Cho hàm số \(y=e^{-x}.\sin x\). Khẳng định nào trong các khẳng định sau đây đúng ?
\(y'+2y"-2y=0\) \(y"+2y'+2y=0\) \(y"-2y'-2y=0\) \(y'-2y"+2y=0\) Hướng dẫn giải:Có \(y=e^{-x}.\sin x\Rightarrow y'=-e^{-x}\sin x+e^{-x}\cos x\)
\(\Rightarrow y"=-y'+\left(e^{-x}\cos x\right)'=e^{-x}\sin x-2e^{-x}\cos x-e^{-x}\sin x=-2e^{-x}\cos x\)
a) \(\left\{{}\begin{matrix}y'=e^{-x}\left(-\sin x+\cos x\right)\\2y"=-4e^{-x}\cos x\\-2y=-2e^{-x}\sin x\end{matrix}\right.\) \(\Rightarrow y'+2y"-2y=-3e^{-x}\left(\sin x+\cos x\right)\ne0\)
b) \(\left\{{}\begin{matrix}y"=-2e^{-x}\cos x\\2y'=-2e^{-x}\sin x+2e^{-x}\cos x\\2y=2e^{-x}\sin x\end{matrix}\right.\) \(\Rightarrow y"+2y'+2y=0\)
c) \(\left\{{}\begin{matrix}y"=-2e^{-x}\cos x\\-2y'=2e^{-x}\sin x-2e^{-x}\cos x\\-2y=-2e^{-x}\sin x\end{matrix}\right.\) \(\Rightarrow y"-2y'-2y=-4e^{-x}\cos x\ne0\)
d) \(\left\{{}\begin{matrix}y'=e^{-x}\left(-\sin x+\cos x\right)\\-2y"=4e^{-x}\cos x\\2y=2e^{-x}\sin x\end{matrix}\right.\) \(\Rightarrow y'-2y"+2y=e^{-x}\sin x+5e^{-x}\cos x\ne0\)
Đáp số: \(y"+2y'+2y=0\)