Question

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

• A. order 1
• B. order 2
• C. degree 3
• D. degree 4

Answer 1

Answer: (C)

Given, $y^2=2c(x+\sqrt{c})...(i)$
On differentiating w.r.t. $x$, we get
$2y\frac{dy}{dx}=2c\Rightarrow c=y\frac{dy}{dx}$
On putting this value of $c$ in Eq. (i), we get
$y^2=2y\frac{dy}{dx}(x+\sqrt{y\frac{dy}{dx}})$
$\Rightarrow y=2\frac{dy}{dx}.x+2y^{1/2}(\frac{dy}{dx}{)}^{3/2}$
$\Rightarrow y-2x\frac{dy}{dx}=2\sqrt{y}(\frac{dy}{dx}{)}^{3/2}$
$\Rightarrow (y-2x\frac{dy}{dx}{)}^2=4y(\frac{dy}{dx}{)}^3$
Therefore, order of this differential equation is $1$ and degree is $3$.