If `y = a cos 2x + b sin 2x` , then

Options

A. `(d^(2) y)/(dx^(2)) + y = 0`
B. `(d^(2) y)/(dx^(2)) + 2y = 0`
C. `(d^(2) y)/(dx^(2)) - 4y = 0`
D. `(d^(2) y)/(dx^(2)) + 4y = 0`

Answer

To solve the problem given \( y = a \cos 2x + b \sin 2x \), we need to find the second derivative of \( y \) and then express it in terms of \( y \) itself to form a differential equation.

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

Given:
\[ y = a \cos 2x + b \sin 2x \]

We differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(a \cos 2x) + \frac{d}{dx}(b \sin 2x) \]

Using the chain rule:
\[ \frac{dy}{dx} = a \cdot (-\sin 2x) \cdot 2 + b \cdot \cos 2x \cdot 2 \]
\[ \frac{dy}{dx} = -2a \sin 2x + 2b \cos 2x \]

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

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

Using the chain rule again:
\[ \frac{d^2y}{dx^2} = -2a \cdot \cos 2x \cdot 2 + 2b \cdot (-\sin 2x) \cdot 2 \]
\[ \frac{d^2y}{dx^2} = -4a \cos 2x - 4b \sin 2x \]

### Step 3: Factor out the common term

Notice that:
\[ \frac{d^2y}{dx^2} = -4(a \cos 2x + b \sin 2x) \]

Since \( y = a \cos 2x + b \sin 2x \), we can substitute \( y \) into the equation:
\[ \frac{d^2y}{dx^2} = -4y \]

### Step 4: Rearranging to form the differential equation

Rearranging gives us the final form of the differential equation:
\[ \frac{d^2y}{dx^2} + 4y = 0 \]

### Final Answer

The differential equation formed from \( y = a \cos 2x + b \sin 2x \) is:
\[ \frac{d^2y}{dx^2} + 4y = 0 \]

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