NCERT Solutions Class-12-Chapter-9-Diifferential Equations
Excercise-9.1
Note:
Order: The order of a differential equation is the order of the highest order derivative appearing in the equation.
Degree: The degree of a differential equation is the power of the highest order derivative, provided the equation is a polynomial equation in derivatives. If terms like \( \sin(y'), \cos(y'), e^{y'} \), etc., are present, the degree is not defined.
Q1
Determine order and degree: \( \frac{d^4y}{dx^4} + \sin(y''') = 0 \)
Solution:
Order: The highest order derivative present is \( \frac{d^4y}{dx^4} \), so the order is 4.
Degree: The given differential equation is not a polynomial equation in its derivatives because of the term \( \sin(y''') \). Therefore, the degree is not defined.
Q2
Determine order and degree: \( y' + 5y = 0 \)
Solution:
Order: The highest order derivative is \( y' \) (i.e., \( \frac{dy}{dx} \)), so the order is 1.
Degree: The exponent of the highest order derivative \( y' \) is 1. So, the degree is 1.
Q3
Determine order and degree: \( \left(\frac{ds}{dt}\right)^4 + 3s \frac{d^2s}{dt^2} = 0 \)
Solution:
Order: The highest order derivative is \( \frac{d^2s}{dt^2} \), so the order is 2.
Degree: The power of the highest order derivative \( \frac{d^2s}{dt^2} \) is 1. So, the degree is 1.
Q4
Determine order and degree: \( \left(\frac{d^2y}{dx^2}\right)^2 + \cos\left(\frac{dy}{dx}\right) = 0 \)
Solution:
Order: The highest order derivative is \( \frac{d^2y}{dx^2} \), so the order is 2.
Degree: The equation involves \( \cos\left(\frac{dy}{dx}\right) \), so it is not a polynomial in derivatives. Thus, the degree is not defined.
Q5
Determine order and degree: \( \frac{d^2y}{dx^2} = \cos 3x + \sin 3x \)
Solution:
Order: The highest order derivative is \( \frac{d^2y}{dx^2} \), so the order is 2.
Degree: The power of the highest order derivative is 1. So, the degree is 1.
Q6
Determine order and degree: \( (y''')^2 + (y'')^3 + (y')^4 + y^5 = 0 \)
Solution:
Order: The highest order derivative is \( y''' \) (3rd derivative), so the order is 3.
Degree: The power of the highest order derivative \( y''' \) is 2. So, the degree is 2.
Q7
Determine order and degree: \( y''' + 2y'' + y' = 0 \)
Solution:
Order: The highest order derivative is \( y''' \), so the order is 3.
Degree: The power of the highest order derivative \( y''' \) is 1. So, the degree is 1.
Q8
Determine order and degree: \( y' + y = e^x \)
Solution:
Order: The highest order derivative is \( y' \), so the order is 1.
Degree: The power of \( y' \) is 1. So, the degree is 1.
Q9
Determine order and degree: \( y'' + (y')^2 + 2y = 0 \)
Solution:
Order: The highest order derivative is \( y'' \), so the order is 2.
Degree: The power of the highest order derivative \( y'' \) is 1. So, the degree is 1.
Q10
Determine order and degree: \( y'' + 2y' + \sin y = 0 \)
Solution:
Order: The highest order derivative is \( y'' \), so the order is 2.
Degree: The power of the highest order derivative \( y'' \) is 1. Note that \( \sin y \) is a function of \( y \), not of a derivative, so it does not affect the definition of degree. Thus, the degree is 1.
Q11
The degree of the differential equation \( \left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + \sin\left(\frac{dy}{dx}\right) + 1 = 0 \) is- 3
- 2
- 1
- not defined
- 3
- 2
- 1
- not defined
Answer: (D)
Solution:
The differential equation contains the term \( \sin\left(\frac{dy}{dx}\right) \). Because of this term, the equation is not a polynomial in its derivatives. Therefore, the degree is not defined.
Q12
The order of the differential equation \( 2x^2 \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + y = 0 \) is- 2
- 1
- 0
- not defined
- 2
- 1
- 0
- not defined
Answer: (A)
Solution:
The highest order derivative present in the equation is \( \frac{d^2y}{dx^2} \). Therefore, the order is 2.