The general second‐order homogeneous linear differential equation has the form

If *a*( *x*), *b*( *x*), and *c*( *x*) are actually constants, *a*( *x*) ≡ *a* ≠ 0, *b*( *x*) ≡ *b*, *c*( *x*) ≡ *c*, then the equation becomes simply

This is the general second‐order homogeneous linear equation **with constant coefficients**.

Theorem A above says that the general solution of this equation is the general linear combination of any two linearly independent solutions. So how are these two linearly independent solutions found? The following example will illustrate the fundamental idea.

**Example 1**: Solve the differential equation *y*″ – *y*′ – 2 *y* = 0.

The trick is to substitute *y* = *e* ^{mx }( *m* a constant) into the equation; you will see shortly why this approach works. If *y* = *e* ^{mx }, then *y*′ = *me* ^{mx }and *y*″ = *m* ^{2} *e* ^{mx }, so the differential equation becomes

The term *e* ^{mx }can be factored out and immediately canceled (since *e* ^{mx }never equals zero):

This quadratic polynomial equation can be solved by factoring:

Now, recall that the solution began by writing *y* = *e* ^{mx }. Since the values of *m* have now been found *m* = −1, *m* = 2), both

Note carefully that the solution of the homogeneous differential equation

**auxiliary polynomial** equation that results from substituting *y* = *e* ^{mx }and then canceling out the *e* ^{mx }term. Once the roots of this auxiliary polynomial equation are found, you can immediately write down the general solution of the given differential equation. Also note that a *second*‐order linear homogeneous differential equation with constant coefficients will always give rise to a *second*‐degree auxiliary polynomial equation, that is, to a *quadratic* polynomial equation.

The roots of any quadratic equation

**quadratic formula**

The quantity under the square root sign, *b* ^{2} – 4 *ac*, is called the **discriminant** of the equation, and its sign determines the nature of the roots. There are exactly three cases to consider.

*Case 1: The discriminant is positive.*

In this case, the roots are real and distinct. If the two roots are denoted *m* _{1} and *m* _{2}, then the general solution of the differential equation is

*Case 2: The discriminant is zero.*

In this case, the roots are real and identical; that is, the polynomial equation has a double (repeated) root. If this double root is denoted simply by *m*, then the general solution of the differential equation is

*Case 3: The discriminant is negative.*

In this case, the roots are distinct conjugate complex numbers, *r* ± *si*. The general solution of the differential equation is then

So here's the process: Given a second‐order homogeneous linear differential equation with constant coefficients ( *a* ≠ 0),

*y*″ by *m* ^{2}, *y*′ by *m*, and *y* by 1). Determine the roots of this quadratic equation, and then, depending on whether the roots fall into Case 1, Case 2, or Case 3, write the general solution of the differential equation according to the form given for that Case.

**Example 2**: Solve the differential equation *y*″ + 3 *y*′ – 10 *y* = 0.

The auxiliary polynomial equation is

This problem falls into Case 1, so the general solution of the differential equation is

**Example 3**: Give the general solution of the differential equation *y*″ – 2 *y*′ + *y* = 0.

The auxiliary polynomial equation is

This problem falls into Case 2, so the general solution of the differential equation is

**Example 4**: Solve the differential equation *y*″ – 6 *y*′ + 25 *y* = 0.

The auxiliary quadratic equation is

This problem falls into Case 3, so the general solution of the differential equation is

**Example 5**: Solve the IVP

First, rewrite the differential equation in standard form:

Next, form the auxiliary polynomial equation

Since the roots are real and distinct, this problem falls into Case 1, and the general solution of the differential equation is therefore

All that remains is to use the two given initial conditions to determine the values of the constants *c* _{1} and *c* _{2}:

These two equations for *c* _{1} and *c* _{2} can be solved by first adding them to yield

*c* _{1} = 2 back into either equation to find *c* _{2} = −1.

From (*), the solution of the given IVP is therefore