Q. 3

# In each of the qu

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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