If  ϕx=∫ϕx−2dx and  ϕ1=0 then ϕx=

ϕx=∫ϕx−2dxDifferentiate with respect to x⇒ϕ1(x)=[ϕ(x)]−2⇒(ϕ(x))2⋅ϕ'(x)=13(ϕ(x))2ϕ1(x)=3 Integrate, we get [ϕ(x)]3=3x+c Since ϕ(1)=0⇒0=3+c⇒c=−3 Then [ϕ(x)]3=3x−3=3(x−1) Hence ϕ(x)=[3(x−1)]1/3