Unit 7: Differential Equations

Modeling Situations with Differential Equations

• Differential equations describe how a quantity changes with respect to another variable, often time, by relating a function to its derivatives. In modeling, the derivative represents the rate of change, and the equation expresses the relationship between the rate and the quantity itself or other variables. For example, population growth, radioactive decay, and velocity under constant acceleration can all be modeled with differential equations.
• To create a model, identify what quantity is changing, determine the variables involved, and establish how the rate of change depends on these variables. This often requires translating a real-world context into mathematical form, such as \( \frac{dy}{dt} = ky \) for exponential growth. Once the equation is formed, solving it will reveal a function that predicts future behavior.
• Understanding the underlying physical, biological, or economic process is critical because it guides whether the model should be linear, exponential, logistic, or something else. The model must also specify constraints like initial values or physical limits. Without realistic assumptions, even a mathematically correct differential equation might produce unrealistic results.
• One of the strengths of differential equations is their flexibility: they can model continuous growth, decay, oscillations, and even chaotic systems. However, the more complex the relationship between variables, the more advanced techniques are needed to solve the equation. For AP Calculus AB, most models are first-order and separable.
• Always check units when creating a differential equation. The units on both sides must match, since a derivative like \( \frac{dy}{dt} \) inherently has “output units per input units.” This consistency ensures the model is dimensionally correct and interpretable.

Verifying Solutions for Differential Equations

• Verifying a solution means confirming that a given function actually satisfies a given differential equation. This is done by substituting the function into the equation and checking whether the equality holds. For example, if given \( y' = 3y \) and a proposed solution \( y = Ce^{3x} \), you differentiate \( y \) and confirm that \( y' \) equals \( 3y \).
• This process is essential because not every function that “looks right” will satisfy the equation exactly. Even small errors in constants, exponents, or coefficients can cause the equation to fail when tested. Always substitute back into the original differential equation, not a manipulated or simplified version.
• When verifying, also check that the solution meets any given initial condition, such as \( y(0) = 5 \). This ensures the solution matches the specific case intended, not just the general solution. Skipping this step may result in a mathematically valid function that is physically irrelevant to the scenario.
• Sometimes, the process of verification reveals that a function is a solution only for specific parameter values. This can occur when constants or coefficients are involved in the equation. Recognizing this dependency can help connect the mathematics back to the real-world interpretation.
• Verifying solutions is also a valuable way to catch algebra or differentiation mistakes early. By systematically substituting and checking, you confirm not only that the function works but also that you applied differentiation rules correctly.

Slope Fields and Solution Curves

• Slope fields (or direction fields) provide a visual representation of a differential equation without solving it analytically. Each point \((x,y)\) in the plane has a small line segment whose slope equals the value of \( \frac{dy}{dx} \) at that point. This gives a picture of the “direction” a solution curve would follow through that point.
• They are especially useful when a differential equation cannot be solved easily, or when multiple solution curves exist depending on initial conditions. By observing the slope field, you can identify general trends, such as whether solutions increase, decrease, or approach asymptotes.
• To sketch a slope field, pick a set of grid points and compute the slope from the differential equation for each point. Then draw a small segment with that slope. Patterns will emerge, such as parallel lines for constant slope or curves bending toward a particular equilibrium.
• Solution curves are continuous paths that move along the slope segments of a slope field. An initial condition \(y(x_0) = y_0\) determines which specific curve the solution follows. This makes slope fields an important bridge between the abstract equation and its concrete behavior.
• Interpreting slope fields connects directly to understanding qualitative behavior. For example, logistic growth slope fields show slopes that flatten at large \(y\), indicating a limiting capacity. Recognizing these patterns helps interpret real-world meaning without explicit formulas.

Separation of Variables to Solve Differential Equations

• Separation of variables is a method used to solve first-order differential equations where the derivative can be expressed as a product of a function of \(x\) and a function of \(y\). The goal is to rewrite the equation so all \(y\)-terms (including \(dy\)) are on one side and all \(x\)-terms (including \(dx\)) are on the other. This rearrangement allows both sides to be integrated independently, producing an implicit or explicit solution.
• After separating, integrate both sides with respect to their respective variables. Often, integration will introduce an arbitrary constant \(C\), representing the family of solutions. If an initial condition is given, substitute it into the equation to determine the specific value of \(C\).
• The method works best when the equation is already separable or can be algebraically manipulated to become separable. If the derivative cannot be split into a product of separate functions of \(x\) and \(y\), this method will not work directly. In AP Calculus AB, most solvable differential equations are intentionally chosen to be separable.
• One advantage of this technique is that it preserves the relationship between variables in the form of an equation rather than an isolated solution, which can make it easier to analyze. This is especially useful for modeling physical or biological systems where the variables have interconnected meaning.
• Even when the integration produces logarithmic, exponential, or trigonometric results, the principle remains the same: isolate the variables, integrate, and apply conditions if provided. This consistency makes separation of variables a powerful tool for solving many real-world problems in calculus.

Particular Solutions Involving Initial Conditions

• When solving a differential equation, the general solution contains an arbitrary constant \(C\) because integration is indefinite. A particular solution is found by using an initial condition, such as \(y(x_0) = y_0\), to determine a specific value for \(C\). This process ensures the solution matches the exact situation described in the problem.
• To find the particular solution, substitute the given \(x_0\) and \(y_0\) into the general solution. Solving for \(C\) fixes the curve so it passes through the given point, which defines the unique solution for that initial condition. This step is essential in applications where a specific starting state is known.
• Particular solutions are often necessary for predictive modeling. For example, in a cooling problem using Newton’s Law of Cooling, the general equation models all possible temperature curves, but the particular solution matches the one specific object cooling from a given temperature.
• When no initial condition is given, the general solution represents a family of curves rather than a single prediction. These curves are all valid mathematically, but they may not be realistic without the context of an initial condition.
• In AP Calculus AB, particular solutions appear frequently in both free-response and multiple-choice questions. They not only test your integration skills but also your ability to interpret and apply given conditions correctly.

Exponential Models with Differential Equations

• Exponential growth and decay models are based on the principle that the rate of change of a quantity is proportional to the quantity itself. This relationship is expressed as \( \frac{dy}{dt} = ky \), where \(k > 0\) for growth and \(k < 0\) for decay. Such models describe phenomena like population growth, radioactive decay, and continuously compounded interest.
• Solving \( \frac{dy}{dt} = ky \) through separation of variables produces \( y = Ce^{kt} \), where \(C\) is the initial amount and \(k\) is the proportionality constant. The sign and magnitude of \(k\) control the speed and direction of change, making it a critical parameter to interpret.
• In growth scenarios, the quantity increases without bound if \(k > 0\), while in decay scenarios, the quantity approaches zero as \(t \to \infty\). This behavior makes exponential models ideal for describing processes that either explode in size or diminish rapidly over time.
• To apply these models, use the given data (often two points in time) to determine \(C\) and \(k\). Once these constants are found, the model can predict future or past values with high accuracy—provided the proportional rate assumption remains valid.
• Exponential models connect directly to logarithmic equations because solving for \(k\) often requires taking the natural log of a ratio. Recognizing this connection makes solving application problems more straightforward.

Interpreting Verbal Descriptions of Change as Separable Differential Equations

• Many problems describe change in words rather than in explicit equations. The key is to translate these descriptions into a mathematical form, often a separable differential equation. For example, "a population grows at a rate proportional to its size" translates to \( \frac{dy}{dt} = ky \).
• Identifying proportional relationships, constants, and variable dependencies in the description is the first step. Words like “proportional to,” “inversely proportional to,” or “directly related to” signal the type of mathematical expression to use.
• Once the verbal statement is translated into \( \frac{dy}{dt} = f(y)g(t) \) form, the separation of variables method can be applied. This requires algebraic manipulation to isolate the dependent and independent variables.
• Clear interpretation of the context ensures the mathematical model matches reality. For example, in a decay problem, the proportionality constant \(k\) should be negative to reflect decreasing behavior.
• This skill is tested frequently in AP problems where the description is intentionally vague, requiring strong translation skills to produce a solvable equation.

Deriving and Applying a Model for Exponential Growth and Decay

• The derivation starts from the assumption that the rate of change is proportional to the quantity itself: \( \frac{dy}{dt} = ky \). This proportionality leads directly to an exponential solution because separating variables and integrating produces \( \ln|y| = kt + C \).
• Exponentiating both sides yields \( y = Ce^{kt} \), where \(C\) is the initial amount. The constant \(k\) determines the growth rate (positive) or decay rate (negative) and is often found using an additional data point.
• Applying the model involves substituting given values into the formula to find unknowns like \(k\) or to make predictions for future times. For example, doubling time and half-life are direct consequences of solving for \(t\) in the equation \( y = Ce^{kt} \).
• Understanding the meaning of \(C\) and \(k\) is essential for interpreting the model’s behavior. \(C\) sets the starting level, while \(k\) controls how quickly the quantity changes.
• Because exponential models are widely used in science, economics, and engineering, mastering their derivation and application ensures a solid foundation for advanced study and real-world problem solving.

Logistic Growth Models

• Logistic growth models describe situations where growth is initially exponential but slows as the quantity approaches a maximum carrying capacity \(L\). The rate of change is proportional both to the current size of the population and to the remaining capacity: \( \frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right) \). This type of model is realistic for populations limited by resources, space, or other environmental constraints.
• The differential equation is separable, and solving it produces \( y = \frac{L}{1 + Ae^{-kt}} \), where \(A\) is determined by the initial condition. This equation predicts that \(y\) will increase quickly at first, then slow down, and eventually level off at \(L\) as \(t \to \infty\).
• The inflection point occurs at \(y = \frac{L}{2}\), meaning the population grows fastest when it is half of the carrying capacity. Before this point, growth accelerates; after it, growth decelerates. This characteristic "S-shaped" (sigmoidal) curve distinguishes logistic growth from exponential growth.
• In AP Calculus AB problems, logistic models often require identifying the carrying capacity, solving for \(k\), or predicting long-term behavior. You may also be asked to interpret the rate of change \( \frac{dy}{dt} \) at specific population levels.
• Because logistic growth accounts for environmental limitations, it is widely used in ecology, epidemiology, and social sciences. Understanding the derivation and solution process ensures you can adapt it to a variety of contexts beyond population modeling.

Slope Fields and Numerical Approximations

• Slope fields (direction fields) are a visual way to represent a differential equation without solving it analytically. Each point \((x, y)\) on the graph has a small segment with slope equal to the value of \(\frac{dy}{dx}\) from the differential equation. By following these slopes, you can sketch possible solution curves that pass through given points, helping visualize the family of solutions.
• To construct a slope field, you calculate the slope from the differential equation at several \((x, y)\) points and draw short, equal-length segments with the correct slope. Patterns emerge, showing how solutions behave globally — for example, converging toward an equilibrium or diverging to infinity.
• Numerical approximation methods, like Euler’s Method, allow you to estimate solution values when an exact solution is difficult or impossible to find. Euler’s Method uses the slope from the differential equation to predict the next \(y\) value: \( y_{n+1} = y_n + h \cdot f(x_n, y_n) \), where \(h\) is the step size.
• The smaller the step size \(h\), the more accurate the approximation, but it also increases computational effort. In AP Calculus AB, problems may ask you to perform several steps of Euler’s Method manually, so practicing with different \(h\) values is important.
• Slope fields and numerical approximations connect visual interpretation and computational estimation, bridging the gap between theory and applied analysis. They are particularly useful in modeling real-world systems where exact equations for \(y(t)\) are either too complex or unknown.

Practice Problems – Differential Equations

Problem 1

A population of bacteria grows at a rate proportional to its current size and doubles every 6 hours. Initially, there are 500 bacteria.
(a) Write a differential equation to model this situation.
(b) Solve the equation to find an explicit formula for \(P(t)\), the population after \(t\) hours.
(c) Find the population after 18 hours.

Solution:

(a) Since growth is proportional to population: \(\frac{dP}{dt} = kP\).
(b) Separate: \(\frac{dP}{P} = k\,dt\), integrate: \(\ln|P| = kt + C\).
Exponentiate: \(P(t) = Ae^{kt}\). Using \(P(0) = 500\), \(A = 500\). Doubling in 6 hours means \(P(6) = 1000 = 500e^{6k} \Rightarrow e^{6k} = 2 \Rightarrow k = \frac{\ln 2}{6}\). Final formula: \(P(t) = 500e^{(\ln 2/6)t}\).
(c) For \(t = 18\): \(P(18) = 500e^{(\ln 2/6)(18)} = 500e^{3\ln 2} = 500 \cdot 2^3 = 4000\) bacteria.

Problem 2

A tank contains 100 liters of water with 20 grams of salt dissolved in it. Brine with a concentration of 1 gram/liter flows in at 5 liters/min, and the mixture is kept uniform and flows out at the same rate. Let \(S(t)\) be the amount of salt in the tank (grams) at time \(t\) (minutes).
(a) Write the differential equation for \(S(t)\).
(b) Solve for \(S(t)\) given the initial condition.
(c) Find the limiting amount of salt in the tank as \(t \to \infty\).

Solution:

(a) Rate in: \(5 \cdot 1 = 5\) g/min. Rate out: \(5 \cdot \frac{S}{100} = \frac{S}{20}\) g/min. DE: \(\frac{dS}{dt} = 5 - \frac{S}{20}\).
(b) Solve: \(\frac{dS}{dt} + \frac{S}{20} = 5\). Integrating factor: \(e^{t/20}\). Multiply: \(e^{t/20} \frac{dS}{dt} + \frac{S}{20}e^{t/20} = 5e^{t/20}\). Integrate: \(S e^{t/20} = 100 e^{t/20} + C\). \(S(t) = 100 + Ce^{-t/20}\). Using \(S(0) = 20\): \(20 = 100 + C \Rightarrow C = -80\). Final: \(S(t) = 100 - 80e^{-t/20}\).
(c) As \(t \to \infty\), \(e^{-t/20} \to 0\), so \(S(t) \to 100\) grams.

Common Misconceptions

• Many students confuse the rate of change with the actual amount of the quantity. In problems like \(\frac{dy}{dt} = ky\), \(y\) is the amount, and \(\frac{dy}{dt}\) is the growth rate. Forgetting this distinction leads to setting up incorrect equations or mixing up variables during separation.
• When working with separable differential equations, students often forget to include the constant of integration \(C\) after integrating. Omitting \(C\) makes it impossible to apply initial conditions correctly, which results in the wrong particular solution.
• Some assume that slope fields give exact solutions. Slope fields are only a graphical representation of the direction of solutions; they help visualize trends but do not replace solving the equation or using numerical approximations for precise values.
• In exponential growth/decay problems, students sometimes incorrectly interpret the constant \(k\) as the rate of change in the same units as the variable. In reality, \(k\) is a proportionality constant that determines the relative growth or decay rate and is tied to the time units in the problem.
• When applying numerical methods like Euler’s Method, students may misapply the formula by using \(y_{n+1} = y_n + h \cdot f(x_{n+1}, y_n)\) instead of \(y_{n+1} = y_n + h \cdot f(x_n, y_n)\). This shifts the computation incorrectly and gives wrong results, especially for larger step sizes.