JEE PYQ: Differential Equation Question 31
Question 31 - 2020 (08 Jan Shift 2)
The differential equation of the family of curves $x^2 = 4b(y+b)$, $b \in \mathbb{R}$, is:
(1) $x(y’)^2 = x + 2yy'$
(2) $x(y’)^2 = 2yy’ - x$
(3) $xy’’ = y'$
(4) $x(y’)^2 = x - 2yy'$
Show Answer
Answer: (1)
Solution
$x^2 = 4by + 4b^2$. Differentiating: $2x = 4by’$, so $b = \frac{x}{2y’}$. Substituting: $x^2 = 4\cdot\frac{x}{2y’}\cdot y + 4\cdot\frac{x^2}{4(y’)^2} = \frac{2xy}{y’} + \frac{x^2}{(y’)^2}$. Multiply by $(y’)^2$: $x^2(y’)^2 = 2xy(y’) + x^2$, so $x(y’)^2 = 2yy’ + x$, i.e. $x(y’)^2 = x + 2yy’$.
BETA
