Learning Objectives
- Recognize homogeneous and nonhomogeneous linear differential equations.
- Determine the characteristic equation of a homogeneous linear equation.
Consider the second-order differential equation
[latex]\large{xy^{\prime\prime}+2x^2y^\prime+5x^3y=0}[/latex].
Notice that [latex]y[/latex] and its derivatives appear in a relatively simple form. They are multiplied by functions of [latex]x[/latex], 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 [latex]y[/latex] or one of its derivatives. There are no terms involving only functions of [latex]x[/latex]. Equations like this, in which every term contains [latex]y[/latex] or one of its derivatives, are called homogeneous.
Not all differential equations are homogeneous. Consider the differential equation
[latex]\large{xy^{\prime\prime}+2x^2y^\prime+5x^3y=x^2}[/latex].
The [latex]x^{2}[/latex] term on the right side of the equal sign does not contain [latex]y[/latex] 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
[latex]\large{a_2(x)y^{\prime\prime}+a_1(x)y^{\prime}+a_0(x)y=r(x)}[/latex],
where [latex]a_2(x)[/latex], [latex]a_1(x)[/latex], [latex]a_0(x)[/latex], and [latex]r(x)[/latex] are real-valued functions and [latex]a_2(x)[/latex] is not identically zero. If [latex]r(x)=0[/latex]—in other words, if [latex]r(x)=0[/latex] for every value of [latex]x[/latex]—the equation is said to be a homogeneous linear equation. If [latex]r(x)\ne0[/latex] for some value of [latex]x[/latex], the equation is said to be a nonhomogeneous linear equation.
Interactive
Visit this website to study more about second-order linear differential equations.
In linear differential equations, [latex]y[/latex] and its derivatives can be raised only to the first power and they may not be multiplied by one another. Terms involving [latex]y^{2}[/latex] or [latex]\sqrt{y^\prime}[/latex] make the equation nonlinear. Functions of [latex]y[/latex] and its derivatives, such as [latex]\sin y[/latex] or [latex]e^{y^\prime}[/latex] 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.
- [latex]y''+3x^{4}y'+x^{2}y^{2}=x^{3}[/latex]
- [latex](\sin x)y''+(\cos x)y'+3y=0[/latex]
- [latex]4t^{2}x''+3txx'+4x=0[/latex]
- [latex]5y''+y=4x^{5}[/latex]
- [latex](\cos x)y''-\sin y'+(\sin x)y-\cos x=0[/latex]
- [latex]8ty''-6t^{2}y'+4ty-3t^2=0[/latex]
- [latex]\sin (x^{2})y''-(\cos x)y'+x^{2}y=y'-3[/latex]
- [latex]y''+5xy'-3y=\cos y[/latex]
Interactive
Visit this website that discusses second-order differential equations.
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.
- [latex](y'')^{2}-y'+8x^{3}y=0[/latex]
- [latex](\sin t)y''+\cos t-3ty'=0[/latex]
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
[latex]x^{2}y''-xy'-3y=0[/latex].
Looking at this equation, notice that the coefficient functions are polynomials, with higher powers of [latex]x[/latex] associated with higher-order derivatives of [latex]y[/latex]. Show that [latex]y=x^{3}[/latex] is a solution to this differential equation.
try it
Show that [latex]y=2x^{2}[/latex] is a solution to the differential equation
[latex]\large{\frac{1}{2}x^{2}y''-xy'+y=0}[/latex].
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 [latex]y_1(x)[/latex] and [latex]y_2(x)[/latex] are solutions to a linear homogeneous differential equation, then the function
[latex]\large{y(x)=c_1y_1(x)+c_2y_2(x)}[/latex],
where [latex]c_1[/latex] and [latex]c_2[/latex] 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
[latex]\large{y^{\prime\prime}-4y^\prime-5y=0}[/latex].
Given that [latex]e^{-x}[/latex] and [latex]e^{5x}[/latex] are solutions to this differential equation, show that [latex]4e^{-x}+e^{5x}[/latex] is a solution.
try it
Consider the differential equation
[latex]y''+5y'+6y=0[/latex].
Given that [latex]e^{-2x}[/latex] and [latex]e^{-3x}[/latex] are solutions to this differential equation, show that [latex]3e^{-2x}+6e^{-3x}[/latex] is a solution.
Watch the following video to see the worked solution to the above Try It
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
[latex]\large{x''+7x'+12x=0}[/latex].
Both [latex]e^{-3t}[/latex] and [latex]2e^{-3t}[/latex] are solutions (check this). However, [latex]x(t)=c_1e^{-3y}+c_2\left(2e^{-3t}\right)[/latex] 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 [latex]e^{-4t}[/latex], 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 [latex]f_1(x),f_2(x),\ldots,f_n(x)[/latex] is said to be linearly dependent if there are constants [latex]c_1,c_2,\ldots c_n[/latex], not all zero, such that [latex]c_1f_1(x)+c_2f_2(x)+\cdots c_nf_n(x)=0[/latex] for all [latex]x[/latex] 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, [latex]f_2(x)\equiv0[/latex]—then choose [latex]c_1=0[/latex] and [latex]c_2=1[/latex], and the condition for linear dependence is satisfied. If, on the other hand, neither [latex]f_1(x)[/latex] nor [latex]f_2(x)[/latex] is identically zero, but [latex]f_1(x)=Cf_2(x)[/latex] for some constant [latex]C[/latex], then choose [latex]c_1=\frac{1}{C}[/latex] and [latex]c_2=-1[/latex], 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 [latex]f_1(x)[/latex] and [latex]f_2(x)[/latex] are linearly independent. Then, there are constants, [latex]c_1[/latex] and [latex]c_2[/latex], not both zero, such that
[latex]\large{c_1f_1(x)+c_2f_2(x)=0}[/latex]
for all [latex]x[/latex] over the interval of interest. Then,
[latex]\large{c_1f_1(x)=-c_2f_2(x)}[/latex].
Now, since we stated that [latex]c_1[/latex] and [latex]c_2[/latex] can’t both be zero, assume [latex]c_2\ne0[/latex]. Then, there are two cases: either [latex]c_1=0[/latex] or [latex]c_1\ne0[/latex]. If [latex]c_1=0[/latex], then
[latex]\begin{aligned} 0&=-c_2f_2(x) \\ 0&=f_2(x) \end{aligned}[/latex],
so one of the functions is identically zero. Now suppose [latex]c_1\ne0[/latex]. Then,
[latex]\large{f_1(x)=\left(-\frac{c_2}{c_1}\right)f_2(x)}[/latex]
and we see that the functions are constant multiples of one another.
theorem: linear dependence of two functions
Two functions, [latex]f_1(x)[/latex] and [latex]f_2(x)[/latex], are said to be linearly dependent if either one of them is identically zero or if [latex]f_1(x)=Cf_2(x)[/latex] for some constant [latex]C[/latex] and for all [latex]x[/latex] 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.
- [latex]f_1(x)=x^{2}[/latex], [latex]f_2(x)=5x^{2}[/latex]
- [latex]f_1(x)=\sin x[/latex], [latex]f_2(x)=\cos x[/latex]
- [latex]f_1(x)=e^{3x}[/latex], [latex]f_2(x)=e^{-3x}[/latex]
- [latex]f_1(x)=3x[/latex], [latex]f_2(x)=3x+1[/latex]
try it
Determine whether the following pairs of functions are linearly dependent or linearly independent: [latex]f_1(x)=e^{x}[/latex], [latex]f_2(x)=3e^{3x}[/latex].
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 [latex]y_1(x)[/latex] and [latex]y_2(x)[/latex] are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by
[latex]y(x)=c_1y_1(x)+c_2y_2(x)[/latex],
where[latex]c_1[/latex] and [latex]c_2[/latex] 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 [latex]y_1(t)=e^{3t}[/latex] and [latex]y_2(t)=e^{-3t}[/latex] are solutions to [latex]y''-9y=0[/latex], what is the general solution?
try it
If [latex]y_1(x)=e^{3x}[/latex] and [latex]y_2(x)=xe^{3x}[/latex] are solutions to [latex]y''-6y'+9y=0[/latex], what is the general solution?
