Question

If $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$ then $x{y}_2+\left(\frac{1}{2}\right)y_1$ is equal to

• A. $y$
• B. $x(e^{\sqrt{x}}+e^{-\sqrt{x}})$
• C. $\left(\frac{1}{4}\right)y$
• D. $y\sqrt{x}$

Answer 1

The correct option is C $\left(\frac{1}{4}\right)y$

Given, $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$
$\Rightarrow y_1=\frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}-\frac{1}{2\sqrt{x}}{e}^{-\sqrt{x}}$
$\Rightarrow$ $2\sqrt{x}{y}_1=e^{\sqrt{x}}-e^{-\sqrt{x}}$.
Differentiating again, we get
$2\sqrt{x}{y}_2+\frac{1}{\sqrt{x}}{y}_1=\frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}+\frac{1}{2\sqrt{x}}{e}^{-\sqrt{x}}=\frac{1}{2\sqrt{x}}y$
$\Rightarrow$ $x{y}_2+\left(\frac{1}{2}\right)y_1=\left(\frac{1}{4}\right)y$