JEE Main · 2004 · 24 Apr 2004 · Offline · Differential Equations

The differential equation for the family of curves

x^{2} + y^{2} - 2 a y = 0

, where

a

is an arbitrary constant is

Answer: (C)

Solution

Solve the family for the constant:

x^{2} + y^{2} - 2 a y = 0

gives

a = \frac{x ^{2} + y ^{2}}{2 y}

. Differentiating

x^{2} + y^{2} - 2 a y = 0

with

a

held constant gives

2 x + 2 y y^{'} - 2 a y^{'} = 0

; substituting

a = \frac{x ^{2} + y ^{2}}{2 y}

and multiplying through by

y

yields

2 x y + 2 y^{2} y^{'} - (x^{2} + y^{2}) y^{'} = 0

, which tidies to

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

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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