{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture_20 (dragged)

lecture_20 (dragged) - MA 36600 LECTURE NOTES WEDNESDAY...

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

View Full Document Right Arrow Icon
MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 4 Variation of Parameters General Method. We explain how to find the general solution of the nonhomogeneous equation a ( t ) y + b ( t ) y + c ( t ) y = f ( t ) . Say that { y 1 , y 2 } is a fundamental set of solutions to the homogeneous equation a ( t ) y + b ( t ) y + c ( t ) y = 0 . Consider the function y ( t ) = u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) . We have the derivatives y = u 1 y 1 + u 2 y 2 y = u 1 y 1 + u 2 y 2 + u 1 y 1 + u 2 y 2 y = u 1 y 1 + u 2 y 2 + u 1 y 1 + u 2 y 2 + u 1 y 1 + u 2 y 2 This gives the expression f ( t ) = a ( t ) y + b ( t ) y + c ( t ) y = u 1 a ( t ) y 1 + b ( t ) y 1 + c ( t ) y 1 + u 2 a ( t ) y 2 + b ( t ) y 2 + c ( t ) y 2 + a ( t ) u 1 y 1 + u 2 y 2 + a ( t ) u 1 y 1 + u 2 y 2 + b ( t ) u 1 y 1 + u 2 y 2 . Since y 1 and y 2 are solutions of a homogeneous equation, the middle row on the right-hand side is zero. We make the assumptions u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) = 0 u 1 ( t ) y 1 ( t ) + u 2 ( t ) y 2 ( t ) = f ( t ) a ( t ) We solve for u 1 = u 1 ( t ) and u 2 = u 2 ( t ) by considering a couple of first order di ff erential equations. Multiply the first equation by y 2 ( t ) (by y 1 ( t )) and the second by y 2 ( t ) (by y 1 ( t ), respectively) to find the following systems of equations: y 1 y 2 u 1 ( t ) + y 2 y 2 u 2 ( t ) = 0 y 1 y 2 u 1 ( t ) + y 2 y 2 u 2 ( t ) = f ( t ) a ( t ) y 2 W u 1 ( t ) = f ( t ) a ( t ) y 2 y 1 y 1 u 1 ( t ) + y 1 y 2
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}