Question

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

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

Answer 1

Answer: (B)

$y^2=2c(x+\sqrt{c})\dots (i)$
Differentiating w.r.t. $x$,
$2y\frac{dy}{dx}=2c$
$\Rightarrow c=y\frac{dy}{dx}$
On putting this value in Eq. (i),
$y^2=2x\cdot y\frac{dy}{dx}+2y\frac{dy}{dx}\sqrt{y\frac{dy}{dx}}$
$2y\frac{dy}{dx}\sqrt{y\frac{dy}{dx}}=y^2-2xy\frac{dy}{dx}$
Squaring on both sides,
$4y^2{\left(\frac{dy}{dx}\right)}^2\left(y\cdot \frac{dy}{dx}\right)={\left\{y^2-2xy\frac{dy}{dx}\right\}}^2$
$4y^3{\left(\frac{dy}{dx}\right)}^3=y^4+4x^2{y}^2{\left(\frac{dy}{dx}\right)}^2-2x{y}^3\left(\frac{dy}{dx}\right)$
$4y{\left(\frac{dy}{dx}\right)}^3-4x^2{\left(\frac{dy}{dx}\right)}^2+2xy\left(\frac{dy}{dx}\right)-y^2=0$
Hence, order $\to 1$, degree $\to 3$