The differential equation representing the family of curves y^2=2c (x+sqrt c ) , where c>0 , is a parameter, is of order and degree as follows :

Differential Equation: The differential equation representing the family of curves y^2=2c… | AIEEE 2005

AIEEE 2005 (Offline)Question 1

The differential equation representing the family of curves $y^{2}=2c\left(x+\sqrt c\right)$, where $c>0$, is a parameter, is of order and degree as follows :

Differential Equation

order $1$, degree $2$order $1$, degree $1$order $1$, degree $3$✓ Correct answerorder $2$, degree $2$

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 $$y^2=2yy'\left(x+\sqrt{yy'}\right)\ \Longrightarrow\ y^2-2xyy'=2yy'\sqrt{yy'}.$$ Step 2 - Clear the radical. Squaring: $\left(y^2-2xyy'\right)^2=4y^3y'^3$. Step 3 - Order and degree. Only $y'$ appears (order $1$), and its highest power is $3$ (degree $3$). Answer: (C)

Differential Equation - JEE Mathematics question (AIEEE 2005 (Offline))

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