`y = 3e^(2x) - 4sin(2x)` Determine whether the function is a solution of the differential equation `y^((4)) - 16y = 0`

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Given,

`y=3e^(2x) -4sin(2x)`

so,

we have to find

`y'=(3e^(2x) -4sin(2x))' =(3e^(2x))'-(4sin(2x))'`

`=3*2 e^(2x)-2*4 cos(2x)`

`=6e^(2x)-8cos(2x)`

similarly

`y'' =(6e^(2x)-8cos(2x))'`

`=6*2 e^(2x)+2*8 sin(2x)`

`=12 e^(2x)+16 sin(2x)`

`y'''=(12 e^(2x)+16 sin(2x))'`

`=12*2 e^(2x)+16*2 cos(2x)`

`=24 e^(2x)+32 cos(2x)`

`y'''' =(24 e^(2x)+32 cos(2x))'`

`=24*2 e^(2x)-32*2 sin(2x)`

`=48 e^(2x)-64sin(2x)`

So lets see whether...

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Given,

`y=3e^(2x) -4sin(2x)`

so,

we have to find

`y'=(3e^(2x) -4sin(2x))' =(3e^(2x))'-(4sin(2x))'`

`=3*2 e^(2x)-2*4 cos(2x)`

`=6e^(2x)-8cos(2x)`

similarly

 

`y'' =(6e^(2x)-8cos(2x))'`

`=6*2 e^(2x)+2*8 sin(2x)`

`=12 e^(2x)+16 sin(2x)`

 

`y'''=(12 e^(2x)+16 sin(2x))'`

`=12*2 e^(2x)+16*2 cos(2x)`

`=24 e^(2x)+32 cos(2x)`

 

`y'''' =(24 e^(2x)+32 cos(2x))'`

`=24*2 e^(2x)-32*2 sin(2x)`

`=48 e^(2x)-64sin(2x)`

 

So lets see whether ` y'''' -16y=0`

=> `48 e^(2x)-64sin(2x) -16(3e^(2x) -4sin(2x))`

`=48 e^(2x)-64sin(2x) -48 e^(2x)+64sin(2x) =0`

so,

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

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