The solution of the differential equation dydx=2xy−3y+2x−3 is
cy+1=ex2−3x
cxy+1=ex2
cyx+1=ey2
ey2−3y=3x+c
dydx=2xy−3y+2x−3=(y+1)(2x−3)dx∫dyy+1=∫(2x−3)dxln|y+1|+lnc=x2−3x⇒ln|c(y+1)|=x2−3x⇒c(y+1)=ex2−3x