IIT-JEE 1999Question 1
The differential equation representing the family of curves $y^{2}=2c\left(x+\sqrt{c}\right)$, where $c$ is a positive parameter, is of
Differential Equation
Solution
Concept. One parameter, so order $1$; remove $c$ and the square root, then read the degree. Formulas used. (1) $y^2=2c\left(x+\sqrt c\right)\Rightarrow2yy'=2c$Step 1 - Eliminate $c$. By (1), $c=yy'$. Then $$y^2-2xyy'=2yy'\sqrt{yy'}.$$ Step 2 - Clear the radical. Squaring and dividing by $y^2$: $$\left(y-2xy'\right)^2=4y\left(y'\right)^3.$$ Step 3 - Read off. Highest derivative $y'$: order $1$. Highest power of $y'$: $3$, so degree $3$. Answer: (A), (C)