Question:
The degree of the differential equation obtained from the equation \((y-a)^2 = 4(x-b)\) [where \(a\) and b are arbitrary constants] is
The degree of the differential equation obtained from the equation \((y-a)^2 = 4(x-b)\) [where \(a\) and b are arbitrary constants] is
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Eliminate both constants by differentiating twice.
Updated On: Oct 1, 2026
- \(1\)
- \(2\)
- \(3\)
- not defined
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Two arbitrary constants mean the differential equation is of order 2. Degree is the power of the highest order derivative after clearing radicals and fractions.
Step 2: Differentiate twice:
\[ 2(y-a)y' = 4 \Rightarrow (y - a)y' = 2 \]
\[ y'^2 + (y-a)y'' = 0 \]
Step 3: Eliminate a:
From the first, \(y - a = \dfrac{2}{y'}\). Substitute:
\[ y'^2 + \frac{2y''}{y'} = 0 \Rightarrow y'^3 + 2y'' = 0 \]
Step 4: Read order and degree:
The highest derivative is \(y''\) (order 2), and it appears to the first power. So the degree is 1. Options (B) and (C) use the power 3 of \(y'\), which is a lower-order derivative.
Final Answer:
The equation y'^3 + 2y'' = 0 has degree 1. \[ \boxed{\text{(A) }1} \]
Two arbitrary constants mean the differential equation is of order 2. Degree is the power of the highest order derivative after clearing radicals and fractions.
Step 2: Differentiate twice:
\[ 2(y-a)y' = 4 \Rightarrow (y - a)y' = 2 \]
\[ y'^2 + (y-a)y'' = 0 \]
Step 3: Eliminate a:
From the first, \(y - a = \dfrac{2}{y'}\). Substitute:
\[ y'^2 + \frac{2y''}{y'} = 0 \Rightarrow y'^3 + 2y'' = 0 \]
Step 4: Read order and degree:
The highest derivative is \(y''\) (order 2), and it appears to the first power. So the degree is 1. Options (B) and (C) use the power 3 of \(y'\), which is a lower-order derivative.
Final Answer:
The equation y'^3 + 2y'' = 0 has degree 1. \[ \boxed{\text{(A) }1} \]
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