In a homogeneous, linear differential equation with constant coefficients, this is the equation you arrive at when you let y=emx. It is also known as the characteristic equation. For example...

```y'' + 2y' - 3y = 0

y=emx
y'=memx
y''=m2emx

m2emx + 2memx - 3emx = 0

Divide by emx (since it cannot equal 0)

m2 + 2m - 3 = 0  <-----auxiliary equation

(m+3)(m-1)=0

m=-3,1

general equation : y=c1e-3x+c2ex

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The problems don't always end up with such a simple general equation for y, but the auxiliary equation is obtained just the same way. It's then factored to solve for m, which in turn solves the differential equation.
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 m2(m+1) (m1 = m2 = 0), then the general solution will look a bit different. This general solution for the repeated root will be of the form:

y = c1emx + c2emxx + c3emx x2 + ... +cnemx xn-1

As you can see, the exponential term remains the same, and the constants may be different, but for each repeated root, xn-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 = c1e-x + c2 + c3x.

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 + bi and a - bi. To get to the point, the general solution will be of the form:

y = c1eaxcos(bx) + c2eaxsin(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 = c1e(a+bi)x + c2e(a-bi)x

Because of the properties of exponents, ea+bi = eaebi. However, ebi can be represented as a series, namely Sigma(n = 0 to infinity, xn/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) + isin(bx), and e raised to a negative imaginary value will be cos(bx) - isin(bx). In the general equation...

y = c1eaxcox(bx) + c2eaxisin(bx) + c3eaxcox(bx) - c4isin(bx)

This can be manipulated to become...

y = (c1 + c3)(eaxcox(bx)) + i(c2 - c4)(eaxsin(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.

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