JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Differential Equations

The differential equation of the family of curves,

x^{2} = 4 b (y + b), b \in R

, is:

Answer: (C)

Solution

Differentiate

x^{2} = 4 b (y + b)

to get

2 x = 4 b y^{'}

, so

b = \frac{x}{2 y ^{'}}

. Substituting back,

x^{2} = 4 \cdot \frac{x}{2 y ^{'}} (y + \frac{x}{2 y ^{'}}) = \frac{2 x y}{y ^{'}} + \frac{x ^{2}}{( y ^{'} ) ^{2}}

. Multiplying through by

(y^{'})^{2}

gives

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

, and dividing by

x

leaves

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

, that is

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

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