The order and degree of the differential equation, of which `xy = ce^x+ be^(-x)+ x^2` is a solution, is:

Options

A. `1,3`
B. `2,1`
C. `3,2`
D. none of these

Answer

To find the order and degree of the differential equation for which the given equation \( xy = ce^x + be^{-x} + x^2 \) is a solution, we will follow these steps:

### Step 1: Differentiate the given equation once
We start with the equation:
\[ xy = ce^x + be^{-x} + x^2 \]
Differentiating both sides with respect to \( x \) using the product rule on the left side:
\[ \frac{d}{dx}(xy) = y + x\frac{dy}{dx} \]
Differentiating the right side:
\[ \frac{d}{dx}(ce^x + be^{-x} + x^2) = ce^x - be^{-x} + 2x \]
Thus, we have:
\[ y + x\frac{dy}{dx} = ce^x - be^{-x} + 2x \]

### Step 2: Differentiate the equation again
Now, we differentiate the equation obtained in Step 1:
\[ \frac{d}{dx}\left(y + x\frac{dy}{dx}\right) = \frac{d}{dx}\left(ce^x - be^{-x} + 2x\right) \]
Using the product rule on the left side:
\[ \frac{dy}{dx} + \left(\frac{dy}{dx} + x\frac{d^2y}{dx^2}\right) = ce^x + be^{-x} + 2 \]
This simplifies to:
\[ x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = ce^x + be^{-x} + 2 \]

### Step 3: Substitute back to eliminate constants
From the original equation, we can express \( ce^x + be^{-x} \) in terms of \( xy \) and \( x^2 \):
\[ ce^x + be^{-x} = xy - x^2 \]
Substituting this back into our differentiated equation gives:
\[ x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy - x^2 + 2 \]

### Step 4: Identify the order and degree
The resulting differential equation is:
\[ x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy - x^2 + 2 \]
- The highest order derivative is \(\frac{d^2y}{dx^2}\), which is a second derivative. Therefore, the **order** of the differential equation is **2**.
- The degree of the differential equation is defined as the power of the highest order derivative present. Here, \(\frac{d^2y}{dx^2}\) is raised to the power of 1. Therefore, the **degree** of the differential equation is **1**.

### Final Answer
The order is **2** and the degree is **1**.