7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables

• What is a particular solution to a differential equation?

  A solution satisfying initial conditions
• The general solution represents a family of functions
• The form of a general solution is \( y(x) = f(x) + C \), while the form of a particular solution is \( y(x) = f(x) + c \), where \( c \) is a constant.


  True
• What are initial conditions used for in differential equations?

  To find the particular solution
• Steps to find a particular solution using initial conditions

  1️⃣ Find the general solution

  2️⃣ Apply the initial conditions

  3️⃣ Determine the value of \( C \)

  4️⃣ Write the particular solution
• A particular solution to a differential equation is a solution that satisfies specific initial conditions.
• What is the particular solution to \( \frac{dy}{dx} = 2x \) with \( y(0) = 3 \)?

  \( y(x) = x^2 + 3 \)
• What is a particular solution to a differential equation?

  A solution satisfying initial conditions
• What is the particular solution for the differential equation \( \frac{dy}{dx} = 2x \) with the initial condition \( y(0) = 3 \)?

  \( y(x) = x^2 + 3 \)
• Initial conditions are usually given at \( x = 0 \).


  True
• What is the form of a separable differential equation?

  \( f(y) \, dy = g(x) \, dx \)
• The differential equation \( \frac{dy}{dx} = 2x \sqrt{y} \) is separable.


  True
• What is the particular solution for \( \frac{dy}{dx} = 2x \sqrt{y} \) with \( y(0) = 1 \)?

  \( y = \left(\frac{x^2 + 2}{2}\right)^2 \)
• Integrating \( \frac{dy}{y} = 2x \, dx \) yields \( \ln|y| = x^2 + C
• A particular solution to a differential equation must satisfy specific initial conditions.

  True
• Initial conditions are used to determine the constant of integration in the general solution.

  True
• A separable differential equation can be expressed in the form \( f(y) \, dy = g(x) \, dx \), where \( f(y) \) is a function of \( y \) only and \( g(x) \) is a function of \( x \) only
• Steps to solve a differential equation using separation of variables:

  1️⃣ Separate variables

  2️⃣ Integrate both sides

  3️⃣ Solve for y

  4️⃣ Find general solution
• After integrating both sides of a separable differential equation, the next step is to solve for the dependent variable
• Steps to solve a separable differential equation using the method of separation of variables.

  1️⃣ Separate variables into the form `f(y) dy = g(x) dx`

  2️⃣ Integrate both sides with respect to `y` and `x`

  3️⃣ Solve for `y` to obtain the general solution `y(x) = f(x, C)`

  4️⃣ Apply initial condition `y(x0) = y0` to find `C`

  5️⃣ Substitute `C` into the general solution to find the particular solution
• When integrating both sides of a separable differential equation, the left side is integrated with respect to `y`.

  True
• If the initial condition is `y(0) = 1`, the value of `A` in the general solution `y = Ae^(x^2)` is 1
• The method of separation of variables requires rearranging the equation into the form `f(y) dy = g(x) dx`.


  True
• After integrating both sides of `dy/y = 2x dx`, the resulting equation is `ln|y| = x^2 + C
• The general solution to a differential equation includes a constant of integration, whereas the particular solution does not.


  True
• What happens to the constant of integration in a particular solution?

  It is determined uniquely
• To find the particular solution of \( \frac{dy}{dx} = 2x \) with \( y(0) = 3 \), the constant \( C \) is equal to 3
• Initial conditions are usually given when \( x = 0 \).


  True
• What is the result of applying initial conditions to a general solution?

  A particular solution
• Match the solution type with its characteristic:

  General Solution ↔️ Family of functions

  Particular Solution ↔️ Unique function
• Initial conditions are required to find a particular solution.


  True
• The constant of integration in a particular solution is determined
• What are initial conditions used for in differential equations?

  Determining the constant C
• Initial conditions provide values such as \( y(x_0) = y_0 \) or \( y'(x_0) = v_0
• Steps to identify if a differential equation is separable

  1️⃣ Isolate \( \frac{dy}{dx} \)

  2️⃣ Separate variables by multiplication

  3️⃣ Verify the form \( f(y) \, dy = g(x) \, dx \)
• Integrating \( \frac{dy}{\sqrt{y}} = 2x \, dx \) yields \( 2\sqrt{y} = x^2 + C
• Steps in the method of separation of variables

  1️⃣ Separate variables

  2️⃣ Integrate both sides

  3️⃣ Solve for y

  4️⃣ Apply initial condition
• What is the particular solution for \( \frac{dy}{dx} = 2xy \) with \( y(0) = 1 \)?

  \( y(x) = e^{x^2} \)
• Initial conditions are specific values of the dependent variable or its derivative at a particular point
• Steps to identify if a differential equation is separable:

  1️⃣ Isolate \( \frac{dy}{dx} \)

  2️⃣ Multiply to separate

  3️⃣ Verify the form \( f(y) \, dy = g(x) \, dx \)