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

The correct option is C degree $3$

$y^2=2d(x+\sqrt{d})$
Differentiating w.r.t. $x$,
$\Rightarrow 2y\frac{dy}{dx}=2d\Rightarrow d=y\frac{dy}{dx}$
Putting the value of $d$ in the above equation,
$\Rightarrow y^2=2y\frac{dy}{dx}(x+\sqrt{y\frac{dy}{dx}})\Rightarrow y=2\frac{dy}{dx}(x+\sqrt{y\frac{dy}{dx}})\Rightarrow y-2x\frac{dy}{dx}=2y^{1/2}{\left(\frac{dy}{dx}\right)}^{3/2}$
Squaring both sides,
$\Rightarrow {(y-2x\frac{dy}{dx})}^2=2y{\left(\frac{dy}{dx}\right)}^3$