5.1. Finding the General Solution for Differential Equations with a Nonhomogeneous erx Term

As you may know from class or from Differential Equations For Dummies, you can't just find the solution to a nonhomogeneous linear second order differential equation that happens to give you g(x) when you plug it in. You have to do some extra work by adding the solution to the homogeneous version of the same differential equation.

Think about it. Say you have this differential equation:

y″ + p(x)y′ + q(x)y = g(x)

and the following solution gives you g(x) when you plug it in:

y = yp(x)

In order for your answer to be correct, you must add in the homogeneous solution; when you plug that into the differential equation, you get 0. So the general solution to the differential equation is

y = c1y1(x) + c2y2(x) + yp(x)

where c1y1(x) + c2y2(x) is the solution of the corresponding homogeneous differential equation:

y″ + p(x)y′ + q(x)y = 0

That is, y1 and y2 are a fundamental set of solutions to the homogeneous differential equation, and yp(x) is a particular (or specific) solution to the nonhomogenuous equation.

NOTE

So to solve a second order differential equation that's both linear and nonhomogeneous, you follow these overall steps:

  1. Find the corresponding homogeneous differential equation by setting g(x) to 0.

  2. Find the general solution, y = c1y1(x) + c2y2(x), of the corresponding homogeneous differential equation.

    This general solution of the homogeneous equation is referred to as yh.

  3. Find a single ...

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