Let `y= e^(-x)(cos2x + sin2x)` Show that  `y'+ y =2e^(-x) (cos2x - sin2x).`

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to differentiate the given function with respect to x, using the product rule and the chain rule, such that:

`(dy)/(dx) = y' = (e^(-x))'(cos 2x + sin 2x) + (e^(-x))(cos 2x + sin 2x)'`

`y' = -e^(-x)(cos 2x + sin 2x) + e^(-x)*(-2sin 2x + 2cos 2x)`

Factoring out `e^(-x)` yields:

`y' = e^(-x)(-cos 2x - sin 2x - 2sin 2x + 2cos 2x)`

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

You need to perform the summation `y + y'` , such that:

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

Factoring out `e^(-x)` yields:

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

`y + y' = e^(-x)(2cos 2x - 2sin 2x)`

Factoring out 2 yields:

`y + y' = 2e^(-x)(cos 2x - sin 2x)`

Hence, evaluating the summation `y + y'` yields `y + y' = 2e^(-x)(cos 2x - sin 2x)` , thus, the identity `y + y' = 2e^(-x)(cos 2x - sin 2x)` holds.

We’ve answered 318,989 questions. We can answer yours, too.

Ask a question