If `y=ae^(mx)+be^(-mx)` then `(d^2y)/(dx^2)` is

Options

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

Answer

To find the second derivative of the function \( y = ae^{mx} + be^{-mx} \), we will follow these steps:

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

Given:
\[ y = ae^{mx} + be^{-mx} \]

To differentiate \( y \) with respect to \( x \):
1. Differentiate \( ae^{mx} \):
- The derivative of \( e^{mx} \) is \( e^{mx} \cdot m \) (using the chain rule).
- Therefore, \( \frac{d}{dx}(ae^{mx}) = a \cdot m e^{mx} \).

2. Differentiate \( be^{-mx} \):
- The derivative of \( e^{-mx} \) is \( e^{-mx} \cdot (-m) \).
- Therefore, \( \frac{d}{dx}(be^{-mx}) = -b \cdot m e^{-mx} \).

Combining these results:
\[ \frac{dy}{dx} = am e^{mx} - bm e^{-mx} \]

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

Now, we differentiate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = am e^{mx} - bm e^{-mx} \]

1. Differentiate \( am e^{mx} \):
- The derivative is \( am \cdot m e^{mx} = am^2 e^{mx} \).

2. Differentiate \( -bm e^{-mx} \):
- The derivative is \( -bm \cdot (-m) e^{-mx} = bm^2 e^{-mx} \).

Combining these results:
\[ \frac{d^2y}{dx^2} = am^2 e^{mx} + bm^2 e^{-mx} \]

### Step 3: Factor out \( m^2 \)

We can factor out \( m^2 \) from the expression:
\[ \frac{d^2y}{dx^2} = m^2 (a e^{mx} + b e^{-mx}) \]

### Step 4: Substitute back for \( y \)

Notice that \( a e^{mx} + b e^{-mx} = y \):
\[ \frac{d^2y}{dx^2} = m^2 y \]

### Final Answer

Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = m^2 y \]

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