The above writeup gives an example of the

auxiliary equation with

rational roots that only appear once. There are two more common patterns that may show up.

**Repeated Roots**

If there exists an m that is a root for more than one factor, such as m^{2}(m+1) (m_{1} = m_{2} = 0), then the general solution will look a bit different. This general solution for the repeated root will be of the form:

y = c_{1}e^{mx}
+ c_{2}e^{mx}x
+ c_{3}e^{mx}
x^{2} + ...
+c_{n}e^{mx}
x^{n-1}

As you can see, the exponential term remains the same, and the constants may be different, but for each repeated root, x^{n-1} is multiplied by the constant and the e.
If the repeated root appears along with another root (including another sequence of repeated roots), the general equation will be written in exactly the same fashion, with only the repeated root portion differing.

Getting back to the example, since our roots are -1 and 0 (double root), the general equation will be as follows:

y = c_{1}e^{-x}
+ c_{2} + c_{3}x.

**Imaginary Roots**

Along with a root being the solution to more than one factor of the auxiliary equation, you can also have the fun of finding a general equation when the auxiliary equation has imaginary roots. Imaginary roots are guarenteed to come in conjugate pairs. They will be of the form a + b*i* and a - b*i*. To get to the point, the general solution will be of the form:

y = c_{1}e^{ax}cos(bx)
+ c_{2}e^{ax}sin(bx)

There is a fairly detailed explanation as to why the equation looks as such, but I will not go into too much depth. As mentioned in the writeup above, the e term will be raised to the m power, but in this case our m is imaginary (complex). So our solution will look something like:

y = c_{1}e^{(a+bi)x}
+ c_{2}e^{(a-bi)x}

Because of the properties of exponents, e^{a+bi} = e^{a}e^{bi}. However, e^{bi} can be represented as a series, namely Sigma(n = 0 to infinity, x^{n}/n!). After a considerable amount of manipulation, it will be easy to see that e raised to a positive imaginary value will be cos(bx) + *i*sin(bx), and e raised to a negative imaginary value will be cos(bx) - *i*sin(bx). In the general equation...

y = c_{1}e^{ax}cox(bx) + c_{2}e^{ax}*i*sin(bx) + c_{3}e^{ax}cox(bx)
- c_{4}*i*sin(bx)

This can be manipulated to become...

y = (c_{1} + c_{3})(e^{ax}cox(bx))
+ *i*(c_{2} - c_{4})(e^{ax}sin(bx)).

Since our c terms are just constants, they can be combined. The same goes for the *i*, this will give us the original equation.