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7.1 Second-order linear equations  (Page 17/15)

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  • Recognize homogeneous and nonhomogeneous linear differential equations.
  • Determine the characteristic equation of a homogeneous linear equation.
  • Use the roots of the characteristic equation to find the solution to a homogeneous linear equation.
  • Solve initial-value and boundary-value problems involving linear differential equations.

When working with differential equations, usually the goal is to find a solution. In other words, we want to find a function (or functions) that satisfies the differential equation. The technique we use to find these solutions varies, depending on the form of the differential equation with which we are working. Second-order differential equations have several important characteristics that can help us determine which solution method to use. In this section, we examine some of these characteristics and the associated terminology.

Homogeneous linear equations

Consider the second-order differential equation

x y ″ + 2 x 2 y ′ + 5 x 3 y = 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

x y ″ + 2 x 2 y ′ + 5 x 3 y = 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

a 2 ( x ) y ″ + a 1 ( x ) y ′ + 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 ) ≠ 0 for some value of x , the equation is said to be a nonhomogeneous linear equation    .

Visit this website to study more about second-order linear differential equations.

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 y ′ make the equation nonlinear. Functions of y and its derivatives, such as sin y or e y ′ , 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.

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.

  1. y ″ + 3 x 4 y ′ + x 2 y 2 = x 3
  2. ( sin x ) y ″ + ( cos x ) y ′ + 3 y = 0
  3. 4 t 2 x ″ + 3 t x x ′ + 4 x = 0
  4. 5 y ″ + y = 4 x 5
  5. ( cos x ) y ″ − sin y ′ + ( sin x ) y − cos x = 0
  6. 8 t y ″ − 6 t 2 y ′ + 4 t y − 3 t 2 = 0
  7. sin ( x 2 ) y ″ − ( cos x ) y ′ + x 2 y = y ′ − 3
  8. y ″ + 5 x y ′ − 3 y = cos y
  1. This equation is nonlinear because of the y 2 term.
  2. 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.
  3. 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.
  4. This equation is linear. Since r ( x ) = 4 x 5 , the equation is nonhomogeneous.
  5. This equation is nonlinear, because of the sin y ′ term.
  6. This equation is linear. Rewriting it in standard form gives
    8 t 2 y ″ − 6 t 2 y ′ + 4 t y = 3 t 2 .

    With the equation in standard form, we can see that r ( t ) = 3 t 2 , 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
    sin ( x 2 ) y ″ − ( cos x + 1 ) y ′ + x 2 y = −3 .

    This equation is, indeed, linear. With r ( x ) = −3 , it is nonhomogeneous.
  8. This equation is nonlinear because of the cos y term.
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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