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lecture_20 - MA 36600 LECTURE NOTES WEDNESDAY MARCH 4...

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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 00 + b ( t ) y 0 + c ( t ) y = f ( t ) . Say that { y 1 , y 2 } is a fundamental set of solutions to the homogeneous equation a ( t ) y 00 + b ( t ) y 0 + 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 0 = ( u 1 y 0 1 + u 2 y 0 2 ) + ( u 0 1 y 1 + u 0 2 y 2 ) y 00 = ( u 1 y 00 1 + u 2 y 00 2 ) + ( u 0 1 y 0 1 + u 0 2 y 0 2 ) + ( u 0 1 y 1 + u 0 2 y 2 ) 0 This gives the expression f ( t ) = a ( t ) y 00 + b ( t ) y 0 + c ( t ) y = u 1 a ( t ) y 00 1 + b ( t ) y 0 1 + c ( t ) y 1 + u 2 a ( t ) y 00 2 + b ( t ) y 0 2 + c ( t ) y 2 + a ( t ) u 0 1 y 1 + u 0 2 y 2 + a ( t ) ( u 0 1 y 1 + u 0 2 y 2 ) 0 + b ( t ) ( u 0 1 y 1 + u 0 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 0 1 ( t ) y 1 ( t ) + u 0 2 ( t ) y 2 ( t ) = 0 u 0 1 ( t ) y 0 1 ( t ) + u 0 2 ( t ) y 0 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 differential equations. Multiply the first equation by y 0 2 ( t ) (by - y 0 1 ( t )) and the second by - y 2 ( t ) (by y 1 ( t ), respectively) to find the following systems of equations: y 1 y 0 2 u 0 1 ( t ) + y 2 y 0 2 u 0 2 ( t ) = 0 - y 0 1 y 2 u 0 1 ( t ) + - y 2 y 0 2 u 0 2 ( t ) = - f ( t ) a ( t ) y 2 W u 0 1 ( t ) = - f ( t ) a ( t ) y 2 - y 1 y 0 1 u 0 1 ( t ) + - y 0 1 y 2 u 0 2 ( t ) = 0 y 1 y 0 1 u 0 1 ( t ) + y 1 y 0 2 u 0 2 ( t ) = f ( t ) a ( t ) y 1 W u 0 2 ( t ) = f ( t ) a ( t ) y 1 where we have denoted the function
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