A Standard Integral Equation Consider the integral equation
$$\phi(x) = x + \lambda\int_0^1 \phi(s)\,ds$$
Integrating with respect to $x$ from $x=0$ to $x=1$:
$$\int_0^1 \phi(x)\,dx = \int_0^1x\,dx + \lambda \int_0^1\Big[\int_0^1\phi(s)\,ds\Big]\,dx$$
which is equivalent to
$$\int_0^1 \phi(x)\,dx = \frac{1}{2} + \lambda \int_0^1\phi(s)\,ds$$
How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank? 
 A: Relabelling the dummy variable $x\mapsto s$ on the LHS of your final equation, $$\int_0^1\phi(s)\,ds-\lambda\int_0^1\phi(s)\,ds=\frac12\\\implies \int_0^1\phi(s)\,ds=\frac1{2(1-\lambda)}$$
Thus $$\phi(x)=x+\frac\lambda{2(1-\lambda)}$$
A: If you are after finding $\phi(x)$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $[0, 1]$:
$$
\phi(x) = \sum_{n \geq 0} a_{n} x^{n}.
$$
Substituting it into your equation, we get:
$$
\sum_{n \geq 0} a_{n} x^{n} = x + \lambda \sum_{n \geq 0}{a_{n} \over n+1}.
$$
Matching up the coefficients of the difference powers of $x$, we get:
$$
a_{n} = 0 \quad \mbox{ for } n \geq 2,
$$
$$
a_{1} = 1,
$$
and
$$
a_{0} = \lambda \left(a_{0} + {a_{1} \over 2}\right).
$$
This gives a relationship between $a_{0}$ and $\lambda$.
A: Note $\int_{0}^{1}{\phi(s)ds}$ is a constant say $a$. Your functional equation (FE) can be rewritten as: $$\phi(x)=x+a\lambda$$
Putting into FE yields:

$$x+a\lambda=x+\lambda\int_{0}^{1}{(s+a\lambda )ds }\iff a\lambda=\lambda\big(\frac{1}{2}+\lambda a\big)$$
If $\lambda=0$ then $\phi(x)=x$
if $\lambda\ne 1$ $a=\frac{1}{2}+\lambda a\iff ( 1-\lambda)a=\frac{1}{2}\iff a=\frac{1}{2-2\lambda}$ and then $\phi(x)=x+\frac{\lambda}{2-2\lambda}$
If $\lambda=1$ there won’t besuch $\phi$.

A: Note that since $\lambda\int_0^1 \phi(s)\,ds$ is a constant (with respect to $x$), then we can write$$\phi(x)=x+a$$and by substitution we conclude that $$x+a=x+\lambda\int _{0}^{1}x+adx\implies\\a=\lambda({1\over 2}+a)\implies\\a={\lambda\over 2-2\lambda}$$ and we obtain$$\phi(x)=x+{\lambda\over 2-2\lambda}\quad,\quad \lambda\ne 1$$The case $\lambda=1$ leads to no solution.
