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. $\frac{1}{4}y$
• D. $\sqrt{x}y$

Answer 1

Answer: (C)

$\frac{1}{4}y$

Given $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$

Differentiating both sides w.r.t $x$, we get

$y_1=\frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}-\frac{1}{2\sqrt{x}}{e}^{-\sqrt{x}}$

$\Rightarrow 2\sqrt{x}{y}_1=e^{\sqrt{x}}-e^{-\sqrt{x}}$

Differentiating again, we get

$2\sqrt{x}{y}_2+\frac{1}{\sqrt{x}}{y}_1=\frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}+\frac{1}{2\sqrt{x}}{e}^{-\sqrt{x}}=\frac{1}{2\sqrt{x}}y$

$\Rightarrow x{y}_2+\left(\frac{1}{2}\right)y_1=\frac{1}{4}y$