Question

If y = 500e^7x + 600e^–7x, show that `(d^2y)/(dx^2)` = 49y.

Answer 1

y = 500e^7x + 600e^–7x  ...(1)

On differentiating with respect to x,

`dy/dx = d/dx (500 e^(7x) + 600 e^(- 7x))`

= `500 d/dx e^(7x) + 600 d/dx e^(- 7x)`

= `500 e^(7x) d/dx (7x) + 600 e^(- 7x) d/dx (-7x)`

= 500e^7x . 7 + 600e^–7x. (−7)

= 3500e^7x − 4200e^−7x

Differentiating again with respect to x,

`(d^2 y)/dx^2 = d/dx [3500e^(7x) − 4200e^(−7x)]`

= `3500 d/dx e^(7x) - 4200 d/dx e^(- 7x)`

= `3500e^(7x) d/dx (7x) - 4200e^(- 7x) d/dx (- 7x)`

= 3500e^7x . 7 − 4200e^−7x . (−7)

= 24500e^7x  − 29400e^−7x

= 500 × 49e^7x + 600 × 49e^−7x

= 49(500e^7x + 600e^−7x)

= 49 y  ...[From equation (1)]

∴ `(d^2y)/dx^2` = 49y