Question

Obtain the differential equation by eliminating the arbitrary constants from the following equation:

y = Ae^5x + Be^-5x 

Answer 1

y = Ae^5x + Be^-5x      ...(1)

Differentiating twice w.r.t. x, we get

`"dy"/"dx" = "Ae"^"5x" xx 5 + "Be"^(- "5x") xx (- 5)`

∴ `"dy"/"dx" = 5"Ae"^"5x" - 5"Be"^(- "5x")`

and `("d"^2"y")/"dx"^2 = "Ae"^"5x" xx 5 + "Be"^(- "5x") xx (- 5)`

`= 25"Ae"^"5x" + 25"Be"^(- "5x")`

`= 25("Ae"^"5x" + "Be"^(- "5x")) = 25"y"`   ....[By(1)]

∴ `("d"^2"y")/"dx"^2 - 25"y" = 0`

This is the required D.E.