The degree of the potential equation `((d^(2)y)/(dx^(2)))^(2)+((dy)/(dx))^(2)=x sin ((dy)/(dx))"is`

Options

A. 1
B. 2
C. 3
D. not defined

Explanation

To determine the degree of the given potential equation

\[ \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = x \sin\left(\frac{dy}{dx}\right), \]

we will follow these steps:

### Step 1: Identify the highest order derivative
The highest order derivative in the equation is \(\frac{d^2y}{dx^2}\), which is a second-order derivative.

### Step 2: Analyze the equation
The equation can be rewritten as:

\[ \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = x \sin\left(\frac{dy}{dx}\right). \]

### Step 3: Expand the sine function
To analyze the degree, we need to consider the term \( \sin\left(\frac{dy}{dx}\right) \). The sine function can be expanded using its Taylor series:

\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]

Substituting \( x \) with \( \frac{dy}{dx} \):

\[ \sin\left(\frac{dy}{dx}\right) = \frac{dy}{dx} - \frac{1}{6}\left(\frac{dy}{dx}\right)^3 + \frac{1}{120}\left(\frac{dy}{dx}\right)^5 - \cdots \]

### Step 4: Analyze the powers of derivatives
In the equation, when we substitute the series expansion of \(\sin\left(\frac{dy}{dx}\right)\) back into the equation, we see that the powers of \(\frac{dy}{dx}\) will keep increasing indefinitely. This means that there is no finite highest power of \(\frac{dy}{dx}\) that can be defined.

### Step 5: Conclusion about the degree
Since the series expansion leads to an infinite number of terms with increasing powers of \(\frac{dy}{dx}\), the degree of the differential equation cannot be defined.

Thus, the degree of the given potential equation is **not defined**.

### Final Answer
The degree of the potential equation is **not defined**.

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