If y=xln|cx|,c∈R is the general solution of the DEdydx=yx+ϕxy, then ϕ(2)+ϕ1(2) is

y=xlog|cx|⇒logc+logx=xy⇒0+1x=y⋅1−xy1y2⇒dydx=yx−y2x2⇒ϕ(t)=−1t2⇒ϕ't=2t3∴ϕ(2)+ϕ'2=-14+14=0