Asked by toby
find the linear differential equation that has y=5-x+2e^xsin3x as a particular solution
Answers
Answered by
Steve
y = 5 - x + 2e^x sin3x
y' = 2e^x sin3x + 6e^x cos3x - 1
y" = 12e^xcos3x - 16e^x sin3x
2y' = 4e^x sin3x + 12e^x cos3x - 2
y" = -16e^x sin3x + 12e^x cos3x
2y'-y" = 20e^x sin3x - 2
10y = 50 - 10x + 20e^x sin3x
2y'-y"-10y = -52+10x
or,
y"-2y'+10y = 52-10x
as usual, check my math
y' = 2e^x sin3x + 6e^x cos3x - 1
y" = 12e^xcos3x - 16e^x sin3x
2y' = 4e^x sin3x + 12e^x cos3x - 2
y" = -16e^x sin3x + 12e^x cos3x
2y'-y" = 20e^x sin3x - 2
10y = 50 - 10x + 20e^x sin3x
2y'-y"-10y = -52+10x
or,
y"-2y'+10y = 52-10x
as usual, check my math
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