Thanks for answering me!

I know that the answer is e^((-9/2)x) (Acos(3√7/2)x) + Bsin(3√7/2)x) ]]>

x = (-b ± √ (b² - 4ac))÷ 2a

x = (-9 ± √ (-63))÷ 2

The negative square root confirms my suspicions.

The roots work out to be -4.5 ± (1.5√7)i, where i is √(-1).

If the solutions to the auxiliary equation are complex, they go into the following equation:

λ = c±di ∴ y = e^c (Acosdx + Bsindx), where A and B are arbitrary constants.

For your example, y = e^-4.5 (Acos(1.5√7)x + Bsin(1.5√7)x)

]]>y + 9y' + 36y= 0

λ^2 + 9λ + 36= 0 → λ= ?

Matilde

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