The differential equation whose solution represents the family `xy= Ae^(ax)+ Be^(-ax)` is

Options

A. `x((d^(2)y)/(dx^(2)))^(2) + 2(dy)/(dx)= xy`
B. `x((d^(2)y)/(dx^(2)))^(2)+ 2(dy)/(dx) = a^(2)xy`
C. `x(d^(2)y)/(dx^(2)) + 2 (dy)/(dx)= a^(2)xy`
D. None of these

Answer

To find the differential equation whose solution represents the family \( xy = Ae^{ax} + Be^{-ax} \), we will follow these steps:

### Step 1: Start with the given equation
We have the equation:
\[ xy = Ae^{ax} + Be^{-ax} \]

### Step 2: Differentiate the equation with respect to \( x \)
Using the product rule on the left-hand side and differentiating the right-hand side:
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(Ae^{ax} + Be^{-ax}) \]
Applying the product rule on the left side:
\[ x \frac{dy}{dx} + y = A \cdot a e^{ax} - B \cdot a e^{-ax} \]
This simplifies to:
\[ x \frac{dy}{dx} + y = a(A e^{ax} - B e^{-ax}) \tag{1} \]

### Step 3: Differentiate the equation again
Now we differentiate equation (1) again:
\[ \frac{d}{dx}\left(x \frac{dy}{dx} + y\right) = \frac{d}{dx}\left(a(A e^{ax} - B e^{-ax})\right) \]
Using the product rule on the left side:
\[ \frac{dy}{dx} + x \frac{d^2y}{dx^2} + \frac{dy}{dx} = a(aA e^{ax} + aB e^{-ax}) \]
This simplifies to:
\[ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = a^2(A e^{ax} + B e^{-ax}) \tag{2} \]

### Step 4: Substitute \( A e^{ax} + B e^{-ax} \) using the original equation
From the original equation, we can express \( A e^{ax} + B e^{-ax} \) as:
\[ A e^{ax} + B e^{-ax} = xy \]
Substituting this into equation (2):
\[ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = a^2 xy \]

### Step 5: Rearranging to form the differential equation
Rearranging gives us the final form of the differential equation:
\[ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - a^2 xy = 0 \]

### Final Answer
The differential equation is:
\[ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = a^2 xy \]