• Identify homogeneous linear equations
• Find and use characteristic equations
• Solve homogeneous equations using characteristic equations
• Solve initial and boundary value problems

Homogeneous Linear Equations

Consider the second-order differential equation:

[latex]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. First-order equations with similar characteristics are linear. The same is true of second-order equations.

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]xy^{\prime\prime}+2x^2y^\prime+5x^3y=x^2[/latex].

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

linear second-order differential equations

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

[latex]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] for every value of [latex]x[/latex], the equation is a homogeneous linear equation. If [latex]r(x)\ne0[/latex] for some value of [latex]x[/latex], the equation is a nonhomogeneous linear equation.

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.

Equations may not always be given in standard form. It can be helpful to rewrite them in that form to decide whether they are linear, or whether a linear equation is homogeneous.

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

1. [latex]y''+3x^{4}y'+x^{2}y^{2}=x^{3}[/latex]
2. [latex](\sin x)y''+(\cos x)y'+3y=0[/latex]
3. [latex]4t^{2}x''+3txx'+4x=0[/latex]
4. [latex]5y''+y=4x^{5}[/latex]
5. [latex](\cos x)y''-\sin y'+(\sin x)y-\cos x=0[/latex]
6. [latex]8ty''-6t^{2}y'+4ty-3t^2=0[/latex]
7. [latex]\sin (x^{2})y''-(\cos x)y'+x^{2}y=y'-3[/latex]
8. [latex]y''+5xy'-3y=\cos y[/latex]

Show Solution

1. This equation is nonlinear because of the [latex]y^{2}[/latex] term.
2. This equation is linear. There is no term involving a power or function of [latex]y[/latex], and the coefficients are all functions of [latex]x[/latex]. The equation is already written in standard form, and [latex]r(x)[/latex] is identically zero, so the equation is homogeneous.
3. This equation is nonlinear. Note that, in this case, [latex]x[/latex] is the dependent variable and [latex]t[/latex] is the independent variable. The second term involves the product of [latex]x[/latex] and [latex]x'[/latex], so the equation is nonlinear.
4. This equation is linear. Since [latex]r(x)=4x^{5}[/latex], the equation is nonhomogeneous.
5. This equation is nonlinear, because of the [latex]\sin y'[/latex] term.
6. This equation is linear. Rewriting it in standard form gives
   [latex]8t^{2}-6t^{2}y'+4ty=3t^{2}[/latex].
   With the equation in standard form, we can see that [latex]r(t)=3t^2[/latex] so the equation is nonhomogeneous.
7. This equation looks like it’s linear, but we should rewrite it in standard form to be sure. We get
   [latex]\sin(x^{2})y''-(\cos x+1)y'+x^{2}y=-3[/latex].This equation is, indeed, linear. With [latex]r(x)=-3[/latex], it is nonhomogeneous.
8. This equation is nonlinear because of the [latex]\cos y[/latex] term.

Before we examine techniques for solving specific types of differential equations, let’s understand 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.

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.

Show Solution

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

[latex]\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}[/latex].