Which of the following is a differential equation of the family of curves `y=Ae^(2x)+Be^(-2x)`

Options

A. `(d^(2)y)/(dx^(2))-2(dy)/(dx)+2y=0`
B. `x(d^(2)y)/(dx^(2))+2(dy)/(dx)-xy+x^(2)-2=0`
C. `(d^(2)y)/(dx^(2))=4y`
D. `((dy)/(dx))^(3)=4y(x(dy)/(dx)-2y)`

Answer

To find the differential equation of the family of curves given by \( y = Ae^{2x} + Be^{-2x} \), we will follow these steps:

### Step 1: Differentiate the given equation
We start by differentiating the equation \( y = Ae^{2x} + Be^{-2x} \) with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(Ae^{2x}) + \frac{d}{dx}(Be^{-2x}) \]
Using the chain rule, we find:
\[ \frac{dy}{dx} = 2Ae^{2x} - 2Be^{-2x} \]

### Step 2: Differentiate again to find the second derivative
Next, we differentiate \( \frac{dy}{dx} \) to find the second derivative \( \frac{d^2y}{dx^2} \).

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2Ae^{2x}) - \frac{d}{dx}(2Be^{-2x}) \]
This gives us:
\[ \frac{d^2y}{dx^2} = 4Ae^{2x} + 4Be^{-2x} \]

### Step 3: Express the second derivative in terms of \( y \)
Notice that we can express \( 4Ae^{2x} + 4Be^{-2x} \) in terms of \( y \):
\[ \frac{d^2y}{dx^2} = 4(Ae^{2x} + Be^{-2x}) = 4y \]

### Step 4: Form the differential equation
From the relationship we derived, we can write the differential equation as:
\[ \frac{d^2y}{dx^2} - 4y = 0 \]

### Conclusion
Thus, the differential equation of the family of curves \( y = Ae^{2x} + Be^{-2x} \) is:
\[ \frac{d^2y}{dx^2} - 4y = 0 \]