Question

If y = e^–x (Acosx + Bsinx), then y is a solution of ______.

Options

• `("d"^2y)/("d"x^2) + 2("d"y)/("d"x)` = 0

• `("d"^2y)/("d"x^2) - 2 ("d"y)/("d"x) + 2y ` = 0

• `("d"^2y)/("d"x^2) + 2 ("d"y)/("d"x) + 2y` = 0

• `("d"^2y)/("d"x^2) + 2y` = 0

Answer 1

If y = e^–x (Acosx + Bsinx), then y is a solution of `("d"^2y)/("d"x^2) + 2 ("d"y)/("d"x) + 2y` = 0.

Explanation:

Given equation is y = e^–x (Acosx + Bsinx)

Differentiating both sides, w.r.t. x, we get

 `("d"y)/("d"x)` = e^–x (–A sin x + B cos x) – e^–x (A cos x + B sin x)

`("d"y)/("d"x)` = e^–x (–A sin x + B cos x) – y

Again differentiating w.r.t. x, we get

`("d"^2y)/("d"x^2) = "e"^-x (-"A" cos x - "B" sin x) - "e"^-x (-"A" sinx + "B"cosx) - ("d"y)/("d"x)`  

⇒ `("d"^2y)/("d"x^2) = -"e"^-x ("A" cosx + "B" sinx) - [("d"y)/("d"x) + y] - ("d"y)/("d"x)`

⇒ `("d"^2y)/("d"x^2) = - y - ("d"y)/("d"x) - y - ("d"y)/("d"x)`

⇒ `("d"^2y)/("d"x^2) = - 2 ("d"y)/("d"x) - 2y`

⇒ `("d"^2y)/("d"x^2) + 2("d"y)/("d"x) + 2y` = 0