There are two definitions of the term “homogeneous differential equation.” One definition calls a first‐order equation of the form

*M* and *N* are both homogeneous functions of the same degree. The second definition — and the one which you'll see much more often—states that a differential equation (of *any* order) is **homogeneous** if once all the terms involving the unknown function are collected together on one side of the equation, the other side is identically zero. For example,

The nonhomogeneous equation

Equation (**) is called the **homogeneous equation corresponding to the nonhomogeneous equation**, (*). There is an important connection between the solution of a nonhomogeneous linear equation and the solution of its corresponding homogeneous equation. The two principal results of this relationship are as follows:

**Theorem A.** If *y* _{1}( *x*) and *y* _{2}( *x*) are linearly independent solutions of the linear homogeneous equation (**), then *every* solution is a linear combination of *y* _{1} and *y* _{2}. That is, the general solution of the linear homogeneous equation is

**Theorem B.** If *x*) is any particular solution of the linear nonhomogeneous equation (*), and if *y* _{h }( *x*) is the general solution of the corresponding homogeneous equation, then the general solution of the linear nonhomogeneous equation is

That is,

[Note: The general solution of the corresponding homogeneous equation, which has been denoted here by *y* _{h }, is sometimes called the **complementary function** of the nonhomogeneous equation (*).] Theorem A can be generalized to homogeneous linear equations of any order, while Theorem **B** as written holds true for linear equations of any order. Theorems A and B are perhaps the most important theoretical facts about linear differential equations—definitely worth memorizing.

**Example 1**: The differential equation

Verify that any linear combination of *y* _{1} and *y* _{2} is also a solution of this equation. What is its general solution?

Every linear combination of *y* _{1} = *e* ^{x }and *y* _{2} = *xe* ^{x }looks like this:

*c* _{1} and *c* _{2}. To verify that this satisfies the differential equation, just substitute. If *y* = *c* _{1} *e* ^{x }+ *c* _{2} *xe* ^{x }, then

*y* _{1} = *e* ^{x }and *y* _{2} = *xe* ^{x }does indeed satisfy the differential equation. Now, since *y* _{1} = *e* ^{x }and *y* _{2} = *xe* ^{x }are linearly independent, Theorem A says that the general solution of the equation is

**Example 2**: Verify that *y* = 4 *x* – 5 satisfies the equation

Then, given that *y* _{1} = *e* ^{− x }and *y* _{2} = *e* ^{− 4x }are solutions of the corresponding homogeneous equation, write the general solution of the given nonhomogeneous equation.

First, to verify that *y* = 4 *x* – 5 is a particular solution of the nonhomogeneous equation, just substitute. If *y* = 4 *x* – 5, then *y*′ = 4 and *y*″ = 0, so the left‐hand side of the equation becomes

Now, since the functions *y* _{1} = *e* ^{− x }and *y* _{2} = *e* ^{− 4x }are linearly independent (because neither is a constant multiple of the other), Theorem A says that the general solution of the corresponding homogeneous equation is

Theorem B then says

**Example 3**: Verify that both *y* _{1} = sin *x* and *y* _{2} = cos *x* satisfy the homogeneous differential equation *y*″ + *y* = 0. What then is the general solution of the nonhomogeneous equation *y*″ + *y* = *x*?

If *y* _{1} = sin *x*, then *y*″ _{1} + *y* _{1} does indeed equal zero. Similarly, if *y* _{2} = cos *x*, then *y*″ _{2} = *y* _{1} = sin *x* and *y* _{2} = cos *x* are linearly independent, Theorem A says that the general solution of the homogeneous equation *y*″ + *y* = 0 is

Now, to solve the given nonhomogeneous equation, all that is needed is any particular solution. By inspection, you can see that *x* satisfies *y*″ + *y* = *x*. Therefore, according to Theorem B, the general solution of this nonhomogeneous equation is