AP Calc separable differential equations growth study guide

What is a separable differential equation?

An equation that can be expressed as $\displaystyle g(y) dy = f(x) dx$, allowing separation of variables.

True or False: All differential equations are separable.

False. Only those that can be rearranged into the form $\displaystyle g(y) dy = f(x) dx$ are separable.

Identify the components: dy/dx = ky.

dy and dx are differentials; k is a constant; y is the dependent variable.

Separate: dy/dx = 3y.

Rearranged: $\displaystyle \\frac{1}{y} dy = 3 dx$.

What steps are needed to solve a separable equation?

1. Separate variables. 2. Integrate both sides. 3. Solve for y, if needed.

Fill in the blank: In a separable equation, we can integrate ___ and ___ separately.

the left side and the right side.

Example problem: Solve dy/dx = 2x.

Separate: $\displaystyle dy = 2x dx$. Integrate: $\displaystyle y = x^2 + C$.

True or False: Separable equations can only have constant coefficients.

False. They can have variable coefficients as long as they can be separated.

What indicates a separable equation can be solved easily?

If the equation can be written as a product of a function of y and a function of x.

Compare: Separable vs. Non-separable equations.

Separable: Can be rearranged into $\displaystyle g(y) dy = f(x) dx$. Non-separable: Cannot be rearranged this way.

What is the general solution form after integrating?

$\displaystyle y = C e^{kt}$, where C is a constant of integration.

If y = 5 when x = 0, find C in y = Ce^{3x}.

5 = Ce^{0} → C = 5. Solution: $\displaystyle y = 5e^{3x}$.

What is a common growth model?

Exponential growth models are common, often represented as: $\displaystyle P(t) = P_0 e^{kt}$, where $\displaystyle P_0$ is the initial population, $\displaystyle k$ is the growth rate.

True or False: Population growth is always linear.

False. Population growth is often exponential, meaning it accelerates over time as resources become available.

Fill in the blank: A population of 100 grows at 5% per year. After one year, it will be ______.

105. This is calculated using $\displaystyle P(1) = 100(1 + 0.05) = 105$.

Compare linear vs exponential growth.

Linear growth adds a constant amount; exponential growth adds a percentage of the current amount, leading to faster increases over time.

What is the formula for continuous compounding?

The formula is $\displaystyle A = Pe^{rt}$, where $\displaystyle A$ is the amount, $\displaystyle P$ is the principal, $\displaystyle r$ is the interest rate, and $\displaystyle t$ is time.

How to solve a separable differential equation for population growth?

Separate variables: $\displaystyle \\frac{dP}{P} = k \, dt$, integrate: $\displaystyle \ln |P| = kt + C$, solve for $\displaystyle P$.

After 3 years, a population grows from 200 to 400. Find the growth rate.

Using $\displaystyle P(t) = P_0 e^{kt}$, solve for $\displaystyle k$ with $\displaystyle 400 = 200 e^{3k}$, yielding $\displaystyle k \approx 0.4055$ (or 40.55%).

True or False: Financial growth is always exponential.

True. Financial investments usually grow exponentially due to continuous compounding of interest.

Which differential equation represents logistic growth?

Logistic growth is modeled by: $\displaystyle \\frac{dP}{dt} = rP(1 - \\frac{P}{K})$, where $\displaystyle K$ is the carrying capacity.

What is a carrying capacity?

The carrying capacity ($\displaystyle K$) is the maximum population size that an environment can sustain indefinitely, impacting growth rates.

Calculate population after 5 years if $\displaystyle P_0 = 1000$, $\displaystyle k = 0.03$.

Using $\displaystyle P(t) = 1000 e^{0.03 \cdot 5}$, compute $\displaystyle P(5) \approx 1000 e^{0.15} \approx 1161.83$.

Cause → Effect: Doubling the initial population size leads to _____ growth.

Faster growth rates due to more individuals reproducing, leading to an exponential increase in population over time.

A population grows at a rate proportional to its size. True or False?

True. This is the fundamental principle of exponential growth modeled by separable differential equations.

Solve: $\displaystyle \frac{dy}{dt} = 4y(1 - \\frac{y}{100})$.

This represents a logistic growth model. Separation of variables and integration gives: $\displaystyle y(t) = \\frac{100}{1 + Ce^{-4t}}$.

In 2010, a store's revenue was $50,000. It grows at 10% per year. Find R(t).

Use $\displaystyle R(t) = 50000e^{0.1t}$. At t=0, revenue is $\displaystyle 50,000; at t=5,$R(5) \approx 81444.66$.

What type of differential equation is $\displaystyle \\frac{dy}{dt} = ky$?

This is a separable differential equation representing exponential growth or decay, depending on the sign of k.

Fill in the blank: The solution to $\displaystyle \\frac{dy}{dt} = -3y$ is ...

The solution is $\displaystyle y(t) = Ce^{-3t}$, indicating exponential decay.

Compare: Exponential growth vs. Logistic growth.

Exponential growth is unbounded, $\displaystyle y(t) = Ce^{kt}$. Logistic growth approaches a carrying capacity, $\displaystyle y(t) = \\frac{K}{1 + Ce^{-rt}}$.

Determine the initial condition for $\displaystyle y' = 5y(1 - \\frac{y}{50})$ at t=0.

If $\displaystyle y(0)=10$, use it in $\displaystyle y(t) = \\frac{50}{1 + 4e^{-5t}}$. This fits the initial condition.

True or False: A separable equation can be solved by direct integration.

True. By separating variables to isolate y and t, and then integrating both sides.

A population starts with 200 and grows at a rate of 50 per year. Find the equation.

The differential equation is $\displaystyle \\frac{dy}{dt} = 50$. Solution: $\displaystyle y(t) = 200 + 50t$.

Solve: $\displaystyle \frac{dy}{dt} = 3y^2$ with $\displaystyle y(0)=1$.

Separate: $\displaystyle \\frac{1}{y^2}dy = 3dt$. Integrate: $\displaystyle -\\frac{1}{y} = 3t + C$. Solve for C using $\displaystyle y(0)=1$.

At what point does $\displaystyle y(t)$ reach 80 in $\displaystyle y(t) = \\frac{100}{1 + Ce^{-kt}}$?

Set $\displaystyle 80 = \\frac{100}{1 + Ce^{-kt}}$; solve for t using the initial conditions and k value.

If a tank leaks at a rate proportional to its content, what's the form?

The equation is $\displaystyle \\frac{dy}{dt} = -ky$, where y is the amount in the tank and k is the leak constant.