The differential equation representing the family of curves $y^{2}=2 c(x+\sqrt{c})$, where $c >0$, is a parameter, is of order and degree as follows:

JEE Main · 2005 · 08 May 2005 · Offline · Differential Equations

The differential equation representing the family of curves

y^{2} = 2 c (x + c)

, where

c > 0

, is a parameter, is of order and degree as follows:

Answer: (C)

Differentiating

y^{2} = 2 c (x + c)

gives

2 y y^{'} = 2 c

, so

c = y y^{'}

. Substituting this back,

y^{2} = 2 y y^{'} (x + y y^{'})

, which rearranges to

y - 2 x y^{'} = 2 y y^{'}

. Squaring to clear the radical gives

(y - 2 x y^{'})^{2} = 4 y y^{'3}

. The highest derivative present is

y^{'}

, so the order is

1

, and once the relation is made polynomial the highest power of

y^{'}

3

, so the degree is

3

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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