Neumann series, successive approximations and resolvent kernel

$x(t) = \int_{0}^{t} \frac{x(s)}{1+s^{2}} d s+\arctan t, t \in [0, T], T > 0 ;$

$x(t) = \int_{0}^{t} e^{-(t-s)} x(s) d s+1, t \in [0,1] ;$

$x(t) = \int_{0}^{t}(t-s) x(s) d s+1, t \in [0,1] ;$

$x(t) = \int_{0}^{1} e^{-t-s} x(s) d s+1, t \in [0,1] ;$

$x(t) = \int_{0}^{1} \frac{x(s)}{1+s^{2}} d s+\arctan t, t \in [0,1] ;$

$x(t) = \int_{0}^{1} \frac{t e^{t s}}{2\left(e^{t}-1\right)} x(s) d s+\frac{1}{2}, t \in [0,1] ;$

$x(t) = \int_{0}^{1} s x(s) d s+1, t \in [0,1] ;$

$x(t) = \int_{1}^{2} \frac{t}{2 s} x(s) d s+1, t \in [1,2]$.

$x(t) = \int_{0}^{1} \frac{x(s)}{(1+s+t)^{2}} d s+1, t \in [0,1] ;$

$x(t) = \int_{0}^{1} \frac{1}{3}\left(t s+s^{2}\right) x(s) d s+t, t \in [0,1] ;$

$x(t) = \int_{0}^{\frac{\pi }{4}} \frac{1}{10} \arctan (t s) x(s) d s+1, t \in \left[0, \frac{\pi }{4}\right]$.

$K(t, s) = e^{t^{2}-s^{2}}, \lambda = 2, y(t) = e^{t^{2}+2 t}$;

$K(t, s) = e^{t-s}, \lambda = 2, y(t) = \sin t$;

$K(t, s) = \frac{1+t^{2}}{1+s^{2}}, \lambda = 1, y(t) = 1+t^{2}$;

$K(t, s) = 3^{t-s}, \lambda = -1, y(t) = t 3^{t}$;

$K(t, s) = \frac{2+\cos t}{2+\cos s}, \lambda = 1, y(t) = e^{t} \sin t$;

$K(t, s) = \frac{\cosh t}{\cosh s}, \lambda = -1, y(t) = t \cosh t$;

$K(t, s) = e^{s-t}, \lambda = 1, y(t) = t e^{\frac{t^{2}}{2}}$.

$K(t, s) = e^{t+s},\left[a, b\right] = \left[0,1\right], \lambda = \frac{1}{e^{2}-1}, y(t) = \sin \pi t$;

$K(t, s) = \sin t \cos s,\left[a, b\right] = \left[0, \frac{\pi }{2}\right], \lambda = -1, y(t) = \cos t$;

$K(t, s) = t e^{s},\left[a, b\right] = \left[-1,1\right], \lambda = \frac{e}{3}, y(t) = \left(3 t^{2}+1\right) e^{-t}$;

$K(t, s) = t^{2} s^{2},\left[a, b\right] = \left[-1,1\right], \lambda = 2, y(t) = e^{t}$;

$K(t, s) = (1+t)(1-s),\left[a, b\right] = \left[-1,0\right], \lambda = 1, y(t) = \pi \cos \pi t$;

$K(t, s) = t s+t^{2} s^{2},\left[a, b\right] = \left[-1,1\right], \lambda = 1, y(t) = 3(t+1)$;

$K(t, s) = 1+(2 t-1)(2 s-1),\left[a, b\right] = \left[0,1\right], \lambda = \frac{1}{2}, y(t) = 3 t^{2}$;

$K(t, s) = \sin t \cos s+\cos 2 t \sin 2 s,\left[a, b\right] = \left[0,2 \pi \right], \lambda = -\frac{1}{\pi }, y(t) = \cos t+\sin t$

$K(t, s) = \sin t \sin s+\cos 2 t \cos 2 s,\left[a, b\right] = \left[0,2 \pi \right], \lambda = \frac{1}{2 \pi }, \quad y(t) = \cos 2 t ;$

$K(t, s) = \cos t \cos s+3 \sin 2 t \sin 2 s,\left[a, b\right] = \left[0,2 \pi \right], \lambda = \frac{1}{6 \pi }, y(t) = 5 \cos t$;

$K(t, s) = 2+\cos t \cos s,\left[a, b\right] = \left[0,2 \pi \right], \lambda = -\frac{1}{8 \pi }, y(t) = 2 t$;

$K(t, s) = e^{t} \cos s,\left[a, b\right] = \left[0, \pi \right], \lambda = \frac{1}{e^{\pi }+1}, y(t) = \frac{t}{4}$.

$R_{\lambda }(t, s) = \lambda \int_{a}^{b} K(t, u) R_{\lambda }(u, s) d u+K(t, s)$;

$R_{\lambda }(t, s) = \lambda \int_{a}^{b} K(u, s) R_{\lambda }(t, u) d u+K(t, s) ;$

$\frac{\partial R_{\lambda }(t, s)}{\partial \lambda } = \int_{a}^{b} R_{\lambda }(t, u) R_{\lambda }(u, s) d u$.

$L_{2}[0,1]$;

$C[0,1]$.