The differential equation for `y=Acos alphax +B sin alphax`, where A and B are arbitary constant is

Options

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

Explanation

To find the differential equation for the function \( y = A \cos(\alpha x) + B \sin(\alpha x) \), where \( A \) and \( B \) are arbitrary constants, we will differentiate the function twice and eliminate the constants. Here are the steps:

### Step 1: Differentiate the function once
We start with the function:
\[ y = A \cos(\alpha x) + B \sin(\alpha x) \]

Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(A \cos(\alpha x)) + \frac{d}{dx}(B \sin(\alpha x)) \]

Using the chain rule:
\[ \frac{dy}{dx} = -A \alpha \sin(\alpha x) + B \alpha \cos(\alpha x) \]

### Step 2: Differentiate the function a second time
Now, we differentiate \( \frac{dy}{dx} \) again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-A \alpha \sin(\alpha x) + B \alpha \cos(\alpha x)) \]

Again using the chain rule:
\[ \frac{d^2y}{dx^2} = -A \alpha^2 \cos(\alpha x) - B \alpha^2 \sin(\alpha x) \]

### Step 3: Express \( \frac{d^2y}{dx^2} \) in terms of \( y \)
Notice that:
\[ \frac{d^2y}{dx^2} = -\alpha^2 (A \cos(\alpha x) + B \sin(\alpha x)) \]
This can be rewritten as:
\[ \frac{d^2y}{dx^2} = -\alpha^2 y \]

### Step 4: Rearranging the equation
Rearranging gives us the differential equation:
\[ \frac{d^2y}{dx^2} + \alpha^2 y = 0 \]

### Final Result
The required differential equation is:
\[ \frac{d^2y}{dx^2} + \alpha^2 y = 0 \]

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