10.06.2022 Views

James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A five star textbook for college calculus

A five star textbook for college calculus

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

1154 Chapter 17 Second-Order Differential Equations

A second-order linear differential equation has the form

1 Psxd d 2 y

dx 2

dy

1 Qsxd

dx

1 Rsxdy − Gsxd

where P, Q, R, and G are continuous functions. We saw in Section 9.1 that equations

of this type arise in the study of the motion of a spring. In Section 17.3 we will further

pursue this application as well as the application to electric circuits.

In this section we study the case where Gsxd − 0, for all x, in Equation 1. Such equations

are called homogeneous linear equations. Thus the form of a second-order linear

homogeneous differential equation is

2 Psxd d 2 y

dx 2

dy

1 Qsxd

dx 1 Rsxdy − 0

If Gsxd ± 0 for some x, Equation 1 is nonhomogeneous and is discussed in Section

17.2.

Two basic facts enable us to solve homogeneous linear equations. The first of these

says that if we know two solutions y 1 and y 2 of such an equation, then the linear combination

y − c 1 y 1 1 c 2 y 2 is also a solution.

3 Theorem If y 1 sxd and y 2 sxd are both solutions of the linear homogeneous

equation (2) and c 1 and c 2 are any constants, then the function

is also a solution of Equation 2.

ysxd − c 1 y 1 sxd 1 c 2 y 2 sxd

Proof Since y 1 and y 2 are solutions of Equation 2, we have

Psxdy0 1 1 Qsxdy 1 9 1 Rsxdy 1 − 0

and Psxdy 2 0 1 Qsxdy 2 9 1 Rsxdy 2 − 0

Therefore, using the basic rules for differentiation, we have

Psxdy0 1 Qsxdy9 1 Rsxdy

− Psxdsc 1 y 1 1 c 2 y 2 d0 1 Qsxdsc 1 y 1 1 c 2 y 2 d9 1 Rsxdsc 1 y 1 1 c 2 y 2 d

− Psxdsc 1 y 1 0 1 c 2 y 2 0d 1 Qsxdsc 1 y 1 9 1 c 2 y 2 9d 1 Rsxdsc 1 y 1 1 c 2 y 2 d

− c 1 fPsxdy 1 0 1 Qsxdy 1 9 1 Rsxdy 1 g 1 c 2 fPsxdy 2 0 1 Qsxdy 2 9 1 Rsxdy 2 g

− c 1 s0d 1 c 2 s0d − 0

Thus y − c 1 y 1 1 c 2 y 2 is a solution of Equation 2.

Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!