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)

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