Q. 3

# In each of the question, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = a e^{3x} + b e^{–2x}

Answer :

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.

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