= Solution
Orient the circle counterclockwise and define its <Cauchy transform>
$$
\Phi(z)=\frac1{2\pi i}\int_C\frac{\phi(\tau)}{\tau-z}\,d\tau.
$$
The inside and outside limits satisfy $\Phi_+-\Phi_-=\phi$ and $\Phi_++\Phi_-=(\pi i)^{-1}\operatorname{PV}\int_C\phi(\tau)/(\tau-t)\,d\tau$. Also $\Phi_-(z)=O(z^{-1})$ at infinity. Applying these formulas to the two <polynomial> coefficients and dividing by $t^2$ gives the Riemann–Hilbert condition
$$
\boxed{(t^2-1)\Phi_+(t)-t\Phi_-(t)=(A-1)t+1.}
$$
The homogeneous jump is $G=t/(t^2-1)$. A <canonical factorization of a rational scalar Riemann-Hilbert jump> is
$$
\boxed{X_+(z)=1,\qquad X_-(z)=z-z^{-1},\qquad G=X_+/X_-.}
$$
The inside factor is analytic and nonvanishing; the exterior factor has no zeros or poles outside the circle and grows like $z$. The jump index is $1-2=-1$, consistently with this growth.
Put $F_+=\Phi_+$ and $F_-=\Phi_-/X_-$. Their additive jump is $R(t)=((A-1)t+1)/(t^2-1)$. Since $R$ is analytic outside the circle and decays there, the functions $F_+$ inside and $F_-+R$ outside glue to an <entire function>. This function tends to zero at infinity, so <Liouville's theorem> gives $F_+=0$, $F_-=-R$. Consequently
$$
\Phi_+=0,\qquad \Phi_-=-\frac{(A-1)z+1}{z}.
$$
The required decay of a <Cauchy transform> forces \b[$A=1$]. Then $\Phi_-=-1/z$, and the jump gives
$$
\boxed{\phi(t)=1/t.}
$$
For the homogeneous problem, $\Phi_+/X_+$ and $\Phi_-/X_-$ glue to an <entire function> vanishing at infinity, so only the zero solution exists; this proves uniqueness. Direct verification uses the principal-value integral $-\pi i/t$: the two terms on the left combine to $(t^4+t^3-t^2-t^4+t^3+t^2)/(2t)=t^2$, exactly the right side at $A=1$.
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