Question

If \[ \frac{d^n y}{dx^n}=y_n \] and \[ y=e^{\sqrt{x}}+e^{-\sqrt{x}}, \] then \[ 4xy_2+2y_1= \]

• A. \(-y\)
• B. \(y\)
• C. \(2y\)
• D. \(-2y\)

Hint:

When a function contains \(\sqrt{x}\), use the substitution \[ t=\sqrt{x} \] so that \[ \frac{d}{dx}=\frac{1}{2t}\frac{d}{dt}. \] This makes differentiation simpler.

Answer 1

The Correct Option is B

Step 1: Recognise the structure of $y$.
$y = e^{\sqrt{x}} + e^{-\sqrt{x}}$. Note that $y = 2\cosh(\sqrt{x})$. This hint can simplify things, but let us use direct differentiation.

Step 2: Differentiate $y$ with respect to $x$ to get $y_1$.
\[ y_1 = \frac{d}{dx}e^{\sqrt{x}} + \frac{d}{dx}e^{-\sqrt{x}} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} + e^{-\sqrt{x}} \cdot \frac{-1}{2\sqrt{x}}. \] \[ y_1 = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}}. \]

Step 3: Multiply both sides by $2\sqrt{x}$.
\[ 2\sqrt{x}\, y_1 = e^{\sqrt{x}} - e^{-\sqrt{x}}. \]

Step 4: Differentiate again using product rule on the left side.
\[ \frac{d}{dx}(2\sqrt{x}\, y_1) = \frac{d}{dx}(e^{\sqrt{x}} - e^{-\sqrt{x}}). \] Left: $\frac{y_1}{\sqrt{x}} + 2\sqrt{x}\, y_2$. Right: $\frac{e^{\sqrt{x}} + e^{-\sqrt{x}}}{2\sqrt{x}} = \frac{y}{2\sqrt{x}}$.

Step 5: Multiply through by $2\sqrt{x}$.
\[ 2y_1 + 4xy_2 = y. \]

Step 6: State the answer.
$4xy_2 + 2y_1 = y$.
\[ \boxed{y} \]