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: Put \(\sqrt{x}=t\).
Let \[ t=\sqrt{x} \] Then, \[ x=t^2 \] and \[ y=e^t+e^{-t} \] Also, \[ \frac{dt}{dx}=\frac{1}{2t} \]

Step 2: Find \(y_1\).
\[ y_1=\frac{dy}{dx} \] \[ \frac{dy}{dt}=e^t-e^{-t} \] Therefore, \[ y_1=\frac{dy}{dt}\cdot \frac{dt}{dx} \] \[ y_1=(e^t-e^{-t})\frac{1}{2t} \] \[ y_1=\frac{e^t-e^{-t}}{2t} \]

Step 3: Find \(y_2\).
\[ y_2=\frac{d}{dx}\left(\frac{e^t-e^{-t}}{2t}\right) \] Since \[ \frac{d}{dx}=\frac{1}{2t}\frac{d}{dt}, \] we get \[ y_2=\frac{1}{2t}\frac{d}{dt}\left(\frac{e^t-e^{-t}}{2t}\right) \] Now, \[ \frac{d}{dt}\left(\frac{e^t-e^{-t}}{2t}\right) = \frac{2t(e^t+e^{-t})-2(e^t-e^{-t})}{4t^2} \] \[ = \frac{t(e^t+e^{-t})-(e^t-e^{-t})}{2t^2} \] Thus, \[ y_2= \frac{t(e^t+e^{-t})-(e^t-e^{-t})}{4t^3} \]

Step 4: Evaluate \(4xy_2+2y_1\).
Since \[ x=t^2, \] we have \[ 4xy_2=4t^2\cdot \frac{t(e^t+e^{-t})-(e^t-e^{-t})}{4t^3} \] \[ = \frac{t(e^t+e^{-t})-(e^t-e^{-t})}{t} \] \[ = e^t+e^{-t}-\frac{e^t-e^{-t}}{t} \] Also, \[ 2y_1=2\cdot \frac{e^t-e^{-t}}{2t} \] \[ 2y_1=\frac{e^t-e^{-t}}{t} \] Therefore, \[ 4xy_2+2y_1 = e^t+e^{-t}-\frac{e^t-e^{-t}}{t} + \frac{e^t-e^{-t}}{t} \] \[ =e^t+e^{-t} \] But, \[ y=e^t+e^{-t} \] Hence, \[ 4xy_2+2y_1=y \]

Step 5: Final conclusion.
Therefore, \[ \boxed{y} \]