If `y =3 cos (log x) + 4 sin (log x),` then `x ^(2) (d ^(2) y )/( dx ^(2)) + x (dy)/(dx) =`

Options

A. y
B. `-y ^(2)`
C. `(1)/(y)`
D. `- y`

Answer

To solve the problem, we need to differentiate the given function \( y = 3 \cos(\log x) + 4 \sin(\log x) \) twice and then substitute it into the expression \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} \).

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

Using the chain rule, we differentiate \( y \):

\[ \frac{dy}{dx} = \frac{d}{dx}[3 \cos(\log x)] + \frac{d}{dx}[4 \sin(\log x)] \]

Using the chain rule, we have:

\[ \frac{d}{dx}[\cos(u)] = -\sin(u) \cdot \frac{du}{dx} \quad \text{and} \quad \frac{d}{dx}[\sin(u)] = \cos(u) \cdot \frac{du}{dx} \]

where \( u = \log x \) and \( \frac{du}{dx} = \frac{1}{x} \).

Thus,

\[ \frac{dy}{dx} = 3 \cdot (-\sin(\log x)) \cdot \frac{1}{x} + 4 \cdot \cos(\log x) \cdot \frac{1}{x} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{-3 \sin(\log x) + 4 \cos(\log x)}{x} \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{-3 \sin(\log x) + 4 \cos(\log x)}{x}\right) \]

Using the quotient rule:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = -3 \sin(\log x) + 4 \cos(\log x) \) and \( v = x \).

Calculating \( \frac{du}{dx} \):

\[ \frac{du}{dx} = -3 \cos(\log x) \cdot \frac{1}{x} + 4 (-\sin(\log x)) \cdot \frac{1}{x} = \frac{-3 \cos(\log x) - 4 \sin(\log x)}{x} \]

And \( \frac{dv}{dx} = 1 \).

Now applying the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{x \left(\frac{-3 \cos(\log x) - 4 \sin(\log x)}{x}\right) - (-3 \sin(\log x) + 4 \cos(\log x)}{x^2}}{x^2} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = \frac{-3 \cos(\log x) - 4 \sin(\log x) + 3 \sin(\log x) - 4 \cos(\log x)}{x^2} \]

Combining terms gives:

\[ \frac{d^2y}{dx^2} = \frac{-7 \cos(\log x) - \sin(\log x)}{x^2} \]

### Step 3: Substitute into the expression

Now we substitute \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the expression \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} \):

\[ x^2 \frac{d^2y}{dx^2} = x^2 \left(\frac{-7 \cos(\log x) - \sin(\log x)}{x^2}\right) = -7 \cos(\log x) - \sin(\log x) \]

And:

\[ x \frac{dy}{dx} = x \left(\frac{-3 \sin(\log x) + 4 \cos(\log x)}{x}\right) = -3 \sin(\log x) + 4 \cos(\log x) \]

Combining these:

\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = (-7 \cos(\log x) - \sin(\log x)) + (-3 \sin(\log x) + 4 \cos(\log x)) \]

This simplifies to:

\[ (-7 + 4) \cos(\log x) + (-1 - 3) \sin(\log x) = -3 \cos(\log x) - 4 \sin(\log x) \]

### Final Result

Thus, we have:

\[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = -y \]