What is an Example of a Particular Solution of a Differential Equation?

A particular solution of a differential equation is a solution that satisfies the equation and contains no arbitrary constants. For example, y = 3x + 3 and y = x² + 11x + 7 are both particular solutions.

Understanding General vs. Particular Solutions

To fully grasp the concept, let's differentiate between general and particular solutions:

• General Solution: The general solution of a differential equation includes arbitrary constants. These constants represent a family of solutions, each differing based on the constant's value.

• Particular Solution: A particular solution is derived from the general solution by applying specific initial conditions or boundary conditions to determine the values of the arbitrary constants. Thus, it's a single, specific solution without any unknown constants.

How to Find a Particular Solution

1. Find the General Solution: Solve the differential equation to obtain the general solution, which will contain one or more arbitrary constants.
2. Apply Initial/Boundary Conditions: Use the given initial conditions (e.g., y(0) = 1) or boundary conditions (e.g., y(0) = 0, y(1) = 2) to substitute into the general solution.
3. Solve for the Constants: Solve the resulting equation(s) for the arbitrary constants.
4. Substitute the Constants: Substitute the values of the constants back into the general solution. The resulting solution is the particular solution.

Example

Let's consider a simple differential equation:

dy/dx = 2x

1. General Solution: Integrate both sides with respect to x:

   y = ∫2x dx = x² + C (where C is an arbitrary constant)

2. Apply Initial Condition: Suppose we have the initial condition y(0) = 1. This means when x=0, y=1.

3. Solve for the Constant: Substitute x=0 and y=1 into the general solution:

   1 = (0)² + C
   C = 1

4. Particular Solution: Substitute C=1 back into the general solution:

   y = x² + 1

Therefore, y = x² + 1 is a particular solution of the differential equation dy/dx = 2x.

Another Example

Suppose the general solution to a differential equation is y = Cx + C², where C is an arbitrary constant. If we are given the initial condition y(1) = 4, then we can find the particular solution.

1. Substitute the initial condition x = 1 and y = 4 into the general solution:

4 = C(1) + C²
C² + C - 4 = 0

2. Using the quadratic formula, we find that C ≈ 1.56 or C ≈ -2.56

3. The particular solutions are therefore approximately:

• y = 1.56x + 1.56² ≈ 1.56x + 2.43
• y = -2.56x + (-2.56)² ≈ -2.56x + 6.55

These are two examples of particular solutions to the same differential equation with different constants based on the initial condition.