Sunday, 2 November 2014

`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))'`


...

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`

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