Finding Particular Solutions Using Initial Conditions and Separation of Variables - AP Calculus AB Study Guide

The Core Idea: Exponential Models with Differential Equations

This topic explores a fundamental connection between differential equations and real-world phenomena that exhibit exponential growth or decay. The core concept is that for many natural processes, the rate at which a quantity changes is directly proportional to the amount of the quantity currently present. For example, the rate at which a population of bacteria grows is proportional to its current size, and the rate at which a radioactive substance decays is proportional to the amount of substance remaining.

This relationship is captured by a specific type of differential equation: $\frac{dy}{dt}=ky$. This equation mathematically states that the rate of change of $y$ with respect to $t$ ($\frac{dy}{dt}$) is equal to $y$ multiplied by a constant of proportionality, $k$. Solving this differential equation gives us the familiar exponential function, which serves as the model for the quantity $y$ over time t`. ## Key Formulas The entire topic is built upon a single differential equation and its corresponding solution. 1. **The Differential Equation for Exponential Models** The differential equation that models exponential growth and decay is: Formula[0] - $y is the quantity that is changing over time.

-   $t$ represents time.

-   $\frac{dy}{dt}$ is the rate of change of the quantity $y$.

-   $k$ is a non-zero constant of proportionality. It is also referred to as the relative growth rate (if $k>0$) or relative decay rate (if $k<0$).

2. The General Solution

   The solution to the differential equation $\frac{dy}{dt}=ky$ is:

   $$y(t)=C{e}^{kt}$$

   • $C$ is a constant that represents the initial value of the quantity $y$, meaning $C=y(0)$.

   • $k$ is the same constant of proportionality from the differential equation.

   • $e$ is the base of the natural logarithm.

Understanding Direct Proportionality

The phrase "the rate of change of $y$ is directly proportional to $y$" is the conceptual foundation of this topic. It's crucial to understand how to translate this phrase into a mathematical equation.

• "The rate of change of $y$" translates to the derivative, $\frac{dy}{dt}$.

• "is directly proportional to" translates to $=k\times$, where $k$ is the constant of proportionality.

• "$y$" simply translates to the variable $y$.

Combining these pieces gives the differential equation: $\frac{dy}{dt}=ky$.

The value of $k$ determines the nature of the model:

• If $k>0$, the rate of change is positive when $y$ is positive. This means the quantity is increasing, leading to exponential growth.

• If $k<0$, the rate of change is negative when $y$ is positive. This means the quantity is decreasing, leading to exponential decay.

The constant $C$ is the value of $y$ when $t=0$. This is because $y(0)=C{e}^{k\cdot 0}=C{e}^0=C\cdot 1=C$. Therefore, $C$ is always the initial amount of the substance or population.

Core Concepts & Rules

• The differential equation $\frac{dy}{dt}=ky$, where $k$ is a non-zero constant, is the mathematical model for exponential growth and decay.

• This equation is used when a quantity's rate of change is directly proportional to its current size.

• The general solution to the differential equation $\frac{dy}{dt}=ky$ is the exponential function $y=C{e}^{kt}$.

• In the solution $y=C{e}^{kt}$, the constant $C$ represents the initial value of $y$, i.e., the value of $y$ at $t=0$.

• The constant $k$ is the constant of proportionality. It is also known as the relative growth or decay rate.

• A positive value for $k$ ($k>0$) signifies exponential growth.

• A negative value for $k$ ($k<0$) signifies exponential decay.

Step-by-Step Example 1: Finding a Particular Solution

Problem: A quantity $y$ changes at a rate given by the differential equation $\frac{dy}{dt}=-0.2y$. If $y(0)=75$, find the particular solution for $y(t)$.

Step 1: Identify the form and the constants.

The differential equation $\frac{dy}{dt}=-0.2y$ is in the form $\frac{dy}{dt}=ky$.

By direct comparison, we can see that the constant of proportionality is $k=-0.2$.

Step 2: Write the general solution.

The general solution for this form of differential equation is $y(t)=C{e}^{kt}$.

Substituting our value for $k$, we get:

$$y(t)=C{e}^{-0.2t}$$

Step 3: Use the initial condition to find $C$

We are given the initial condition $y(0)=75$. This means that when $t=0$, $y=75$. We substitute these values into our general solution.

$$75=C{e}^{-0.2(0)}$$

$$75=C{e}^0$$

$$75=C\cdot 1$$

$$C=75$$

Step 4: Write the particular solution.

Now that we have found both $C$ and $k$, we can write the final particular solution by substituting $C=75$ back into the equation from Step 2.

$$y(t)=75e^{-0.2t}$$

Step-by-Step Example 2: Exam-Style Application

Problem: The population of a certain species of fish in a lake, $P$, is increasing at a rate directly proportional to the current population. At time $t=0$ years, the population is 600. At $t=4$ years, the population is 1800. Find an equation for the population $P(t)$ at any time $t$.

Step 1: Translate the problem into a differential equation.

The phrase "increasing at a rate directly proportional to the current population" translates to:

$$\frac{dP}{dt}=kP$$

where $k$ is a positive constant because the population is increasing.

Step 2: Write the general solution and use the initial condition.

The general solution is $P(t)=C{e}^{kt}$.

We are given the initial condition $P(0)=600$. We use this to find $C$.

$$600=C{e}^{k(0)}$$

$$600=C$$

So, our model is now $P(t)=600e^{kt}$.

Step 3: Use the second data point to find $k$

We are given that at $t=4$, the population is $P=1800$. We substitute these values into our current model.

$$1800=600e^{k(4)}$$

Step 4: Solve for $k$

To solve for $k$, we first isolate the exponential term.

$$\frac{1800}{600}=e^{4k}$$

$$3=e^{4k}$$

Now, take the natural logarithm of both sides to eliminate $e$.

$$\ln (3)=\ln (e^{4k})$$

$$\ln (3)=4k$$

Finally, solve for $k$.

$$k=\frac{\ln (3)}{4}$$

Step 5: Write the final particular solution.

Substitute the value of $k$ back into the model from Step 2.

$$P(t)=600e^{\left(\frac{\ln 3}{4}\right)t}$$

Using Your Calculator

Solving differential equations of the form $\frac{dy}{dt}=ky$ is an analytical process based on knowing the form of the solution. A calculator is not used to find the symbolic solution $y=C{e}^{kt}$.

However, a calculator is useful for two main purposes:

1. Numerical Calculations: In Example 2, we found $k=\frac{\ln (3)}{4}$. A calculator can be used to find a decimal approximation for $k$ if needed ($k\approx 0.27465$). It can also be used to evaluate the final model at a specific time, for example, to find the population at $t=10$.

   • 600 * e^((ln(3)/4) * 10) 2. **Checking Your Answer:** After finding a model like $P(t) = 600e^{\left(\frac{\ln 3}{4}\right)t}`, you can use your calculator's graphing feature to verify that it passes through the given points.

   • Enter the function into Y1.

   • Check the table (TBLSET and TABLE) to confirm that when $X=0$, $Y=600$ and when $X=4$, $Y=1800$.

   • Graph the function and use the CALC: value` feature to check the points. ## AP Exam Quick Hit ### Common Question Types - **Finding the Model from Context:** A word problem describes a scenario of exponential growth or decay (e.g., population growth, radioactive decay, Newton's Law of Cooling) and provides two data points. You must set up the differential equation, find $C and $k$, and write the specific model. (This is identical to Example 2 above).

• Interpreting a Given Solution: You are given the solution, such as $A(t)=200e^{-0.05t}$, which models the amount of a substance. You might be asked to identify the differential equation that produced this model.

  • Example: The solution is $A(t)=200e^{-0.05t}$. We identify $k=-0.05$. The differential equation must be $\frac{dA}{dt}=-0.05A$.

• Solving for Time: You are given the model and asked to find the time $t$ at which the quantity reaches a certain value.

  • Example: Using $P(t)=600e^{\left(\frac{\ln 3}{4}\right)t}$, find the time $t$ when the population reaches 5400. You would solve the equation $5400=600e^{\left(\frac{\ln 3}{4}\right)t}$ for $t$.

Common Mistakes

• Algebraic Errors when Solving for $k$: A very common mistake is incorrect application of logarithms. When solving $A=B{e}^{kt}$, students must first isolate the exponential by dividing ($A/B=e^{kt}$) before taking the natural log. A frequent error is to incorrectly take the log of both sides first, leading to an incorrect expression like $\ln (A)=\ln (B)+kt$.

• Mixing up $C$ and $k$: Confusing the roles of the initial value ($C$) and the constant of proportionality ($k$). Remember, $C$ is the value at $t=0$, while $k$ is part of the exponent and defines the rate.

• Incorrectly Handling Initial Conditions: Forgetting that $C$ is the initial value at $t=0$. If an "initial" condition is given at a time other than $t=0$ (e.g., $y(1)=50$), you must solve for $C$ algebraically rather than assuming $C=50$.

• Sign Errors with $k$: Using a positive $k$ for a decay problem (e.g., "half-life," "depreciates," "decreases") or a negative $k$ for a growth problem. The sign of $k$ must match the context of the problem.