If `y=ae^(mx)+be^(-mx)," then "y_(2)=`

Options

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

Answer

To solve the problem, we need to find the second derivative of the function \( y = ae^{mx} + be^{-mx} \).

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = ae^{mx} + be^{-mx} \]

2. **First Derivative**:
We differentiate \( y \) with respect to \( x \).
\[ \frac{dy}{dx} = \frac{d}{dx}(ae^{mx}) + \frac{d}{dx}(be^{-mx}) \]
Using the chain rule:
- The derivative of \( ae^{mx} \) is \( ame^{mx} \) (where \( m \) is a constant).
- The derivative of \( be^{-mx} \) is \( -bme^{-mx} \) (again, using the chain rule).

Thus, we have:
\[ \frac{dy}{dx} = ame^{mx} - bme^{-mx} \]

3. **Second Derivative**:
Now, we differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(ame^{mx}) + \frac{d}{dx}(-bme^{-mx}) \]
Again, applying the chain rule:
- The derivative of \( ame^{mx} \) is \( am^2e^{mx} \).
- The derivative of \( -bme^{-mx} \) is \( bm^2e^{-mx} \).

Therefore, we have:
\[ \frac{d^2y}{dx^2} = am^2e^{mx} + bm^2e^{-mx} \]

4. **Factor out \( m^2 \)**:
We can factor out \( m^2 \) from the expression:
\[ \frac{d^2y}{dx^2} = m^2(ae^{mx} + be^{-mx}) \]

5. **Final Expression**:
Since \( y = ae^{mx} + be^{-mx} \), we can substitute back:
\[ \frac{d^2y}{dx^2} = m^2y \]

### Conclusion:
Thus, the second derivative \( y_{(2)} \) is given by:
\[ y_{(2)} = m^2y \]