If $k > 0$, the quantity $y$ is growing exponentially. If $k < 0$, the quantity $y$ is decaying exponentially.

It provides a specific value of the function at a particular point, allowing us to determine the constant of integration and find a unique solution.

Exponential growth occurs when the rate of increase of a quantity is proportional to the quantity itself. This leads to rapid increases over time.

Separate the variables, integrate both sides with respect to their respective variables, and solve for the function.

It represents the family of solutions that satisfy the differential equation; it is determined by initial conditions.

They are used to model phenomena such as population growth, radioactive decay, compound interest, and the spread of diseases.

Exponential decay occurs when the rate of decrease of a quantity is proportional to the quantity itself. This leads to rapid decreases over time.

To isolate each variable on one side of the equation, allowing for direct integration and solving.

It represents the function that satisfies the relationship between the function and its derivatives.

The natural logarithm is the inverse function of the exponential function with base $e$, allowing us to isolate variables in exponential expressions.

1. Substitute the given values into $y = y_0 e^{kt}$. 2. Divide both sides by $y_0$. 3. Take the natural logarithm of both sides. 4. Solve for $k$.

1. Write the equation $y = y_0 e^{kt}$. 2. Determine $y_0$ (initial population). 3. Find $k$ (growth rate). 4. Substitute $y_0$ and $k$ into the equation.

1. Determine the exponential model $y = y_0 e^{kt}$. 2. Substitute the known values of $y_0$, $k$, and $t$ (future time). 3. Calculate $y$ to find the future population.

1. Identify the initial amount $y_0$. 2. Use the given information to find the decay constant $k$. 3. Substitute $y_0$ and $k$ into the equation $y = y_0 e^{kt}$. 4. Solve for the desired variable.

1. Substitute the given values into $y = y_0 e^{kt}$. 2. Divide both sides by $y_0$. 3. Take the natural logarithm of both sides. 4. Solve for $t$ using $t = \frac{\ln(\frac{y}{y_0})}{k}$.

1. Use the half-life formula $k = \frac{\ln(2)}{T}$, where $T$ is the half-life. 2. Substitute the given half-life value into the formula. 3. Calculate $k$.

1. Substitute the given values into $y = y_0 e^{kt}$. 2. Solve for $y_0$ using $y_0 = \frac{y}{e^{kt}}$.

1. Set $y = 2y_0$ in the equation $y = y_0 e^{kt}$. 2. Simplify to $2 = e^{kt}$. 3. Take the natural logarithm of both sides. 4. Solve for $t$ using $t = \frac{\ln(2)}{k}$.

1. Identify the initial amount $A_0$. 2. Use the given information to find the constant $k$. 3. Write the equation $A(t) = A_0 e^{kt}$.

1. Assign an initial value to the bacteria, $y_0$. 2. Note that after 5 hours, $y = 2y_0$. 3. Use the exponential growth equation to find $k$. 4. Use the exponential growth equation to find $t$ when $y = 4y_0$.

$y = y_0 e^{kt}$

$k = \frac{1}{t} \ln(\frac{y(t)}{y_0})$

$\ln|y| + C$

$y = y_0 cdot e^{kt}$

Using the formula $k = \frac{\ln(\frac{y}{y_0})}{t}$, where $y$ is the population at time $t$, and $y_0$ is the initial population.

$A(t) = A_0 e^{kt}$, where $A_0$ is the initial amount and $k$ is the decay constant.

$\frac{dy}{dt}$

$\frac{dy}{dt} = ky$

$y = e^{kt+C} = e^C e^{kt}$

$e^C = |y_0|$