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Unformatted text preview: MAP 2302 Exam III, Spring 2011 1. Determine the undetermined coefficient form of a particular solution to y 00 6 y + 5 y = e 5 t + 2 t sin( t ) 3 cos( t ) . (Compare with problem 31, Section 4.5.) First solve the homogeneous equation, y 00 6 y + 5 y = 0 . It is linear homogeneous with constant coefficients, and so we find the roots of the char acteristic equation, λ 2 6 λ + 5 = ( λ 5)( λ 1). The (distinct real) roots are r 1 = 1 and r 2 = 5. Hence { e t ,e 5 t } is a fundamental set of solutions to the homogeneous equation. To find the form of a particular solution y p , we first find the form of a solution y 1 of y 00 6 y + 5 y = e 5 t . Because e 5 t is already a solution of the homogeneous equation, y 1 = Ate 5 t , for some A . Next we note that the form of a particular solution y 2 of y 00 6 y + 5 y = 2 t sin( t ) 3 cos( t ) is y 2 = ( Bt + C ) sin( t ) + ( Dt + E ) cos( t ) ....
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This note was uploaded on 12/15/2011 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.
 Spring '08
 TUNCER

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