Need help solving an Integral Equation Need help solving:
$$
f(x) = x + \lambda \int_{0}^{1}y(x+y)f(y)dy
$$
keeping terms through $\lambda^{2}$,
(a) by using the Fredholm method
(b) by using the Neumann method
 A: Ok, I guess I'll try to help with the Neumann method. (Hopefully it's not too late for you.) First, note that if $\lambda$ was $0$, then $f(x) = x$. If both $\lambda$ and $x$ were small, then we'd hope that $f(x) \approx x$. So let's plug $f(x)\approx x$ into the left-hand side of your equation, to get a new approximation $f_{1}(x)$ to $f(x)$:
\begin{align*}
f_{1}(x) 
&= x+\lambda \int_{0}^{1} y(x+y)y\, \mathrm{d}y\\
&= x+\lambda\left(\frac{1}{4}+\frac{x}{3}\right) \\
&= \left(1+\frac{\lambda}{3}\right)x + \frac{\lambda}{4}
\end{align*}
Ok, so now let's plug $f_{1}(x)$ into the rhs of your equation to get a better approximation $f_{2}$:
\begin{align*}
f_{2}(x) 
&= x+\lambda \int_{0}^{1} y(x+y)f_{1}(y)\, \mathrm{d}y\\
&= x+\lambda \int_{0}^{1} y(x+y)\left[ \left(1+\frac{\lambda}{3}\right)x + \frac{\lambda}{4}  \right] \, \mathrm{d}y  \\
&=x+\lambda \left[\frac{1}{4}+\frac{x}{3} + \frac{\lambda}{6} + \frac{17 \lambda}{72}x   \right] \\
&=\frac{\lambda}{4} + \frac{\lambda^{2}}{6}+ \left(1+\frac{\lambda}{3}+ \frac{17\lambda^{2}}{72}  \right)x 
\end{align*}
Ok, so we have terms of order $\lambda^{2}$; let's make sure this doesn't change when we refine the approximation:
\begin{align*}
f_{3}(x) 
&= x + \lambda \int_{0}^{1} y(x+y)\left[\frac{\lambda}{4} + \frac{\lambda^{2}}{6}+ \left(1+\frac{\lambda}{3}+ \frac{17\lambda^{2}}{72}  \right)y   \right] \, \mathrm{d}y\\
&=
x+\lambda\left[\frac{11 \lambda ^2}{96}+\frac{\lambda }{6}+\frac{35 \lambda ^2 x}{216}+\frac{17 \lambda  x}{72}+\frac{x}{3}+\frac{1}{4}  \right] \\
&=\frac{\lambda}{4} +\frac{\lambda^{2}}{6} + \left(1+ \frac{\lambda}{3} + \frac{17\lambda^{2}}{72} \right)x +\mathcal{O}(\lambda^{3}) \quad \text{as } \lambda \to 0.
\end{align*}
So we're good. I'm going to venture a guess that you can obtain the exact solution by assuming $f(x) = c_{0} + c_{1}x$ and then solving for $c_{0}$ and $c_{1}$ in terms of $\lambda$. Perhaps that will help with the Neumann method. When I carried out this procedure, I obtained
\begin{align*}
f(x) = -\frac{18\lambda}{-72+48\lambda +\lambda^{2}}+ \frac{24(\lambda-3)}{-72+48\lambda + \lambda^{2}} x
\end{align*}
