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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
Solution
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
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