Question:
The difference between degree and order of a differential equation that represents the family of curves given by \(y^2=a\left(x+\frac{\sqrt{a}}{2}\right)\), \(a\gt0\) is _________.
The difference between degree and order of a differential equation that represents the family of curves given by \(y^2=a\left(x+\frac{\sqrt{a}}{2}\right)\), \(a\gt0\) is _________.
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Differentiate once to get \(a=2yy'\), eliminate \(a\), rationalise and read off order and degree.
Updated On: Nov 4, 2026
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Correct Answer: 2
Solution and Explanation
Differentiate \(y^2=ax+\frac{a^{3/2}}{2}\) w.r.t. \(x\): \(2yy'=a\). Only one arbitrary constant, so the order is 1.
Substitute \(a=2yy'\) into the curve: \(y^2=2yy'\,x+\frac{(2yy')\sqrt{2yy'}}{2}=2xyy'+yy'\sqrt{2yy'}\).
So \(y^2-2xyy'=yy'\sqrt{2yy'}\). Squaring to remove the radical: \((y^2-2xyy')^2=2y^3(y')^3\).
Highest derivative \(y'\) appears with highest power 3, so degree \(=3\) and order \(=1\).
Difference \(=3-1=2\).
Answer: 2
Substitute \(a=2yy'\) into the curve: \(y^2=2yy'\,x+\frac{(2yy')\sqrt{2yy'}}{2}=2xyy'+yy'\sqrt{2yy'}\).
So \(y^2-2xyy'=yy'\sqrt{2yy'}\). Squaring to remove the radical: \((y^2-2xyy')^2=2y^3(y')^3\).
Highest derivative \(y'\) appears with highest power 3, so degree \(=3\) and order \(=1\).
Difference \(=3-1=2\).
Answer: 2
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