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!
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm x}\pars{t} -3\int_{0}^{1}\pars{s + t}{\rm x}\pars{s}\,\dd s ={\rm y}\pars{t}}$
\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}
$$ \left.\begin{array}{rcrcl} -\,\half\,\mu & - & \nu & = & \phi \\ -3\mu & - & \half\,\nu & = & \varphi \end{array}\right\rbrace\qquad\imp\qquad\color{#66f}{\large% \left\lbrace\begin{array}{rcr} \mu & = & {2 \over 11}\,\pars{\phi - 2\varphi} \\[2mm] \nu & = & -\,{2 \over 11}\,\pars{6\phi - \varphi} \end{array}\right.} $$
