DE Tutorial - Part II: Nonhomogenous Second Order Equations and Their Applications

**The DE Tutorial is currently being split up into different threads to make editing these posts easier.**

**Second Order NON-Homogeneous Differential Equations**

Ah...the dreadful second order non-homogeneous differential equation has the form:

I will go through three techniques on how to solve these nasties:- Method of Undetermined Coefficients
- The "Annihilator" Method (Similar to #1)
- Variation of Parameters

__Technique 1 : Method of Undetermined Coefficients__:

In the case we have a differential equation like , we can guess what the particular solution to a DE may be, depending on what is. For example, let us say that . We would assume that a particular solution to the DE would be . To find the Undetermined Coefficients, plug back into the original DE.

If , we assume that the particular solution to the DE would have the form of . We too would substitute into the DE to find the unknown coefficient value.

If , we assume that the particular solution to the DE would have the form of . Again, to find the unknown coefficients, substitute into the original DE.

Sometimes, the guess of isn't that obvious. Try to think outside the box when solving these problems! :D

Note that the solution to the non-homogeneous DE is a linear combination of the **complimentary solution** (solution to the homogeneous equation) and the **particular solution** (solution to the non-homogeneous equation)

__Example 16__:

Solve ; ; .

Solve the homogeneous equation first.

.

Thus, .

Now solve the non-homogeneous equation.

.

Using the method of Undetermined Coefficients, we guess and assume that the particular solution will have the form:

.

Substituting these values into the original DE, we get:

Now compare the coefficients (like in Partial Fractions)

Solving for the constants, we get , and (verify).

Thus, .

Therefore, the solution is

.

Find so we can apply the initial conditions:

.

Apply the initial conditions.

.

Solving for , get (verify)

Therefore, the solution to the initial value problem is:

.

I will discuss the next two sometime later...