Find the differential equation of the family of curves `y=A e^(2x)+B e^(-2x)` , where A and B are arbitrary constants.

Explanation

To find the differential equation of the family of curves given by \( y = A e^{2x} + B e^{-2x} \), where \( A \) and \( B \) are arbitrary constants, we will follow these steps:

### Step 1: Differentiate the equation once
We start with the equation:
\[ y = A e^{2x} + B e^{-2x} \]
Now, we differentiate this equation with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(A e^{2x}) + \frac{d}{dx}(B e^{-2x}) \]
Using the chain rule, we get:
\[ \frac{dy}{dx} = A \cdot 2 e^{2x} + B \cdot (-2) e^{-2x} \]
This simplifies to:
\[ \frac{dy}{dx} = 2A e^{2x} - 2B e^{-2x} \]

### Step 2: Differentiate the first derivative
Next, we differentiate \( \frac{dy}{dx} \) again with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2A e^{2x}) + \frac{d}{dx}(-2B e^{-2x}) \]
Applying the chain rule again, we find:
\[ \frac{d^2y}{dx^2} = 2A \cdot 2 e^{2x} - 2B \cdot (-2) e^{-2x} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = 4A e^{2x} + 4B e^{-2x} \]

### Step 3: Express \( A e^{2x} + B e^{-2x} \) in terms of \( y \)
From the original equation, we know:
\[ y = A e^{2x} + B e^{-2x} \]
We can express \( 4A e^{2x} + 4B e^{-2x} \) in terms of \( y \):
\[ \frac{d^2y}{dx^2} = 4(A e^{2x} + B e^{-2x}) = 4y \]

### Step 4: Write the final differential equation
Thus, we arrive at the differential equation:
\[ \frac{d^2y}{dx^2} - 4y = 0 \]

### Summary
The differential equation of the family of curves \( y = A e^{2x} + B e^{-2x} \) is:
\[ \frac{d^2y}{dx^2} - 4y = 0 \]