Question

The differential equation representing the family of curves $y^2=2c(x+\sqrt{c})$, where $c>0$, is a parameter, is of order and degree as follows:

• A. order 1, degree 2
• B. order 1, degree 1
• C. order 1, degree 3
• D. order 2, degree 2

Answer 1

The correct option is C order 1, degree 3

Given, $y^2=2c(x+\sqrt{c})$ ...(1)
$2y{y}^{\prime }=2c.1\Rightarrow y{y}^{\prime }=c$ ...(2)
$\Rightarrow y^2=2y{y}^{\prime }(x+\sqrt{y{y}^{\prime }})$
On putting value of $c$ from (2) in (1)

$\Rightarrow y^2-2y{y}^{\prime }x=\left(2y{y}^{\prime }\sqrt{y{y}^{\prime }}\right)$

squaring both sides, we get

$\left[y\right(y-2y^{\prime }x){]}^2=4y^2{y}^{\prime 2}y{y}^{\prime }$

$y^2(y-2y^{\prime }x{)}^2=4y^3{y}^{\prime 3}$
On simplifying, we get
${(y-2x{y}^{\prime })}^2=4y{y^{\prime }}^3$ ...(3)
Hence, equation $3$ is of order $1$ and degree $3$.