`y=e^(-x)\ sin x`

so

`(dy)/(dx)=e^(-x)\ cos x+sin x(-e^(-x))`

`=e^(-x)[cos x-sin x]`

and

`(d^2y)/(dx^2)=e^(-x)[-sin x-cos x]+` `[cos x-sin x][-e^(-x)]`

`=e^(-x)[-2\ cos x]`

So

`"LHS"=(d^2y)/(dx^2)+2(dy)/(dx)+2y`

`=e^(-x)[-2\ cos x]+` `2[e^(-x)(cos x-sin x)]+` `2[e^(-x)\ sin x]`

`=e^(-x)[-2\ cos x+2\ cos x -` `{:2\ sin x+2\ sin x]`

`=0`

`="RHS"`