Find the order and degree ( if defined) of each of the following equations :
`(y'')^(2)+cos y'=0`.

Answer

To find the order and degree of the given differential equation \( (y'')^2 + \cos(y') = 0 \), we will follow these steps:

### Step 1: Identify the highest derivative
The equation involves derivatives of \( y \):
- \( y' \) is the first derivative of \( y \) with respect to \( x \).
- \( y'' \) is the second derivative of \( y \) with respect to \( x \).

The highest derivative present in the equation is \( y'' \).

### Step 2: Determine the order
The order of a differential equation is defined as the highest order of derivative present in the equation. Since the highest derivative here is \( y'' \), which is a second derivative, the order of the equation is:

**Order = 2**

### Step 3: Determine the degree
The degree of a differential equation is defined as the power of the highest derivative when the equation is expressed as a polynomial in derivatives.

In our equation, we have:
- \( (y'')^2 \) which is a polynomial in \( y'' \) of degree 2.
- However, the term \( \cos(y') \) involves a trigonometric function of \( y' \), which is not a polynomial expression.

Since the equation is not a polynomial in derivatives due to the presence of the \( \cos(y') \) term, the degree of the equation is not defined.

**Degree = Not defined**

### Final Answer
- Order: 2
- Degree: Not defined