The differential equation for `y=Ae^(3x)+Be^(-3x)` is

Options

A. `(d^(2)y)/(dx^(2))-9y=0`
B. `(d^(2)y)/(dx^(2))-3y=0`
C. `(d^(2)y)/(dx^(2))+9y=0`
D. `(d^(2)y)/(dx^(2))+3y=0`

Answer

To find the differential equation for \( y = A e^{3x} + B e^{-3x} \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)
Given:
\[ y = A e^{3x} + B e^{-3x} \]
Differentiating \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}(A e^{3x}) + \frac{d}{dx}(B e^{-3x}) \]
Using the chain rule:
\[ \frac{dy}{dx} = A \cdot 3 e^{3x} + B \cdot (-3) e^{-3x} \]
Thus,
\[ \frac{dy}{dx} = 3A e^{3x} - 3B e^{-3x} \]

### Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)
Now, we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(3A e^{3x}) + \frac{d}{dx}(-3B e^{-3x}) \]
Applying the chain rule again:
\[ \frac{d^2y}{dx^2} = 3A \cdot 3 e^{3x} - 3B \cdot (-3) e^{-3x} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = 9A e^{3x} + 9B e^{-3x} \]

### Step 3: Relate \( \frac{d^2y}{dx^2} \) to \( y \)
Notice that:
\[ y = A e^{3x} + B e^{-3x} \]
Thus, we can express \( \frac{d^2y}{dx^2} \) in terms of \( y \):
\[ \frac{d^2y}{dx^2} = 9(A e^{3x} + B e^{-3x}) = 9y \]

### Step 4: Form the differential equation
Rearranging gives us:
\[ \frac{d^2y}{dx^2} - 9y = 0 \]
This is the required differential equation.

### Final Answer
The differential equation for \( y = A e^{3x} + B e^{-3x} \) is:
\[ \frac{d^2y}{dx^2} - 9y = 0 \]