Question 10 of 96beginner💡 UnderstandVery Short1 mark

Determine order and degree (if defined) of differential equations given in Exercises 1 to 10.

y′′+2y′+sin⁡y=0y'' + 2y' + \sin y = 0
Correct Answer

Order = 2, Degree = 1

Exercise: EXERCISE 9.1 | Q: 10 | (Page 4)
For More Understanding

Explanation

According to the provided context (Example 1), the order is determined by the highest order derivative present, which is y′′y'' (second derivative), making the order 2. The degree is defined only if the differential equation is a polynomial equation in its derivatives. In the given equation y′′+2y′+sin⁡y=0y'' + 2y' + \sin y = 0, the derivatives y′′y'' and y′y' appear as polynomials (power 1). The term sin⁡y\sin y is a function of the dependent variable yy, not a derivative, so it does not prevent the equation from being a polynomial in derivatives. Thus, the degree is the highest power of the highest derivative, which is 1.

Solution Steps

  1. Step 1: Identify the highest order derivative. The given differential equation is y′′+2y′+sin⁡y=0y'' + 2y' + \sin y = 0. The highest order derivative is y′′y'' (second derivative). Therefore, the order is 2.

  2. Step 2: Check if the degree is defined. The equation must be a polynomial equation in derivatives. Here, y′′y'' and y′y' are polynomial terms (power 1). The term sin⁡y\sin y involves yy, not a derivative.

  3. Step 3: Determine the degree. Since the equation is a polynomial in derivatives, the degree is defined. The highest power of the highest order derivative (y′′y'') is 1. Therefore, the degree is 1.