Question

If y = 500e^7x + 600e^-7x, then show that y₂ – 49y = 0.

Answer 1

y = 500e^7x + 600e^-7x 

`y_1 = "dy"/"dx" = 500 "d"/"dx" (e^(7x)) + 600 "d"/"dx" (e^(-7x))`

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

`y_2 = ("d"^2"y")/"dx"^2 = 500xx7 "d"/"dx" (e^(7x)) + 600(-7) "d"/"dx" (e^(-7x))`

`= 500 xx 7(7e^(7x)) + 600 xx (-7)(-7) e^(-7x)`

`= 500 xx 49e^7x + 600 xx 49e^(-7x)`

`y_2 = 49 [500 r^(7x) + 600e^(-7x)]` = 49y

(or) y₂ – 49y = 0