Cho hàm số \(y=e^{-x}.\sin x\). Hệ thức nào đúng ?
\(y'+2y"-2y=0\) \(y"+2y'+2y=0\) \(y"-2y'-2y=0\) \(y'-2y"+2y=0\) Hướng dẫn giải:\(y=e^{-x}\sin x,y'=-e^{-x}\sin x+e^{-x}\cos x,y"=e^{-x}\sin x-e^{-x}\cos x-e^{-x}\cos x-e^{-x}\sin x=-2e^{-x}\cos x\)
a) \(y'+2y"-2y=\left(-e^{-x}\sin x+e^{-x}\cos x\right)+2\left(-2e^{-x}\cos x\right)-2\left(e^{-x}\sin x\right)=-3e^{-x}\left(\sin x+\cos x\right)\ne0\)
b) \(y"+2y'+2y=\left(-2e^{-x}\cos x\right)+2\left(-e^{-x}\sin x+e^{-x}\cos x\right)+2\left(e^{-x}\sin x\right)=0\)
c) \(y"-2y'-2y=\left(-2e^{-x}\cos x\right)-2\left(-e^{-x}\sin x+e^{-x}\cos x\right)-2\left(e^{-x}\sin x\right)=-4e^{-x}\left(\cos x+\sin x\right)\ne0\)
d) \(y'-2y"+2y=\left(-e^{-x}\sin x+e^{-x}\cos x\right)-2\left(-2e^{-x}\cos x\right)+2\left(e^{-x}\sin x\right)=e^{-x}\left(\sin x-3\cos x\right)\ne0\)
Đáp số: \(y"+2y'+2y=0\)