Question

If $y=e^{\sqrt{x}}+e^{-\sqrt{x}},$ then $x{y}^{\prime \prime }+\frac{1}{2}{y}^{\prime }=$

• A. $\frac{1}{4}y$
• B. $-\frac{1}{4}y$
• C. $0$
• D. $y$

Answer 1

The correct option is A $\frac{1}{4}y$

$y=e^{\sqrt{x}}+e^{-\sqrt{x}}\Rightarrow y^{\prime }=e^{\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)-e^{-\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)\Rightarrow 2\sqrt{x}{y}^{\prime }=e^{\sqrt{x}}-e^{-\sqrt{x}}\Rightarrow 2\sqrt{x}{y}^{\prime \prime }+\frac{y^{\prime }}{\sqrt{x}}=\frac{e^{\sqrt{x}}}{2\sqrt{x}}+\frac{e^{-\sqrt{x}}}{2\sqrt{x}}\Rightarrow 2x{y}^{\prime \prime }+y^{\prime }=\frac{y}{2}\therefore x{y}^{\prime \prime }+\frac{y^{\prime }}{2}=\frac{y}{4}$