Answer :

It is given that y = ae3x + be-2x --------(1)

Now, differentiating both side we get,


y’ = 3ae3x - 2be-2x --------(2)


Now, again differentiating both sides, we get,


y’’ = 9ae3x + 4be-2x -------(3)


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


(2ae3x + 2be-2x) + (3ae3x - 2be-2x) = 2y – y’


5ae3x = 2y + y’



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


(3ae3x + 3be-2x) - (3ae3x - 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.


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