Find the order and degree, if defined, of the following differential equation :
`y'''+y^(2)+e^(x)=0`

Answer

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

### Step 1: Identify the highest order derivative
The given differential equation can be rewritten in a more recognizable form:
\[ y''' + y^2 + e^x = 0 \]
Here, \( y''' \) represents the third derivative of \( y \) with respect to \( x \).

### Step 2: Determine the order
The order of a differential equation is defined as the highest order of the derivative present in the equation. In this case, the highest order derivative is \( y''' \), which is the third derivative. Therefore, the order of the differential equation is:
\[ \text{Order} = 3 \]

### Step 3: Determine the degree
The degree of a differential equation is defined as the power of the highest order derivative when the equation is a polynomial in derivatives. In our case, the highest order derivative \( y''' \) has a power of 1 (since it is not raised to any power other than 1). Therefore, the degree of the differential equation is:
\[ \text{Degree} = 1 \]

### Final Answer
Thus, the order and degree of the given differential equation \( y''' + y^2 + e^x = 0 \) are:
- **Order:** 3
- **Degree:** 1