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July 30, 2016
Posted by **Sara** on Tuesday, July 2, 2013 at 11:34pm.

- Calculus -
**Steve**, Wednesday, July 3, 2013 at 4:52amwhat's the troub;e? Just plug and chug

y1 = e^(-2x)

y1' = -2e^(-2x)

y1" = 4e^(-2x)

y"+4y'+4 = 0

y2 = xe^(-2x)

y2' = (1-2x) e^(-2x)

y2" = (4x-4) e^(-2x)

y"+4y'+4y = 0