Solving exponential differential equations:

• What is the general form of exponential differential equations?

  $\frac{dy}{dt} =$$ky$
• What are the steps in the method of separation of variables for exponential differential equations?

  1️⃣ Rearrange the equation to isolate dy on one side and dt on the other: dy/y = k dt

  2️⃣ Integrate both sides: ∫ dy/y = ∫ k dt

  3️⃣ Evaluate the integrals: ln(y) = kt + C

  4️⃣ Solve for y: y = Ce^(kt)
• To integrate both sides of the equation `dy/y = k dt`, we get ln(y) on the left side.
• Exponential differential equations model the rate of change of y proportional to its current value.


  True
• In exponential decay, the value of k is negative.


  True
• What constant replaces e^C in the final solution y = Ae^(kt)?

  A
• In the exponential differential equation dy/dt = ky, the constant k represents the rate of growth or decay
• The first step in the method of separation of variables for the equation dy/dt = ky is to rearrange the equation.
• What is the integral of dy/y?

  ln|y| + C1
• Exponentiating both sides of $\ln(y) =$$kt +$$C$ is a valid step in solving for y.


  True
• What are initial conditions typically denoted as?

  $y(0) =$$y_{0}$
• Steps to apply initial conditions $y(t_{0}) =$$y_{0}$ to find the constant A in $y =$$Ae^{kt}$


  1️⃣ Apply initial condition: $y_{0} =$$Ae^{kt_{0}}$

  2️⃣ Solve for A: $A =$$\frac{y_{0}}{e^{kt_{0}}}$

  3️⃣ Substitute A back: $y =$$\frac{y_{0}}{e^{kt_{0}}} e^{kt}$

  4️⃣ Simplify: $y =$$y_{0}e^{k(t - t_{0})}$
• What is the value of A when the initial condition is $y(0) =$$y_{0}$?


  A = y_{0}</latex>
• In the general form of exponential differential equations, the constant `k` represents the rate of growth or decay.
• In exponential growth, the value of `k` is positive.
• To solve for y in ln(y) = kt + C, we exponentiate both sides.


  True
• After integrating both sides of the equation ln(y) = kt + C, the general solution for y is y = Ae^(kt).

  True
• A negative k value in the equation dy/dt = ky indicates exponential decay.


  True
• After solving for y in the equation ln(y) = kt + C, the simplified general solution is y = Ce^(kt)
• What is the simplified form of $y =$$e^{kt} *$$e^{C}$?


  $y =$$Ae^{kt}$
• Steps to solve the differential equation $\frac{dy}{dt} =$$ky$


  1️⃣ Separate variables: $\frac{dy}{y} =$$k dt$

  2️⃣ Integrate both sides: $\ln(y) =$$kt +$$C$

  3️⃣ Exponentiate both sides: $y =$$e^{kt + C}$

  4️⃣ Combine constants: $y =$$Ae^{kt}$
• The specific solution to \frac{dy}{dt} = 2y</latex> with $y(0) =$$5$ is $y =$$5e^{2t}$.


  True
• The steps to solve $\frac{dy}{dt} =$$ky$ include separating variables, integrating both sides, and combining constants.


  True
• In the final solution $y =$$y_{0}e^{kt}$, $y_{0}$ represents the initial value of y at t = 0
• Steps in solving exponential differential equations using separation of variables

  1️⃣ Rearrange the equation to isolate variables: dy/y = k dt

  2️⃣ Integrate both sides: ∫ dy/y = ∫ k dt

  3️⃣ Evaluate the integrals: ln(y) = kt + C

  4️⃣ Solve for y: y = e^(kt + C)

  5️⃣ Simplify: y = Ce^(kt)
• What type of rate does a positive k value represent in the equation dy/dt = ky?

  Growth
• What is the result of integrating both sides of the equation dy/y = k dt?

  ln(y) = kt + C
• Match the step with its outcome in the method of separation of variables:

  Rewrite the equation ↔️ ∫ dy/y = ∫ k dt

  Integrate both sides ↔️ ln|y| = kt + C

  Exponentiate both sides ↔️ y = Ae^(kt)
• The constant $e^{C}$ can be replaced with another constant, say A
• In the example $\frac{dy}{dt} =$$2y$ and $y(0) =$$5$, the first step is to separate the variables
• The final solution to the differential equation $\frac{dy}{dt} =$$ky$ is $y =$$Ae^{kt}$, where k is the constant rate of growth or decay
• The final solution to $\frac{dy}{dt} =$$ky$ with initial condition $y(0) =$$y_{0}$ is $y =$$y_{0}e^{kt}$.


  True