Question

The function $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$, 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 $\frac{1}{4}y$

$y^1=e^{\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)-e^{-\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)$

$y^{{}^{\prime }}=\frac{1}{2\sqrt{x}}(e^{\sqrt{x}}-e^{-\sqrt{x}})$

$y^{{}^{\prime \prime }}=\frac{1}{2\sqrt{x}}(\frac{e^{\sqrt{x}}}{2\sqrt{x}}+\frac{e^{-\sqrt{x}}}{2\sqrt{x}})+\frac{-1}{4(x{)}^{\frac{3}{2}}}(e^{\sqrt{x}}-e^{-\sqrt{x}})$

$y^{{}^{\prime \prime }}=\frac{y}{4x}-\frac{y^1}{2x}$

$x{y}^{\prime \prime }+y^{{}^{\prime }}=\frac{y}{4}$