346_pdfsam_math 54 differential equation solutions odd

346_pdfsam_math 54 differential equation solutions odd - 8...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 6 Thus p 1 ( x ) 0, p 2 ( x ) = 1 / ( x x + 1), p 3 ( x ) = 1 / x + 1, and g ( x ) 0. Functions p 1 ( x ) and q ( x ) are continuous on whole real line; p 3 ( x ) is defined and continuous for x > 1; p 2 ( x ) is defined and continuous for x > 1 and x = 0. Therefore, all these function is continuous on ( 1 , 0) and (0 , ). The initial point lies on (0 , ), and so, by Theorem 1, the given initial value problem has a unique solution on (0 , ). 7. Assume that c 1 , c 2 , and c 3 are constants for which c 1 e 3 x + c 2 e 5 x + c 3 e x 0 on ( −∞ , ) . (6.1) If we show that this is possible only if c 1 = c 2 = c 3 = 0, then linear independence will follow. Evaluating the linear combination in (6.1) at x = 0, x = ln 2, and x = ln 2, we find that constants c 1 , c 2 , and c 3 satisfy c 1 + c 2 + c 3 = 0 , 8 c 1 + 32 c 2 + 1 2 c 3 = 0 , 1 8 c 1 + 1 32 c 2 + 2 c 3 = 0 . This system is a homogeneous system of linear equations whose determinant
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 8 1 / 32 2 ± ± ± ± ± ± ± ± = ± ± ± ± ± 32 1 / 2 1 / 32 2 ± ± ± ± ± − ± ± ± ± ± 8 1 / 2 1 / 8 2 ± ± ± ± ± + ± ± ± ± ± 8 3 2 1 / 8 1 / 32 ± ± ± ± ± = 2827 64 6 = 0 . Hence it has the unique trivial solution, that is, c 1 = c 2 = c 3 = 0. 9. Let y 1 = sin 2 x , y 2 = cos 2 x , and y 3 = 1. We want to fnd c 1 , c 2 , and c 3 , not all zero, such that c 1 y 1 + c 2 y 2 + c 3 y 3 = c 1 sin 2 x + c 2 cos 2 x + c 3 · 1 = 0 , ±or all x in the interval ( −∞ , ∞ ). Since sin 2 x + cos 2 x = 1 ±or all real numbers x , we can choose c 1 = 1, c 2 = 1, and c 3 = − 1. Thus, these ±unctions are linearly dependent. 342...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern