Question

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

• A. order $1$
• B. order $3$
• C. degree $1$
• D. degree $3$

Answer 1

Answer: (A), (D)

degree $3$

We have,
$y^2=2d(x+\sqrt{d})\cdots \left(i\right)$
$\Rightarrow 2y\frac{dy}{dx}=2d\Rightarrow d=\frac{ydy}{dx}$
Now, from equation $\left(i\right),$
$y^2=2y\frac{dy}{dx}(x+\sqrt{y\frac{dy}{dx}})$
$y^2=2xy\frac{dy}{dx}+2y^{3/2}{\left(\frac{dy}{dx}\right)}^{3/2}$
${(y^2-2xy\frac{dy}{dx})}^2=4y^3{\left(\frac{dy}{dx}\right)}^3$
Therefore order is $1$ and degree is $3$