Show existence and uniqueness of integral equality with neumann-series

Question

I want to show that for

$$x(s)-\int_0^12rs\cdot x(r)dr=\sin(\pi s)$$

there exists exactly one solution $x \in C^0([0,1],\mathbb R)$.

Answer 1

Hint:(Much easier and with neumann-series)

Write your integral equation as $(I-T)x=c$.

Show now that $(I-T)$ invertible by compution its inverse function. Then the integral equation has an unique solution for every choice of c.

Answer 2

Assume that $x$ is a solution of $$ x(s)-2s\int_{0}^{1}rx(r)\,dr = \sin(\pi s). $$ Let $g(r)=r$. In terms of an inner product, $(\cdot,\cdot)$. $$ x(s)=2(x,g)g(s)+\sin(\pi s) $$ Take the inner-product of the above with $g$. Because $1\ne 2(g,g)$, $$ (x,g) = 2(x,g)(g,g)+(\sin(\pi s),g),\\ (x,g) = \frac{(\sin(\pi s),g)}{1-2(g,g)}. $$ So the solution is unique if it exists, and it must equal $$ x=2g(x,g)+\sin(\pi s)=\frac{2(\sin(\pi s),g)}{1-2(g,g)}g+\sin(\pi s),\\ x = \frac{2(\sin(\pi s),g)}{1-2(g,g)}g+\sin(\pi s). $$ Now test this solution: $$ \begin{align} x -2g(x,g) & =\left[\frac{2(\sin(\pi s),g)}{1-2(g,g)}g+\sin(\pi s)\right] -2g\left(\frac{2(\sin(\pi s),g)}{1-2(g,g)}g+\sin(\pi s),g\right) \\ & = \left[\frac{2(\sin(\pi s),g)}{1-2(g,g)}(1-2(g,g))\right]g+\sin(\pi s)-2(\sin(\pi s),g)g \\ & = \sin(\pi s). \end{align} $$ So, the proposed solution checks, and it is unique.