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Unformatted text preview: y ( t ) = 5 y 1 ( t ) = 5 cos t is a solution to y y + y = 5 sin t . (b) We can express sin t 3 e 2 t = g 1 ( t ) 3 g 2 ( t ). So, by the superposition principle the desired solution is y ( t ) = y 1 ( t ) 3 y 2 ( t ) = cos t e 2 t . (c) Since 4 sin t + 18 e 2 t = 4 g 1 ( t ) + 18 g 2 ( t ), the function y ( t ) = 4 y 1 ( t ) + 18 y 2 ( t ) = 4 cos t + 6 e 2 t is a solution to the given equation. 196...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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