If `y = a cos (ln x)+ b sin (ln x)`, then `x^(2)(d^(2)y)/(dx^(2))+x(dy)/(dx)=`

Options

A. 0
B. y
C. `-y`
D. None of these

Answer

To solve the problem, we need to differentiate the function \( y = a \cos(\ln x) + b \sin(\ln x) \) and then evaluate the expression \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} \).

### Step 1: Differentiate \( y \) with respect to \( x \)

Given:
\[ y = a \cos(\ln x) + b \sin(\ln x) \]

Using the chain rule, we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}[a \cos(\ln x)] + \frac{d}{dx}[b \sin(\ln x)] \]

Using the derivative of \( \cos(u) \) and \( \sin(u) \) where \( u = \ln x \):
\[ \frac{dy}{dx} = -a \sin(\ln x) \cdot \frac{1}{x} + b \cos(\ln x) \cdot \frac{1}{x} \]
\[ \frac{dy}{dx} = \frac{1}{x}(-a \sin(\ln x) + b \cos(\ln x)) \]

### Step 2: Multiply \( \frac{dy}{dx} \) by \( x \)

Now we compute \( x \frac{dy}{dx} \):
\[ x \frac{dy}{dx} = x \cdot \frac{1}{x}(-a \sin(\ln x) + b \cos(\ln x)) = -a \sin(\ln x) + b \cos(\ln x) \]

### Step 3: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \) again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x}(-a \sin(\ln x) + b \cos(\ln x))\right) \]

Using the product rule:
\[ \frac{d^2y}{dx^2} = -\frac{1}{x^2}(-a \sin(\ln x) + b \cos(\ln x)) + \frac{1}{x} \left(-a \cos(\ln x) \cdot \frac{1}{x} + b (-\sin(\ln x) \cdot \frac{1}{x})\right) \]
\[ = -\frac{1}{x^2}(-a \sin(\ln x) + b \cos(\ln x)) + \frac{1}{x^2}(-a \cos(\ln x) - b \sin(\ln x)) \]

Combining the terms:
\[ \frac{d^2y}{dx^2} = \frac{1}{x^2} \left( a \sin(\ln x) - b \cos(\ln x) - a \cos(\ln x) - b \sin(\ln x) \right) \]
\[ = \frac{1}{x^2} \left( (a - b) \sin(\ln x) - (a + b) \cos(\ln x) \right) \]

### Step 4: Compute \( x^2 \frac{d^2y}{dx^2} \)

Now, we compute \( x^2 \frac{d^2y}{dx^2} \):
\[ x^2 \frac{d^2y}{dx^2} = (a - b) \sin(\ln x) - (a + b) \cos(\ln x) \]

### Step 5: Combine \( x^2 \frac{d^2y}{dx^2} \) and \( x \frac{dy}{dx} \)

Now we add \( x^2 \frac{d^2y}{dx^2} \) and \( x \frac{dy}{dx} \):
\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = \left[(a - b) \sin(\ln x) - (a + b) \cos(\ln x)\right] + \left[-a \sin(\ln x) + b \cos(\ln x)\right] \]

Combining these:
\[ = (a - b - a) \sin(\ln x) + (- (a + b) + b) \cos(\ln x) \]
\[ = -b \sin(\ln x) - a \cos(\ln x) \]

### Final Result

Thus, we find that:
\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = -y \]