`y = c_(1) e^(2x) +c_(2)e^(-2x)`

Answer

To form a differential equation from the given function \( y = c_1 e^{2x} + c_2 e^{-2x} \), we will follow these steps:

### Step 1: Multiply the entire equation by \( e^{2x} \)
We start with the equation:
\[ y = c_1 e^{2x} + c_2 e^{-2x} \]
Multiplying both sides by \( e^{2x} \):
\[ y e^{2x} = c_1 e^{4x} + c_2 \]

### Step 2: Differentiate the equation with respect to \( x \)
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y e^{2x}) = \frac{d}{dx}(c_1 e^{4x} + c_2) \]
Using the product rule on the left side:
\[ \frac{dy}{dx} e^{2x} + y \cdot 2 e^{2x} = 4c_1 e^{4x} \]
This simplifies to:
\[ \frac{dy}{dx} e^{2x} + 2y e^{2x} = 4c_1 e^{4x} \]

### Step 3: Isolate the terms involving \( c_1 \)
We can rearrange the equation:
\[ \frac{dy}{dx} e^{2x} = 4c_1 e^{4x} - 2y e^{2x} \]

### Step 4: Multiply by \( e^{-2x} \) to eliminate the exponentials
Now, we multiply the entire equation by \( e^{-2x} \):
\[ \frac{dy}{dx} = 4c_1 e^{2x} - 2y \]

### Step 5: Differentiate again
Now we differentiate the equation again:
\[ \frac{d^2y}{dx^2} = 4c_1 \cdot 2 e^{2x} - 2 \frac{dy}{dx} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = 8c_1 e^{2x} - 2 \frac{dy}{dx} \]

### Step 6: Substitute \( c_1 \) back in terms of \( y \)
From our earlier steps, we know:
\[ 4c_1 = \frac{dy}{dx} + 2y e^{-2x} \]
Substituting \( c_1 \) into our second derivative equation gives:
\[ \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - 8y = 0 \]

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

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