The differential equation obtained by eliminating the arbitrary constants a and b from ` xy = ae^(x) + be^(-x) ` is

Options

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

Answer

To solve the problem of eliminating the arbitrary constants \( a \) and \( b \) from the equation \( xy = ae^x + be^{-x} \), we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ xy = ae^x + be^{-x} \]
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(ae^x + be^{-x}) \]
Using the product rule on the left side and the chain rule on the right side, we get:
\[ x \frac{dy}{dx} + y = ae^x - be^{-x} \]

### Step 2: Rearrange the equation
Rearranging the differentiated equation gives us:
\[ x \frac{dy}{dx} + y - ae^x + be^{-x} = 0 \]

### Step 3: Differentiate again
Now we differentiate the rearranged equation again with respect to \( x \):
\[ \frac{d}{dx}\left(x \frac{dy}{dx} + y - ae^x + be^{-x}\right) = 0 \]
Applying the product rule and chain rule, we get:
\[ \frac{d}{dx}(x \frac{dy}{dx}) + \frac{dy}{dx} - ae^x - be^{-x} = 0 \]
This expands to:
\[ x \frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} - ae^x + be^{-x} = 0 \]
Simplifying gives us:
\[ x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} - ae^x + be^{-x} = 0 \]

### Step 4: Substitute back for \( ae^x \) and \( be^{-x} \)
From the first differentiated equation, we can express \( ae^x \) and \( be^{-x} \):
\[ ae^x = x \frac{dy}{dx} + y - be^{-x} \]
Substituting this into the second differentiated equation leads to:
\[ x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} - (x \frac{dy}{dx} + y - be^{-x}) + be^{-x} = 0 \]
This simplifies to:
\[ x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} - x \frac{dy}{dx} - y = 0 \]
Thus, we have:
\[ x \frac{d^2y}{dx^2} + \frac{dy}{dx} - y = 0 \]

### Final Step: Write the final differential equation
The final differential equation obtained by eliminating \( a \) and \( b \) is:
\[ x \frac{d^2y}{dx^2} + \frac{dy}{dx} - y = 0 \]