Linear Algebra Solutions 65

Linear Algebra Solutions 65 - has the solution xn yn x0 y0...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: has the solution xn yn x0 y0 = An =P P P −1 λn 0 1 0 λn 2 x0 y0 = [X1 |X2 ] λn 0 1 0 λn 2 = [X1 |X2 ] λn α 1 λn β 2 where α β n λ1 0 0 λ2 = P −1 α β = λn αX1 + λn βX2 , 1 2 = P −1 x0 y0 . x . We substitute y ˙ (b) In matrix form, the system is X = AX , where X = X = P Y , where Y = [x1 , y1 ]t . Then ˙ ˙ X = P Y = AX = A(P Y ), so x1 y1 λ1 0 0 λ2 ˙ Y = (P −1 AP )Y = . Hence x1 = λ1 x1 and y˙1 = λ2 y1 . Then ˙ x1 = x1 (0)eλ1 t But x(0) y (0) so x1 (0) y1 (0) and =P y1 = y1 (0)eλ2 t . x1 (0) y1 (0) x(0) y (0) = P −1 = , α β . Consequently x1 (0) = α and y1 (0) = β and x y =P x1 y1 x0 y0 = [X1 |X2 ] = αeλ1 t X1 + βeλ2 t X2 . 68 α eλ1 t βeλ2 t ...
View Full Document

This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

Ask a homework question - tutors are online