Find Particular Solution

find Particular solution differential equations
Contents:
What is a Particular Solution?
Find particular solution: Example

What is a Particular Solution?

A particular solution requires you to find a single solution that meets the constraints of the question. A problem that asks you to find a series of functions has a general solution as the answer—a solution that contains a constant (+ C), which could represent one of a possibly infinite number of functions.

For example, a problem with the differential equation
dydv x3 + 8
requires a general solution with a constant for the answer, while the differential equation
dydv x3 + 8; f(0) = 2
requires a particular solution, one that fits the constraint f(0) = 2.

Watch this 5 minute video showing the difference between particular and general, or read on below for how to find particular solution differential equations.

Differential Equations: General Solutions vs. Particular Solutions

Find Particular solution: Example

Example problem #1: Find the particular solution for the differential equation dydx = 5, where y(0) = 2.

Step 1: Rewrite the equation using algebra to move dx to the right (this step makes integration possible):

  • dy = 5 dx

Step 2: Integrate both sides of the equation to get the general solution differential equation. Need to brush up on the rules? See: Common integration rules.

  • ∫ dy = ∫ 5 dx →
  • ∫ 1 dy = ∫ 5 dx →
  • y = 5x + C

Step 3: Rewrite the general equation to satisfy the initial condition, which stated that when x = 0, y = 2:

  • 2 = 5(0) + C
  • C = 2

The differential equation particular solution is y = 5x + 2

Particular solution differential equations, Example problem #2:
Find the particular solution for the differential equation dydx= 18x, where y(5) = 230.

Step 1: Rewrite the equation using algebra to move dx to the right:

  • dy = 18x dx

Step 2: Integrate both sides of the equation:

  • ∫ dy = ∫ 18x dx →
  • ∫ 1 dy = ∫ 18x dx →
  • y = 9x2 + C

Step 3: Rewrite the general equation to satisfy the initial condition, which stated that when x = 5, y = 230:

  • 230 = 9(5)2 + C
  • C = 5

The differential equation particular solution is y = 5x + 5

That’s it!

References

4.5 The Superposition Principle and Undetermined Coefficients Revisited.


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