7.6 Finding Particular Solutions Using Initial Conditions

• What does a differential equation relate?

  A function to its derivatives
• A particular solution to a differential equation is unique because of the specific initial conditions
• The general solution of $\frac{dy}{dx} =$$3x^{2}$ is y = x^3 + C</latex>
• Steps to find a particular solution using initial conditions

  1️⃣ Find the general solution of the differential equation

  2️⃣ Apply the initial conditions to solve for $C$

  3️⃣ Write the particular solution using the determined value of $C$
• Steps to find the particular solution for $\frac{dy}{dx} =$$2x$ with initial condition $y(1) =$$3$


  1️⃣ Find the general solution: $y =$$x^{2} +$$C$

  2️⃣ Apply the initial condition: $3 =$$(1)^{2} +$$C$

  3️⃣ Solve for $C$: $C =$$2$

  4️⃣ State the particular solution: $y =$$x^{2} +$$2$
• What is the general solution of the differential equation $\frac{dy}{dx} =$$2x$?


  $y =$$x^{2} +$$C$
• If $y(0) =$$2$, the value of C in the general solution $y =$$2x^{2} +$$3x +$$C$ is 2
• The particular solution of the differential equation $\frac{dy}{dx} =$$2x$ with initial condition $y(1) =$$3$ is $y =$$x^{2} +$$2$.


  True
• The general solution of a differential equation is unique.

  False
• Match the solution type with its characteristic:

  General Solution ↔️ Non-unique due to constant C

  Particular Solution ↔️ Unique because constants are determined by initial conditions
• The general solution of a differential equation includes an arbitrary constant
• What kind of solution do initial conditions lead to?

  Unique solution
• Match the feature with the correct type of solution:

  Includes an arbitrary constant (C) ↔️ General Solution

  Constant is uniquely determined ↔️ Particular Solution
• A particular solution is a specific function obtained by using initial conditions.


  True
• Initial conditions are values of the function and its derivatives at a specific point.


  True
• Steps to find the value of \( C \) using the initial condition $y(0) =$$5$


  1️⃣ Substitute $x =$$0$ and $y =$$5$ into the general solution

  2️⃣ Simplify the equation to solve for $C$

  3️⃣ Find the value of $C$
• A differential equation relates a function to its derivatives
• What are initial conditions used for in differential equations?

  Determine particular solutions
• What is the general solution of the differential equation \frac{dy}{dx} = 2e^{2x}</latex>?

  $y =$$e^{2x} +$$C$
• Match the solution type with its description:

  General Solution ↔️ Represents all possible functions that satisfy the differential equation.

  Particular Solution ↔️ A specific function obtained using initial conditions.
• Why are initial conditions necessary to find a particular solution?

  To determine constants
• The first step to find a particular solution is to find the general solution
• A general solution is unique without initial conditions.

  False
• Steps to find particular solutions using initial conditions

  1️⃣ Find the general solution of the differential equation

  2️⃣ Apply the initial condition to solve for the constant C

  3️⃣ Write the particular solution using the determined value of C
• What is the general solution of the differential equation \frac{dy}{dx} = 4x + 3</latex>?

  $y =$$2x^{2} +$$3x +$$C$
• Using the initial condition $y(1) =$$3$ for the differential equation \frac{dy}{dx} = 2x</latex>, the value of C is 2
• What distinguishes a particular solution from a general solution?

  Initial conditions
• Initial conditions are used to determine the constant C
• The particular solution $y =$$x^{2} +$$2$ satisfies both the differential equation $\frac{dy}{dx} =$$2x$ and the initial condition $y(1) =$$3$.


  True
• What is the particular solution of the differential equation $\frac{dy}{dx} =$$2e^{2x}$ with the initial condition $y(0) =$$4$?


  $y =$$e^{2x} +$$3$
• What is the general solution of the differential equation $\frac{dy}{dx} =$$2x$?


  $y =$$x^{2} +$$C$
• The particular solution of a differential equation is derived from the general solution using initial conditions
• What is the particular solution of \frac{dy}{dx} = 3x^{2}</latex> with the initial condition $y(0) =$$5$?


  $y =$$x^{3} +$$5$
• Match the solution type with its description:

  General Solution ↔️ Represents all possible solutions

  Particular Solution ↔️ Solved using initial conditions
• Initial conditions specify the function and derivative values at a single point.


  True
• What is the particular solution of the differential equation $\frac{dy}{dx} =$$2x$ with the initial condition $y(1) =$$3$?


  $y =$$x^{2} +$$2$
• If $y(0) =$$4$, the value of C in the general solution $y =$$e^{2x} +$$C$ is 3
• The general solution of a differential equation represents all possible functions that satisfy the equation
• What are initial conditions used for in the context of differential equations?

  To find particular solutions
• Substituting initial conditions into the general solution allows us to find the value of $C$.


  True