7}\right)\) so \(\boxed{t=20\ln\!\left(\frac{7}{6}\right)\approx 3.08\ \text{min}}\). The short time makes sense: inflow adds salt quickly (12 g/min) while early outflow removal is weak because the tank is initially dilute.
• (d) \(\displaystyle \int_0^{10}Q(t)\,dt=\int_0^{10}\!\big(240-210e^{-t/20}\big)\,dt=240(10)-210\!\left[-20e^{-t/20}\right]_0^{10}\). This equals \(2400+4200\big(e^{-1/2}-1\big)\approx 2400+4200(0.60653-1)\approx 747.43\ \text{g·min}\). Average: \(\boxed{\frac{1}{10}\int_0^{10}Q(t)\,dt\approx 74.743\ \text{g}}\), consistent with rising toward 240 g but still early in the process.
• Checks & connections: The DE is separable/linear, the limit \(240\) g is the Unit 7 equilibrium, and the average value computation mirrors Unit 6. A common mistake is to treat the outflow concentration as constant; it must be \(Q/120\), otherwise the model is dimensionally and conceptually incorrect.

Problem 2 (Logistic Modeling with Parameter Fit and Euler’s Method)

• A population \(P(t)\) follows logistic dynamics with carrying capacity \(M=1000\). Assume the AB-friendly form \(\displaystyle \frac{dP}{dt}=rP\!\left(1-\frac{P}{M}\right)\) with unknown growth parameter \(r\), and initial value \(P(0)=80\). A measurement gives \(P(5)=120\). Determine \(r\), predict when \(P\) reaches \(600\), and approximate \(P(2)\) via Euler’s Method with step \(h=1\).
• (a) Use the explicit logistic solution \(P(t)=\displaystyle \frac{M}{1+A e^{-rt}}\) with \(A=\frac{M-P_0}{P_0}\). Compute \(A\) from the initial condition and solve for \(r\) using the data at \(t=5\), showing clean algebra and logarithms.
• (b) Solve for the time \(t\) when \(P(t)=600\). Invert the logistic formula to isolate the exponential and apply the value of \(r\) from part (a). Comment briefly on why the time to 600 is much larger than to 120 given logistic slowing near \(M\).
• (c) Perform two Euler steps of size \(h=1\) starting at \(t=0\) to approximate \(P(2)\). Use the update \(P_{n+1}=P_n+h\cdot rP_n\!\left(1-\frac{P_n}{M}\right)\) and keep three decimal places during intermediate calculations to control rounding error.
• (d) Decide whether the Euler estimate is an over- or under-approximation at early times by reasoning about concavity. Justify your claim using the sign of the second derivative for the logistic model or a qualitative slope-field argument.

Solution to Problem 2

• (a) With \(M=1000\) and \(P_0=80\), \(A=\dfrac{1000-80}{80}=11.5\). Using \(P(5)=120\): \(120=\dfrac{1000}{1+11.5e^{-5r}}\Rightarrow 1+11.5e^{-5r}=\dfrac{1000}{120}=\dfrac{25}{3}\). Thus \(11.5e^{-5r}=\dfrac{25}{3}-1=\dfrac{22}{3}\Rightarrow e^{-5r}=\dfrac{22}{3\cdot 11.5}=\dfrac{22}{34.5}\approx 0.637682\), so \(\boxed{r=\frac{-\ln(0.637682)}{5}\approx 0.0900\ \text{per time unit}}\).
• (b) Set \(600=\dfrac{1000}{1+11.5e^{-rt}}\Rightarrow 1+11.5e^{-rt}=\dfrac{5}{3}\Rightarrow 11.5e^{-rt}=\dfrac{2}{3}\Rightarrow e^{-rt}=\dfrac{2}{3\cdot 11.5}\approx 0.057971\). Then \(t=\dfrac{-\ln(0.057971)}{r}\approx \dfrac{2.847}{0.0900}\approx \boxed{31.6\ \text{time units}}\). The time is much longer than to reach 120 because crowding slows growth as \(P\) moves toward \(M\).
• (c) Euler with \(h=1\): \(f(P)=rP\!\left(1-\frac{P}{1000}\right)\), \(r\approx 0.09\), \(P_0=80\). Step to \(t=1\): \(f(P_0)=0.09\cdot 80\cdot(1-0.08)=0.09\cdot 80\cdot 0.92=6.624\Rightarrow P_1=80+1\cdot 6.624=86.624\). Step to \(t=2\): \(f(P_1)=0.09\cdot 86.624\cdot\left(1-0.086624\right)\approx 0.09\cdot 86.624\cdot 0.913376\approx 7.127\Rightarrow \boxed{P_2\approx 93.751}\).
• (d) Early-time concavity for logistic is upward because \(y''=r\!\left(1-\frac{2P}{M}\right)y'\) and initially \(P\ll \frac{M}{2}\) makes \(1-\frac{2P}{M}>0\). A tangent line lies below a concave-up curve, so forward Euler underestimates the true solution for small \(t\). Therefore \(P_2\approx 93.751\) is a slight under-approximation of the exact \(P(2)\).
• Checks & connections: Part (a) used parameter fitting from Unit 7 modeling, part (b) inverted the explicit logistic to solve for time, and part (c) numerically approximated the DE with Euler’s Method. The average/accumulation ideas from Unit 6 support interpreting total change, while Units 8 applications echo the same modeling logic in new physical contexts.