Homogeneous, Particular, and General Solutions

In this section is a review of what homogeneous, particular, and general solutions are, as well as the general procedure for finding the general solution.

The Essentials

A linear differential equation of nth order fits into this form:

\[ y^{(n)} + p_{n-1}(x) y^{(n-1)} + \cdots + p_1(x) y' + p_0(x) y = f(x) \]

If \( f(x) = 0 \), then it is called a homogeneous equation. When \( f(x) \neq 0 \), the equation is called non-homogeneous. Nonhomogeneous equations have a particular solution, and a homogeneous solution (aka null solution). The particular solution solves the equation, and the homogeneous solution solves the corresponding homogeneous equation (where \( f(x) \) is set to 0). The general solution is the expressions that contain all possible solutions to the equation. It takes the form of the particular solution added to a linear combination of the homogeneous solutions. After some thought, this makes sense because, in the end, the terms from the homogeneous solution don’t contribute anything to the answer, the result of the homogeneous solution is zero so adding the homogeneous solution to the particular solution doesn’t change the validity of the particular solution.

Procedure: To find the general solution:

  • First, if the equation is of the 2nd order and one of the solutions to the homogeneous equation is given, the reduction of order method may be used whether the equation is homogeneous or non-homogeneous.
  • To find the homogeneous solution, the method of constant coefficients and the Cauchy-Euler method can be used.
  • If the equation is non-homogeneous, first find the homogeneous solution, then use the method of undetermined coefficients or the method of variation of parameters to find the particular solution.
  • Alternatively, Laplace transforms can be used to find both the general solution without finding the homogeneous solutions first.

Example

Consider the example from the reduction of order page:

\[ x^2 y'' - 4xy' + 6y = \sqrt{x} \]

We were given that one term of the homogeneous solution is \( y_1(x) = x^2 \). The general solution turns out to be:

\[ y(x) = \frac{4}{15} \sqrt{x} + C_1 x^3 + C_2 x^2 \]

With the particular and homogeneous solutions:

\[ y_p(x) = \frac{4}{15} \sqrt{x} \] \[ y_h(x) = C_1 x^3 + C_2 x^2 \]

The particular solution gives the result of \( \sqrt{x} \) . From the homogeneous solution, it can be tested that \( x^3 \) and \( x^2 \) both make the equation equal 0, along with any linear combination of those two values ( \( 3x^3 \) or \( 5x^2 + 97x^3 \), for example). The general solution, therefore, has to include the linear combination of the homogeneous solutions in order to include every possible term that will solve the equation. If initial conditions are given then the unknown constants in the homogeneous solution have specific values and it isn’t just a linear combination of the terms. When finding these coefficients, the initial conditions must be applied to the general solution, not the homogeneous solution.