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, $y^2=2d(x+\sqrt{d})\dots$(i)
$\Rightarrow 2y{y}_1=2d$
$\Rightarrow d=y{y}_1$
From Eq. (i),
$y^2=2y{y}_1\left(x+\sqrt{y{y}_1}\right)$
$\Rightarrow y^2-2y{y}_{1}x=\sqrt{y{y}_1}\cdot 2y{y}_1$
$\Rightarrow {\left(y^2-2y{y}_{1}x\right)}^2=4{\left(y{y}_1\right)}^3$
So, degree of above equation is $3$ .