Q. The equation of family of a curve is $y^{2} = 4a(x + a)$ then differential equation of the family is
A) $y = y^{'} + x$
B) $y = y^{''} + x$
C) $y = 2y^{'}x + yy^{'2}$
D) $y^{''} + y^{'} + y^{2} = 0$
Explanation
(c)
Given, $y^{2} = 4ax + 4a^{2}$ …(i)
$\Rightarrow$ $2yy^{'} = 4a$
On putting the value of $4a$ in eq(i), we get
$y^{2} = 2yy^{'}x + 4.\ \frac{y^{2}y^{'2}}{4}$
$\Rightarrow$ $y = 2y^{'}x + yy^{'2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.