Question

The differential equation representing the family of curves $y^2 = 2d(x + \sqrt{d})$ where $d$ is a parameter, is of

• A. order 2
• B. degree 2
• C. degree 3
• D. degree 4

Answer 1

Answer: (C)

Given, $y^{2}=2 d(x+\sqrt{d}) \ldots$(i)
$\Rightarrow 2 y y_{1}=2 d $
$\Rightarrow d=y y_{1}$
From Eq. (i),
$y^{2}=2 y y_{1}\left(x+\sqrt{y y_{1}}\right)$
$\Rightarrow y^{2}-2 y y_{1} x=\sqrt{y y_{1}} \cdot 2 y y_{1}$
$\Rightarrow \left(y^{2}-2 y y_{1} x\right)^{2}=4\left(y y_{1}\right)^{3}$
So, degree of above equation is $3$ .