It is given that y = ae^3x + be^-2x --------(1)

Now, differentiating both side we get,

y’ = 3ae^3x - 2be^-2x --------(2)

Now, again differentiating both sides, we get,

y’’ = 9ae^3x + 4be^-2x -------(3)

Now, let us multiply equation (1) with 2 and then adding it to equation (2), we get,

(2ae^3x + 2be^-2x) + (3ae^3x - 2be^-2x) = 2y – y’

⇒ 5ae^3x = 2y + y’

Now, let us multiply equation (1) with 3 and subtracting equation (2), we get

(3ae^3x + 3be^-2x) - (3ae^3x - 2be^-2x) = 3y – y’

⇒ 5be^-2x = 3y - y’

y” = 9. +4

⇒ y” = 6y + y’

⇒ y” – y’ - 6y = 0

Therefore, the required differential equation is y” – y’ - 6y = 0.