KEAM 2021 Math Solved Paper

© examsiri.com

Question : 82 of 120

Marks: +1, -0

The general solution of the differential equation 4xy+12x+(2x2+3)y′=0 is

2x2+3
y+3
=C
y−3
2x2+3
=C
y+2
2x2+3
=C
(y−3)(2x2+3)=C
(y+3)(2x2+3)=C

Go to Question:

Prev Question

Next Question