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: (A), (C)

Degree 3

Given curve is $y^2=2c(x+\sqrt{c})$.
Differentiate w.r.t. x, $2y\frac{dy}{dx}=2c\Rightarrow c=y\frac{dy}{dx}$
Hence differential equation is
$y^2=2y\frac{dy}{dx}(x+\sqrt{y\frac{dy}{dx}})\Rightarrow \frac{y}{\frac{2dy}{dx}}-x=\sqrt{y\frac{dy}{dx}}$
Squaring and multiplying by ${\left(\frac{dy}{dx}\right)}^2$
$y{\left(\frac{dy}{dx}\right)}^3-x^2{\left(\frac{dy}{dx}\right)}^2+xy\left(\frac{dy}{dx}\right)-\frac{y^2}{4}=0$
Hence order is 1 and degree is 3.