Cho \(y=\left(x+\sqrt{x^2+1}\right)^3\). Chọn khẳng định đúng.
\(\left(1+x^2\right)y'+xy"-9y=0\).\(\left(1+x^2\right)y"+xy'-9y=0\).\(\left(1+x^2\right)y"-xy'-9y=0\).\(\left(1+x^2\right)y"+xy-9y'=0\).Hướng dẫn giải:\(y'=3\left(x+\sqrt{x^2+1}\right)^2\left(x+\sqrt{x^2+1}\right)'=3\left(1+\dfrac{x}{\sqrt{x^2+1}}\right)\left(x+\sqrt{x^2+1}\right)^2=\dfrac{1}{\sqrt{x^2+1}}\left(x+\sqrt{x^2+1}\right)^3\Rightarrow y'\sqrt{x^2+1}=y\) (1)
\(\Rightarrow\left(y'\sqrt{x^2+1}\right)'=y'\Rightarrow y".\sqrt{x^2+1}+y'.\dfrac{x}{\sqrt{x^2+1}}=y'\Rightarrow y".\left(x^2+1\right)+y'x=y'\sqrt{x^2+1}\) (2)
Thế (1) vào (2) ta được \(\Rightarrow y".\left(x^2+1\right)+y'x=y\)