If `y=x+sqrt(x^(2)-1)," then "(x^(2)-1)y_(2)+xy_(1)=`

Options

A. 0
B. 1
C. y
D. `-y`

Answer

To solve the problem, we need to find the expression \( (x^2 - 1)y_2 + xy_1 \), where \( y = x + \sqrt{x^2 - 1} \), \( y_1 = \frac{dy}{dx} \), and \( y_2 = \frac{d^2y}{dx^2} \).

### Step 1: Find \( y_1 = \frac{dy}{dx} \)

Given:
\[ y = x + \sqrt{x^2 - 1} \]

To differentiate \( y \) with respect to \( x \):
\[ y_1 = \frac{dy}{dx} = 1 + \frac{1}{2\sqrt{x^2 - 1}} \cdot (2x) = 1 + \frac{x}{\sqrt{x^2 - 1}} \]

### Step 2: Find \( y_2 = \frac{d^2y}{dx^2} \)

Now, we differentiate \( y_1 \):
\[ y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}\left(1 + \frac{x}{\sqrt{x^2 - 1}}\right) \]

Using the quotient rule for \( \frac{x}{\sqrt{x^2 - 1}} \):
Let \( u = x \) and \( v = \sqrt{x^2 - 1} \).
Then,
\[ y_2 = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \frac{x}{\sqrt{x^2 - 1}} \).

Calculating \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{x}{\sqrt{x^2 - 1}} \]

Now substituting:
\[ y_2 = \frac{\sqrt{x^2 - 1} \cdot 1 - x \cdot \frac{x}{\sqrt{x^2 - 1}}}{x^2 - 1} \]
\[ = \frac{\sqrt{x^2 - 1} - \frac{x^2}{\sqrt{x^2 - 1}}}{x^2 - 1} \]
\[ = \frac{(x^2 - 1) - x^2}{(x^2 - 1)\sqrt{x^2 - 1}} = \frac{-1}{(x^2 - 1)\sqrt{x^2 - 1}} \]

### Step 3: Substitute \( y_1 \) and \( y_2 \) into the expression

Now we substitute \( y_1 \) and \( y_2 \) into the expression \( (x^2 - 1)y_2 + xy_1 \):
\[ (x^2 - 1)y_2 + xy_1 = (x^2 - 1)\left(-\frac{1}{(x^2 - 1)\sqrt{x^2 - 1}}\right) + x\left(1 + \frac{x}{\sqrt{x^2 - 1}}\right) \]
\[ = -\frac{1}{\sqrt{x^2 - 1}} + x + \frac{x^2}{\sqrt{x^2 - 1}} \]
\[ = x - \frac{1}{\sqrt{x^2 - 1}} + \frac{x^2}{\sqrt{x^2 - 1}} \]
\[ = x + \frac{x^2 - 1}{\sqrt{x^2 - 1}} \]

### Final Expression

Thus, the final expression is:
\[ = x + \frac{x^2 - 1}{\sqrt{x^2 - 1}} \]