Q. $y = cx - c^{2}$, is the general solution of the differential equation
A) $\left( y^{'} \right)^{2} - xy^{'} + y = 0$
B) $y^{''} = 0$
C) $y^{'} = c$
D) $\left( y^{'} \right)^{2} + xy^{'} + y = 0$
Explanation
(a)
We have, $y = cx - c^{2}$
On differentiating $w.r.t\ 'x'$ , we get
$y^{'} = c$
On putting this value in Eq. (i), we get
$y = x\left( y^{'} \right) - \left( y^{'} \right)^{2}$
$\Rightarrow$ $\left( y^{'} \right)^{2} - xy^{'} + y = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.