# `y = C_1e^2x + C_2e^(-2x) + C_3sin(2x) + C_4cos(2x)` Determine whether the function is a solution of the differential equation `y^((4)) - 16y = 0`

Given,

`y = C_1e^2x + C_2e^(-2x) + C_3sin(2x) + C_4cos(2x)`

let us find

`y'=(C_1e^2x + C_2e^(-2x) + C_3sin(2x) + C_4cos(2x))'`

`= 2 C_1e^2x +(-2) C_2e^(-2x) +2 C_3cos(2x) -2 C_4 sin(2x)`

`y''=(2 C_1e^2x +(-2) C_2e^(-2x) +2 C_3cos(2x) -2 C_4 sin(2x))'`

`=4 C_1e^2x +(4) C_2e^(-2x) -4 C_3sin(2x) -4 C_4 cos(2x)`

`y'''=(4 C_1e^2x +(4) C_2e^(-2x) -4 C_3sin(2x) -4 C_4 cos(2x))'`

`=8 C_1e^2x +(-8) C_2e^(-2x) -8 C_3cos(2x) +8 C_4 sin(2x)`

`y''''=(8 C_1e^2x +(-8) C_2e^(-2x) -8 C_3cos(2x) +8 C_4 sin(2x))'`

`=(16 C_1e^2x +(-8)(-2) C_2e^(-2x) -8(-2) C_3sin(2x) +8(2) C_4 cos(2x))`

`=(16C_1e^2x +16 C_2e^(-2x) +16 C_3sin(2x) + 16C_4cos(2x))`

So lets check whether `y'''' -16 y =0` or not

`(16C_1e^2x +16 C_2e^(-2x) +16 C_3sin(2x) + 16C_4cos(2x))-16(C_1e^2x + C_2e^(-2x) + C_3sin(2x) + C_4cos(2x))`

=`(16C_1e^2x +16 C_2e^(-2x) +16 C_3sin(2x) + 16C_4cos(2x))-(16C_1e^2x +16 C_2e^(-2x) +16 C_3sin(2x) + 16C_4cos(2x))`

`=0`

so,

`y'''' -16 y =0`