Posted by Sara on Tuesday, July 2, 2013 at 11:34pm.
what'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