Homogeneous Linear Equations

Learning Objectives

Recognize homogeneous and nonhomogeneous linear differential equations.
Determine the characteristic equation of a homogeneous linear equation.

Consider the second-order differential equation

$\large{xy^{\prime\prime}+2x^2y^\prime+5x^3y=0}$ .

Notice that $y$ and its derivatives appear in a relatively simple form. They are multiplied by functions of $x$ , but are not raised to any powers themselves, nor are they multiplied together. As discussed in Introduction to Differential Equations, first-order equations with similar characteristics are said to be linear. The same is true of second-order equations. Also note that all the terms in this differential equation involve either $y$ or one of its derivatives. There are no terms involving only functions of $x$ . Equations like this, in which every term contains $y$ or one of its derivatives, are called homogeneous.

Not all differential equations are homogeneous. Consider the differential equation

$\large{xy^{\prime\prime}+2x^2y^\prime+5x^3y=x^2}$ .

The $x^{2}$ term on the right side of the equal sign does not contain $y$ or any of its derivatives. Therefore, this differential equation is nonhomogeneous.

definition

A second-order differential equation is linear if it can be written in the form

$\large{a_2(x)y^{\prime\prime}+a_1(x)y^{\prime}+a_0(x)y=r(x)}$ ,

where $a_2(x)$ , $a_1(x)$ , $a_0(x)$ , and $r(x)$ are real-valued functions and $a_2(x)$ is not identically zero. If $r(x)=0$ —in other words, if $r(x)=0$ for every value of $x$ —the equation is said to be a homogeneous linear equation. If $r(x)\ne0$ for some value of $x$ , the equation is said to be a nonhomogeneous linear equation.

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In linear differential equations, $y$ and its derivatives can be raised only to the first power and they may not be multiplied by one another. Terms involving $y^{2}$ or $\sqrt{y^\prime}$ make the equation nonlinear. Functions of $y$ and its derivatives, such as $\sin y$ or $e^{y^\prime}$ are similarly prohibited in linear differential equations.

Note that equations may not always be given in standard form (the form shown in the definition). It can be helpful to rewrite them in that form to decide whether they are linear, or whether a linear equation is homogeneous.

Example: Classifying second-order equations

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.

$y''+3x^{4}y'+x^{2}y^{2}=x^{3}$
$(\sin x)y''+(\cos x)y'+3y=0$
$4t^{2}x''+3txx'+4x=0$
$5y''+y=4x^{5}$
$(\cos x)y''-\sin y'+(\sin x)y-\cos x=0$
$8ty''-6t^{2}y'+4ty-3t^2=0$
$\sin (x^{2})y''-(\cos x)y'+x^{2}y=y'-3$
$y''+5xy'-3y=\cos y$

Show Solution

This equation is nonlinear because of the $y^{2}$ term.
This equation is linear. There is no term involving a power or function of $y$ , and the coefficients are all functions of $x$ . The equation is already written in standard form, and $r(x)$ is identically zero, so the equation is homogeneous.
This equation is nonlinear. Note that, in this case, $x$ is the dependent variable and $t$ is the independent variable. The second term involves the product of $x$ and $x'$ , so the equation is nonlinear.
This equation is linear. Since $r(x)=4x^{5}$ , the equation is nonhomogeneous.
This equation is nonlinear, because of the $\sin y'$ term.
This equation is linear. Rewriting it in standard form gives
With the equation in standard form, we can see that $r(t)=3t^2$ so the equation is nonhomogeneous.
This equation looks like it’s linear, but we should rewrite it in standard form to be sure. We get
$\sin(x^{2})y''-(\cos x+1)y'+x^{2}y=-3$ .This equation is, indeed, linear. With $r(x)=-3$ , it is nonhomogeneous.
This equation is nonlinear because of the $\cos y$ term.

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try it

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.

$(y'')^{2}-y'+8x^{3}y=0$
$(\sin t)y''+\cos t-3ty'=0$

Show Solution

Later in this section, we will see some techniques for solving specific types of differential equations. Before we get to that, however, let’s get a feel for how solutions to linear differential equations behave. In many cases, solving differential equations depends on making educated guesses about what the solution might look like. Knowing how various types of solutions behave will be helpful.

Example: verifying a solution

Consider the linear, homogeneous differential equation

$x^{2}y''-xy'-3y=0$ .

Looking at this equation, notice that the coefficient functions are polynomials, with higher powers of $x$ associated with higher-order derivatives of $y$ . Show that $y=x^{3}$ is a solution to this differential equation.

Show Solution

Let $y=x^{3}$ . Then $y'=3x^{2}$ and $y''=6(x)$ . Substituting into the differential equation, we see that

$\begin{aligned} x^{2}y''-xy'-3y&=x^2(6x)-x(3x^2)-3(x^3) \\ &=6x^3-3x^3-3x^3 \\ &=0 \end{aligned}$ .

try it

Show that $y=2x^{2}$ is a solution to the differential equation

$\large{\frac{1}{2}x^{2}y''-xy'+y=0}$ .

Show Solution

Let $y=2x^2$ . Then $y'=4x$ and $y''=4$ . Substituting into the differential equation, we see that

$\begin{aligned} \frac{1}{2}x^{2}y''-xy'+y&=\frac{1}{2}x^2(4)-x(4x)+(2x^2) \\ &=2x^2-4x^2+2x^2 \\ &=0 \end{aligned}$ .

Although simply finding any solution to a differential equation is important, mathematicians and engineers often want to go beyond finding one solution to a differential equation to finding all solutions to a differential equation. In other words, we want to find a general solution. Just as with first-order differential equations, a general solution (or family of solutions) gives the entire set of solutions to a differential equation. An important difference between first-order and second-order equations is that, with second-order equations, we typically need to find two different solutions to the equation to find the general solution. If we find two solutions, then any linear combination of these solutions is also a solution. We state this fact as the following theorem.

Theorem: superposition principle

If $y_1(x)$ and $y_2(x)$ are solutions to a linear homogeneous differential equation, then the function

$\large{y(x)=c_1y_1(x)+c_2y_2(x)}$ ,

where $c_1$ and $c_2$ are constants, is also a solution.

The proof of this superposition principle theorem is left as an exercise.

Example: verifying the superposition principle

Consider the differential equation

$\large{y^{\prime\prime}-4y^\prime-5y=0}$ .

Given that $e^{-x}$ and $e^{5x}$ are solutions to this differential equation, show that $4e^{-x}+e^{5x}$ is a solution.

Show Solution

We have

$y(x)=4e^{-x}+e^{5x},\text{ so }y^\prime(x)=-4e^{-x}+5e^{5x}\text{ and }y^{\prime\prime}=4e^{-x}+25e^{5x}$ .

Then

$\begin{aligned} y^{\prime\prime}-4y^\prime-5y&=\left(4e^{-x}+25e^{5x}\right)-4\left(-4e^{-x}+5e^{5x}\right)-5\left(4e^{-x}+e^{5x}\right) \\ &=4e^{-x}+25e^{5x}+16e^{-x}-20e^{5x}-20e^{-x}-5e^{5x} \\ &=0 \end{aligned}$ .

Thus, $y(x)=4e^{-x}+e^{5x}$ is a solution.

try it

Consider the differential equation

$y''+5y'+6y=0$ .

Given that $e^{-2x}$ and $e^{-3x}$ are solutions to this differential equation, show that $3e^{-2x}+6e^{-3x}$ is a solution.

Show Solution

We have $y(x)=3e^{-2x}+6e^{-3x}$ , so $y'(x)=-6e^{-2x}-18e^{-3x}$ and $y''(x)=12e^{-2x}+54e^{-3x}$ .

Then

$\begin{aligned} y^{\prime\prime}+5y^\prime+6y&=\left(12e^{-2x}+54e^{-3x}\right)+5\left(-6e^{-2x}-18e^{-3x}\right)+6\left(3e^{-2x}+6e^{-3x}\right) \\ &=12e^{-2x}+54e^{-3x}-30e^{-2x}-90e^{-3x}+18e^{-2x}+36e^{-3x} \\ &=0 \end{aligned}$ .

Thus, $y(x)=3e^{-2x}+6e^{-3x}$ is a solution.

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 7.3” here (opens in new window).

Unfortunately, to find the general solution to a second-order differential equation, it is not enough to find any two solutions and then combine them. Consider the differential equation

$\large{x''+7x'+12x=0}$ .

Both $e^{-3t}$ and $2e^{-3t}$ are solutions (check this). However, $x(t)=c_1e^{-3y}+c_2\left(2e^{-3t}\right)$ is not the general solution. This expression does not account for all solutions to the differential equation. In particular, it fails to account for the function $e^{-4t}$ , which is also a solution to the differential equation.

It turns out that to find the general solution to a second-order differential equation, we must find two linearly independent solutions. We define that terminology here.

definition

A set of functions $f_1(x),f_2(x),\ldots,f_n(x)$ is said to be linearly dependent if there are constants $c_1,c_2,\ldots c_n$ , not all zero, such that $c_1f_1(x)+c_2f_2(x)+\cdots c_nf_n(x)=0$ for all $x$ over the interval of interest. A set of functions that is not linearly dependent is said to be linearly independent.

In this chapter, we usually test sets of only two functions for linear independence, which allows us to simplify this definition. From a practical perspective, we see that two functions are linearly dependent if either one of them is identically zero or if they are constant multiples of each other.

First we show that if the functions meet the conditions given previously, then they are linearly dependent. If one of the functions is identically zero—say, $f_2(x)\equiv0$ —then choose $c_1=0$ and $c_2=1$ , and the condition for linear dependence is satisfied. If, on the other hand, neither $f_1(x)$ nor $f_2(x)$ is identically zero, but $f_1(x)=Cf_2(x)$ for some constant $C$ , then choose $c_1=\frac{1}{C}$ and $c_2=-1$ , and again, the condition is satisfied.

Next, we show that if two functions are linearly dependent, then either one is identically zero or they are constant multiples of one another. Assume $f_1(x)$ and $f_2(x)$ are linearly independent. Then, there are constants, $c_1$ and $c_2$ , not both zero, such that

$\large{c_1f_1(x)+c_2f_2(x)=0}$

for all $x$ over the interval of interest. Then,

$\large{c_1f_1(x)=-c_2f_2(x)}$ .

Now, since we stated that $c_1$ and $c_2$ can’t both be zero, assume $c_2\ne0$ . Then, there are two cases: either $c_1=0$ or $c_1\ne0$ . If $c_1=0$ , then

$\begin{aligned} 0&=-c_2f_2(x) \\ 0&=f_2(x) \end{aligned}$ ,

so one of the functions is identically zero. Now suppose $c_1\ne0$ . Then,

$\large{f_1(x)=\left(-\frac{c_2}{c_1}\right)f_2(x)}$

and we see that the functions are constant multiples of one another.

theorem: linear dependence of two functions

Two functions, $f_1(x)$ and $f_2(x)$ , are said to be linearly dependent if either one of them is identically zero or if $f_1(x)=Cf_2(x)$ for some constant $C$ and for all $x$ over the interval of interest. Functions that are not linearly dependent are said to be linearly independent.

Example: testing for linear dependence

Determine whether the following pairs of functions are linearly dependent or linearly independent.

$f_1(x)=x^{2}$ , $f_2(x)=5x^{2}$
$f_1(x)=\sin x$ , $f_2(x)=\cos x$
$f_1(x)=e^{3x}$ , $f_2(x)=e^{-3x}$
$f_1(x)=3x$ , $f_2(x)=3x+1$

Show Solution

$f_2(x)=5f_1(x)$ , so the functions are linearly dependent.
There is no constant $C$ such that $f_1(x)=Cf_2(x)$ , so the functions are linearly independent.
There is no constant $C$ such that $f_1(x)=Cf_2(x)$ , so the functions are linearly independent. Don’t get confused by the fact that the exponents are constant multiples of each other. With two exponential functions, unless the exponents are equal, the functions are linearly independent.
There is no constant $C$ such that $f_1(x)=Cf_2(x)$ , so the functions are linearly independent.

try it

Determine whether the following pairs of functions are linearly dependent or linearly independent: $f_1(x)=e^{x}$ , $f_2(x)=3e^{3x}$ .

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If we are able to find two linearly independent solutions to a second-order differential equation, then we can combine them to find the general solution. This result is formally stated in the following theorem.

theorem: general solution to a homogeneous equation

If $y_1(x)$ and $y_2(x)$ are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by

$y(x)=c_1y_1(x)+c_2y_2(x)$ ,

where $c_1$ and $c_2$ are constants.

When we say a family of functions is the general solution to a differential equation, we mean that (1) every expression of that form is a solution and (2) every solution to the differential equation can be written in that form, which makes this theorem extremely powerful. If we can find two linearly independent solutions to a differential equation, we have, effectively, found all solutions to the differential equation—quite a remarkable statement. The proof of this theorem is beyond the scope of this text.

Example: writing the general solution

If $y_1(t)=e^{3t}$ and $y_2(t)=e^{-3t}$ are solutions to $y''-9y=0$ , what is the general solution?

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Note that $y_1$ and $y_2$ are not constant multiples of one another, so they are linearly independent. Then, the general solution to the differential equation is $y(t)=c_1e^{3t}+c_2e^{-3t}$ .

try it

If $y_1(x)=e^{3x}$ and $y_2(x)=xe^{3x}$ are solutions to $y''-6y'+9y=0$ , what is the general solution?

Show Solution

$y(x)=c_1e^{3t}+c_2xe^{3t}$

Module 7: Second-Order Differential Equations