Solve the equation: $x(t)-3\int_0^1(s+t)x(s)ds=y(t)$ Given $y\in L^2[0,1]$, Solve the equation:
$$x(t)-3\int_0^1(s+t)x(s)ds=y(t)$$
I have noticed that the equation is $(I-K)(x(t))=y(t)$, where $K(f(t))=\int_0^13(s+t)f(s)ds$ is a compact integral operator in $L^2[0,1]$, so Fredholm alternative is an idea.
I don't know how to continue with the homogeneous equation $(I-K)(x(t))=0$. How can I get a solution to the inhomogeneous equation?
Thanks!
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$\ds{{\rm x}\pars{t} -3\int_{0}^{1}\pars{s + t}{\rm x}\pars{s}\,\dd s
     ={\rm y}\pars{t}}$

\begin{align}
\color{#66f}{\large{\rm x}\pars{t}}&=
\color{#66f}{\large 3\mu + 3\nu t + {\rm y}\pars{t}}\,,\qquad
\mu \equiv \int_{0}^{1}s{\rm x}\pars{s}\,\dd s\,,\quad
\nu\equiv\int_{0}^{1}{\rm x}\pars{s}\,\dd s
\end{align}

\begin{align}
\mu &=\int_{0}^{1}s\bracks{3\mu + 3\nu s + {\rm y}\pars{s}}\,\dd s
={3 \over 2}\,\mu + \nu + \phi\,,\qquad
\color{#66f}{\large\phi}\equiv
\color{#66f}{\large\int_{0}^{1}s{\rm y}\pars{s}\,\dd s}
\\[3mm]
\nu &=\int_{0}^{1}\bracks{3\mu + 3\nu s + {\rm y}\pars{s}}\,\dd s
=3\mu + {3 \over 2}\,\nu + \varphi\,,\qquad
\color{#66f}{\large\varphi}\equiv\color{#66f}{\large%
\int_{0}^{1}{\rm y}\pars{s}\,\dd s}
\end{align}

$$
\left.\begin{array}{rcrcl}
-\,\half\,\mu & - & \nu & = & \phi
\\
-3\mu & - & \half\,\nu & = & \varphi
\end{array}\right\rbrace\qquad\imp\qquad\color{#66f}{\large%
\left\lbrace\begin{array}{rcr}
\mu & = & {2 \over 11}\,\pars{\phi - 2\varphi}
\\[2mm]
\nu & = & -\,{2 \over 11}\,\pars{6\phi - \varphi}
\end{array}\right.}
$$

