Proving solutions of Linear Variational Problem Let $$a(v,w) = \int_0^1 a_o(x)v(x)w(x)dx + \lambda \int_0^1 v(y)w(x)dxdy, \; \; \forall v,w \in L^2(0,1)$$ be a bilinear, continuous, elliptic functional with $a_0(x) \in C^0[0,1], \; \;a_0(x) > 0$ on $[0,1]$ and $\lambda \in \mathbb{R}.$ Suppose that $f \in L^2(0,1)$ and consider the variational problem:
$$u \in L^2(0,1) \; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$
$$ a(u,v) = \int_0^1f(x)v(x)dx, \; \; \forall v \in L^2(0,1)  $$
(i) Suppose that $a_0 = 1$. Show that if $\lambda \ne -1$ then this problem has a closed form solution and find it.
(ii) if $\lambda = -1$, the problem has no solutions unless $\int_0^1f(y)dy = 0$ and if this holds, then it actually has infinitely many solutions of the form $u = f + C$ where $C$ is a constant.
What I have so far:
I have manipulated the formulation of the problem to arrive at:
$$ a_0(x)u(x) + \lambda \int_0^1u(y)dy = f(x), x \in (0,1) $$
Substituting the given conditions, this becomes:
$$u(x) + \lambda\int_0^1u(y)dy = f(x), x \in (0,1) $$
I'm kind of stuck here. I was thinking that this could be differentiated somehow to turn it into an ODE that can be shown to be solvable/not solvable based on what lambda is, but I'm not sure how to make that work... or if it's even a valid idea.
Thanks for your input!

i) integrate $u(x) + \lambda \int_{0}^{1} u(y) dy = f(x)$ from $0$ to $1$ to get
\begin{equation}
\int_{0}^{1} u(x) dx + \lambda \int_{0}^{1} \int_{0}^{1} u(y) dy dx = \int_{0}^{1} f(x) dx
\end{equation}
now giving $\int_{0}^{1} u(y) dy$ a name, say, $\int_{0}^{1} u(y) dy = q$, we have
\begin{equation}
q + \lambda \int_{0}^{1} q dx = q (1 + \lambda)  = \int_{0}^{1} f(x) dx
\end{equation}
so, provided, $\lambda \neq -1$, we have 
\begin{equation}
\int_{0}^{1} u(y) dy = q =  \frac{1}{(1 + \lambda)}\int_{0}^{1} f(x) dx
\end{equation}
plugging this back into the original equation gives
\begin{equation}
u(x) = f(x) - \lambda q 
\end{equation}
or simply
\begin{equation}
u(x) = f(x) - \frac{\lambda}{(1 + \lambda)}\int_{0}^{1} f(x) dx
\end{equation}
ii)
falls out directly from the Fredholm alternative:
https://en.wikipedia.org/wiki/Fredholm_alternative
