If ` y=a cos ( log x )+ bsin (log x) where a ,b are ` parameters ,then ` x^(2) y''+ xy '=`

Options

A. ` -2y `
B. ` 2y `
C. ` -y`
D. ` y`

Answer

To solve the problem \( x^2 y'' + x y' \) where \( y = a \cos(\log x) + b \sin(\log x) \), we will follow these steps:

### Step 1: Find \( y' \)

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

Using the chain rule for differentiation:
\[ y' = \frac{dy}{dx} = \frac{d}{dx}[a \cos(\log x)] + \frac{d}{dx}[b \sin(\log x)] \]

Calculating each derivative:
1. For \( a \cos(\log x) \):
\[ \frac{d}{dx}[a \cos(\log x)] = -a \sin(\log x) \cdot \frac{d}{dx}[\log x] = -a \sin(\log x) \cdot \frac{1}{x} \]

2. For \( b \sin(\log x) \):
\[ \frac{d}{dx}[b \sin(\log x)] = b \cos(\log x) \cdot \frac{d}{dx}[\log x] = b \cos(\log x) \cdot \frac{1}{x} \]

Combining these results:
\[ y' = -\frac{a \sin(\log x)}{x} + \frac{b \cos(\log x)}{x} \]
Thus,
\[ y' = \frac{b \cos(\log x) - a \sin(\log x)}{x} \]

### Step 2: Find \( y'' \)

Now we differentiate \( y' \):
\[ y' = \frac{b \cos(\log x) - a \sin(\log x)}{x} \]

Using the quotient rule:
\[ y'' = \frac{d}{dx}\left(\frac{b \cos(\log x) - a \sin(\log x)}{x}\right) \]
Let \( u = b \cos(\log x) - a \sin(\log x) \) and \( v = x \).

Using the quotient rule:
\[ y'' = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \( \frac{du}{dx} \):
1. For \( b \cos(\log x) \):
\[ \frac{d}{dx}[b \cos(\log x)] = -b \sin(\log x) \cdot \frac{1}{x} \]

2. For \( -a \sin(\log x) \):
\[ \frac{d}{dx}[-a \sin(\log x)] = -a \cos(\log x) \cdot \frac{1}{x} \]

Combining these:
\[ \frac{du}{dx} = -\frac{b \sin(\log x)}{x} - \frac{a \cos(\log x)}{x} \]
Thus,
\[ \frac{du}{dx} = -\frac{b \sin(\log x) + a \cos(\log x)}{x} \]

Now substituting back into the quotient rule:
\[ y'' = \frac{x \left(-\frac{b \sin(\log x) + a \cos(\log x)}{x}\right) - (b \cos(\log x) - a \sin(\log x)) \cdot 1}{x^2} \]
This simplifies to:
\[ y'' = \frac{-b \sin(\log x) - a \cos(\log x) - b \cos(\log x) + a \sin(\log x)}{x^2} \]
Thus,
\[ y'' = \frac{(a - b) \sin(\log x) - (a + b) \cos(\log x)}{x^2} \]

### Step 3: Calculate \( x^2 y'' + x y' \)

Now we substitute \( y' \) and \( y'' \) into the expression:
\[ x^2 y'' + x y' = x^2 \left(\frac{(a - b) \sin(\log x) - (a + b) \cos(\log x)}{x^2}\right) + x \left(\frac{b \cos(\log x) - a \sin(\log x)}{x}\right) \]

This simplifies to:
\[ (a - b) \sin(\log x) - (a + b) \cos(\log x) + (b \cos(\log x) - a \sin(\log x)) \]

Combining like terms:
\[ = (a - b - a) \sin(\log x) + (b - (a + b)) \cos(\log x) \]
\[ = -b \sin(\log x) - a \cos(\log x) \]

### Final Result

Thus, the expression simplifies to:
\[ = - (a \cos(\log x) + b \sin(\log x)) = -y \]

### Conclusion
The final result is:
\[ x^2 y'' + x y' = -y \]