Q80 / 225 · JEE Main 2005 · Mathematics

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:

  • A
    order 1 , degree 2
  • B
    order 1, degree 1
  • Correct answer
    C
    order 1, degree 3
  • D
    order 2, degree 2

Solution

$$ \begin{aligned} & y^2=2 c(x+\sqrt{c} ) .....(i)\\ & 2 y y^{\prime}=2 c \cdot 1 \text { or } y y^{\prime}=c ....(ii)\\ & \Rightarrow y^2=2 y y^{\prime}\left(x+\sqrt{y y^{\prime}}\right) \end{aligned} $$ [on putting value of c from (ii) in (i)] On simplifying, we get $$ \left(y-2 x y^{\prime}\right)^2=4 y y^{\prime 3} .....(iii) $$ Hence equation (iii) is of order 1 and degree 3.

Hence, the correct answer is option: 'C'.

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