{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midterm_2_review (dragged) 5

midterm_2_review (dragged) 5 - 6 MA 36600 MIDTERM#2 REVIEW...

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

View Full Document Right Arrow Icon
6 MA 36600 MIDTERM #2 REVIEW Chapter 4 § 4.1: General Theory of n th Order Equations. An n th order linear di ff erential equation is an equation of the form P 0 ( t ) d n y dt n + P 1 ( t ) d n 1 y dt n 1 + · · · + P n 1 ( t ) dy dt + P n ( t ) y = G ( t ) . We wish to solve the initial value problem n i =1 P n j ( t ) y ( j ) = G ( t ) where y ( j 1) ( t 0 ) = y ( j 1) 0 , i = 1 , 2 , . . . , n. The sum L [ y ] = P 0 ( t ) d n y dt n + P 1 ( t ) d n 1 y dt n 1 + · · · + P n 1 ( t ) dy dt + P n ( t ) y is a linear operator . That is, given functions f = f ( t ) and g = g ( t ) as well as constants c 1 and c 2 , we have L c 1 f + c 2 g = c 1 L [ f ] + c 2 L [ g ]. In particular, if { y 1 , y 2 , . . . , y m } is a set of functions satisfying L [ y i ] = 0, then y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + · · · + c m y m ( t ) also satisfies L [ y ] = 0. Say that we can find ( n + 1) functions y 1 = y 1 ( t ) , . . . , y n ( t ) and Y = Y ( t ) such that each y i = y i ( t ) is a solution to the homogeneous equation n j =0 P n j ( t ) y ( j ) i = 0; the Wronskian is nonzero: W y 1 , y 2 , . . . , y n ( t ) = y 1 ( t ) y 2 ( t ) · · · y n ( t ) y (1) 1 ( t ) y (1) 2 ( t ) · · · y (1) n ( t ) .
Image of page 1
This is the end of the preview. Sign up to access the rest of the 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