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