If `y=acosmx+bsinmx," then "y_(2)=`

Options

A. `-m^(2)y`
B. `m^(2)y`
C. `-my`
D. my

Answer

To find the second derivative \( y_2 \) of the function \( y = a \cos(mx) + b \sin(mx) \), we will follow these steps:

### Step 1: Differentiate \( y \) to find \( y_1 \)
Given:
\[ y = a \cos(mx) + b \sin(mx) \]

We will differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = y_1 = \frac{d}{dx}(a \cos(mx)) + \frac{d}{dx}(b \sin(mx)) \]

Using the chain rule:
\[ \frac{d}{dx}(\cos(mx)) = -m \sin(mx) \quad \text{and} \quad \frac{d}{dx}(\sin(mx)) = m \cos(mx) \]

Thus, we have:
\[ y_1 = a(-m \sin(mx)) + b(m \cos(mx)) \]
\[ y_1 = -am \sin(mx) + bm \cos(mx) \]

### Step 2: Differentiate \( y_1 \) to find \( y_2 \)
Now we differentiate \( y_1 \) to find \( y_2 \):
\[ y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}(-am \sin(mx)) + \frac{d}{dx}(bm \cos(mx)) \]

Using the chain rule again:
\[ \frac{d}{dx}(-am \sin(mx)) = -am \cdot m \cos(mx) = -am^2 \cos(mx) \]
\[ \frac{d}{dx}(bm \cos(mx)) = bm \cdot (-m \sin(mx)) = -bm^2 \sin(mx) \]

Thus, we have:
\[ y_2 = -am^2 \cos(mx) - bm^2 \sin(mx) \]

### Step 3: Factor out common terms
We can factor out \(-m^2\):
\[ y_2 = -m^2(a \cos(mx) + b \sin(mx)) \]

### Step 4: Substitute back for \( y \)
Since \( y = a \cos(mx) + b \sin(mx) \), we can substitute:
\[ y_2 = -m^2 y \]

### Final Result
Thus, the second derivative is:
\[ y_2 = -m^2 y \]

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