Question

Find the differential equation of the family of curves y = Ae^2x + B.e^–2x.

Answer 1

y = Ae^2x + B.e^–2x.

`("d"y)/("d"x) = 2"Ae"^(2x) - 2"B"e"^(-2x)` and `("d"^2y)/("dx"^2) = 4"Ae"^(2x) + 4"Be"^(-2x)`

Thus `("d"^2y)/("dx"^2) = 4y`

i.e., `("d"^2y)/("dx"^2) - 4y` = 0.