{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture_18 (dragged) 2

# lecture_18 (dragged) 2 - MA 36600 LECTURE NOTES FRIDAY...

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

MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 3 Then the general solution to the nonhomogeneous equation is the function y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + Y ( t ) for some constants c 1 and c 2 . We remark that typically finding y 1 and y 2 is easy: when the coe cients a ( t ), b ( t ), and c ( t ) are constant functions, one can compute y 1 and y 2 by considering roots of the characteristic equation a r 2 + b r + c = 0. We will spend the next couple of lectures discussing how to find a solution Y . Note that once one solution Y is found, we can compute all solutions y . We give a proof of the claim above. Say that y = y ( t ) is any solution to the nonhomogeneous equation, and let Y = Y ( t ) be as in assumption (iii). Denote the di ff erence u ( t ) = y ( t ) Y ( t ) . We see that u = u ( t ) is a solution to the homogeneous equation: a ( t ) u + b ( t ) u + c ( t ) u = a ( t ) y Y + b ( t ) y Y + c ( t ) y Y = a ( t ) y + b ( t ) y + c ( t ) y a ( t ) Y + b ( t ) Y + c ( t ) Y = f ( t ) f ( t ) = 0 . By assumption (i), both y 1 and y 2 are also solutions to a ( t ) u + b ( t ) u + c ( t ) u = 0; and by assumption (ii) these functions are linearly independent. Hence
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}