Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.1

Question 1. For each of the following equations, determine its order, degree (if exists)
(i) \( \frac { dy }{ dx } + xy = \cot x \)
(ii) \( (\frac { d^3y }{ dx^3 })^{2/3} – 3 \frac { d^2y }{ dx^2 } + 5\frac { dy }{ dx } + 4 = 0 \)
(iii) \( (\frac { d^2y }{ dx^2 })^2 + (\frac { dy }{ dx })^2 = x \sin (\frac { d^2y }{ dx^2 }) \)
(iv) \( \sqrt{\frac { dy }{ dx }} – 4 \frac { dy }{ dx } – 7x = 0 \)
(v) \( y(\frac { dy }{ dx }) = \frac { x }{ (\frac { dy }{ dx })+(\frac { dy }{ dx })^3 } \)
(vi) \( x^2 \frac { d^2y }{ dx^2 } + [1 + (\frac { dy }{ dx })^2]^{1/2} = 0 \)
(vii) \( (\frac { d^2y }{ dx^2 })^3 = \sqrt{1+(\frac { dy }{ dx })} \)
(viii) \( \frac { d^2y }{ dx^2 } = xy + \cos (\frac { dy }{ dx }) \)
(ix) \( \frac { d^2y }{ dx^2 } + 5 \frac { dy }{ dx } + \int ydx = x^3 \)
(x) \( x = e^{xy(\frac { dy }{ dx })} \)
Answer:
(i) \( \frac { dy }{ dx } + xy = \cot x \)
The highest order derivative in this equation is \( \frac { dy }{ dx } \). Its power is 1. A differential equation's order is the highest derivative present, and its degree is the power of that highest derivative after making the equation polynomial in terms of derivatives.
Thus, the order of the equation is 1 and the degree is 1.
(ii) \( (\frac { d^3y }{ dx^3 })^{2/3} – 3 \frac { d^2y }{ dx^2 } + 5\frac { dy }{ dx } + 4 = 0 \)
To find the degree, we need to remove the fractional exponent. We raise both sides of the equation to the power of 3:
\( (\frac { d^3y }{ dx^3 })^{2/3} = 3 \frac { d^2y }{ dx^2 } - 5\frac { dy }{ dx } - 4 \)
\( ((\frac { d^3y }{ dx^3 })^{2/3})^3 = (3 \frac { d^2y }{ dx^2 } - 5\frac { dy }{ dx } - 4)^3 \)
\( (\frac { d^3y }{ dx^3 })^2 = (3 \frac { d^2y }{ dx^2 } - 5\frac { dy }{ dx } - 4)^3 \)
In this new equation, the highest order derivative is \( \frac { d^3y }{ dx^3 } \). Its power is 2.
Therefore, the order of the equation is 3 and the degree is 2.
(iii) \( (\frac { d^2y }{ dx^2 })^2 + (\frac { dy }{ dx })^2 = x \sin (\frac { d^2y }{ dx^2 }) \)
The highest order derivative in this equation is \( \frac { d^2y }{ dx^2 } \). So, the order is 2. However, the term \( \sin (\frac { d^2y }{ dx^2 }) \) means the equation cannot be written as a polynomial in terms of its derivatives. When the derivative appears inside a trigonometric, exponential, or logarithmic function, the degree is undefined.
Hence, the order is 2, but the degree is not defined.
(iv) \( \sqrt{\frac { dy }{ dx }} – 4 \frac { dy }{ dx } – 7x = 0 \)
To find the degree, we need to remove the square root. First, isolate the square root term:
\( \sqrt{\frac { dy }{ dx }} = 4 \frac { dy }{ dx } + 7x \)
Now, square both sides:
\( (\sqrt{\frac { dy }{ dx }})^2 = (4 \frac { dy }{ dx } + 7x)^2 \)
\( \frac { dy }{ dx } = 16 (\frac { dy }{ dx })^2 + 49x^2 + 56x \frac { dy }{ dx } \)
The highest order derivative in this equation is \( \frac { dy }{ dx } \). Its highest power is 2.
Thus, the order of the equation is 1 and the degree is 2.
(v) \( y(\frac { dy }{ dx }) = \frac { x }{ (\frac { dy }{ dx })+(\frac { dy }{ dx })^3 } \)
First, rearrange the equation by multiplying both sides by the denominator on the right side:
\( y(\frac { dy }{ dx }) [(\frac { dy }{ dx })+(\frac { dy }{ dx })^3] = x \)
\( y(\frac { dy }{ dx })^2 + y(\frac { dy }{ dx })^4 = x \)
The highest order derivative in this equation is \( \frac { dy }{ dx } \). Its highest power is 4.
Therefore, the order of the equation is 1 and the degree is 4.
(vi) \( x^2 \frac { d^2y }{ dx^2 } + [1 + (\frac { dy }{ dx })^2]^{1/2} = 0 \)
To find the degree, we must remove the fractional exponent. Isolate the term with the fractional exponent:
\( x^2 \frac { d^2y }{ dx^2 } = -[1 + (\frac { dy }{ dx })^2]^{1/2} \)
Now, square both sides:
\( (x^2 \frac { d^2y }{ dx^2 })^2 = (-[1 + (\frac { dy }{ dx })^2]^{1/2})^2 \)
\( x^4 (\frac { d^2y }{ dx^2 })^2 = 1 + (\frac { dy }{ dx })^2 \)
The highest order derivative is \( \frac { d^2y }{ dx^2 } \). Its power is 2.
So, the order of the equation is 2 and the degree is 2.
(vii) \( (\frac { d^2y }{ dx^2 })^3 = \sqrt{1+(\frac { dy }{ dx })} \)
To find the degree, we must remove the square root. Square both sides of the equation:
\( ((\frac { d^2y }{ dx^2 })^3)^2 = (\sqrt{1+(\frac { dy }{ dx })})^2 \)
\( (\frac { d^2y }{ dx^2 })^6 = 1+(\frac { dy }{ dx }) \)
The highest order derivative in this equation is \( \frac { d^2y }{ dx^2 } \). Its power is 6.
Thus, the order of the equation is 2 and the degree is 6.
(viii) \( \frac { d^2y }{ dx^2 } = xy + \cos (\frac { dy }{ dx }) \)
The highest order derivative in this equation is \( \frac { d^2y }{ dx^2 } \). So, the order is 2. However, the term \( \cos (\frac { dy }{ dx }) \) makes it impossible to express the equation as a polynomial in terms of its derivatives.
Therefore, the order is 2, but the degree is not defined.
(ix) \( \frac { d^2y }{ dx^2 } + 5 \frac { dy }{ dx } + \int ydx = x^3 \)
To find the order and degree, we must eliminate the integral. Differentiate the entire equation with respect to x:
\( \frac { d }{ dx } (\frac { d^2y }{ dx^2 }) + \frac { d }{ dx } (5 \frac { dy }{ dx }) + \frac { d }{ dx } (\int ydx) = \frac { d }{ dx } (x^3) \)
\( \frac { d^3y }{ dx^3 } + 5 \frac { d^2y }{ dx^2 } + y = 3x^2 \)
In this new equation, the highest order derivative is \( \frac { d^3y }{ dx^3 } \). Its power is 1.
Thus, the order of the equation is 3 and the degree is 1.
(x) \( x = e^{xy(\frac { dy }{ dx })} \)
The highest order derivative in this equation is \( \frac { dy }{ dx } \). So, the order is 1. However, the derivative \( \frac { dy }{ dx } \) appears in the exponent of an exponential term. This means the equation cannot be written as a polynomial in terms of its derivatives.
Therefore, the order is 1, but the degree is not defined.
In simple words: The 'order' of a differential equation is the highest derivative it contains (like \( \frac{dy}{dx} \) or \( \frac{d^2y}{dx^2} \)). The 'degree' is the highest power of that highest derivative, but only if you can write the whole equation without roots or special functions (like sin, cos, exp) around the derivatives. If derivatives are inside those functions, the degree does not exist.

🎯 Exam Tip: Remember to simplify the differential equation by removing any fractional powers or radicals involving derivatives before determining the degree. Also, if derivatives are within transcendental functions (like sin, cos, log, exp), the degree is always undefined.