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Answer 1 · 2017-05-12T23:19:22

$y'' - y' - 2y =-12(x + 2)$

Given: $y = e^(2x) + 3(2x+3) = e^(2x) +6x + 9$

Find the first derivative using $(e^u)' = u' e^u$ :

Let $u = 2x; " " u' = 2$

$y ' = 2e^(2x) + 6$

$y'' = 4 e^(2x)$

Find $" "y'' - y' - 2y$ :

$y'' - y' - 2y = 4 e^(2x) - ( 2e^(2x) + 6) - 2(e^(2x) +6x + 9)$

Distribute the negatives:

$y'' - y' - 2y = 4 e^(2x) - 2e^(2x) - 6 -2e^(2x) - 12x -18$

Simplify by combining like-terms:

$y'' - y' - 2y = -6 - 12x - 18 = -24 -12x$

Factor: $" "y'' - y' - 2y =-12(x + 2)$