Volterra integral equation with variable boundaries $$\phi (x)=x+\lambda \int_{a}^{x}(x-y)\phi (y)dy$$
I'm also Trying to solve this integral equation like she does
https://math.stackexchange.com/questions/404959/solving-an-integral-equation-with-a-separable-kernel/408757#408757
and I also have some doubts about it. What to do? It's the first time I'm solving something like this, I only learned the matrix and determinant method for usual equation integrals.  Can you explain how to solve this equation integrals? I searched the web but didn't get much anything to help me solve this. Thank a lot
 A: Related problem: (I), (II). You can use the method of successive approximation which can be summarized as
$$ \phi_{n} (x)=x+\lambda \int_{a}^{x}(x-y)\phi_{n-1} (y)dy,$$
where $\phi_0(x)$ can be any real valued function. For simplicity, you can choose $\phi_0(x)=0$, then start to find $\phi_1,\phi_2,\dots$. For instance, 
$$ \phi_1(x)=x+\lambda \int_{a}^{x}(x-y)(0)dy = x.$$
$$ \phi_{2} (x)=x+\lambda \int_{a}^{x}(x-y)\phi_{1} (y)dy $$
$$ \phi_{2} (x)=x+\lambda \int_{a}^{x}(x-y)(y)dy = \dots.$$
If you can get a general formula for $\phi_n(x)$, then the solution $\phi(x)$ will be 
$$ \phi(x) = \lim_{n\to \infty} \phi_n(x). $$
Off course this process will impose some conditions on $\lambda$. 
Another approach: Here is another approach. Differentiating the integral equation, we get
$$ \phi (x)=x+\lambda \int_{a}^{x}(x-y)\phi (y)dy \implies \phi'(x)= 1+\lambda\int_{a}^{x}\phi(y) dy. $$
Differentiating again, we get the differential equation
$$ \phi''(x) = \lambda \phi(x). $$
Solving the ode subject to the initial conditions $\phi(a)=a$ and $\phi'(a)=1$ gives the desired solution
$$ \phi \left( x \right) = \frac{1}{2}\,{\frac { \left( \sqrt {\lambda}a+1
 \right) {{\rm e}^{\sqrt {\lambda}x}}}{{{\rm e}^{\sqrt {\lambda}a}}
\sqrt {\lambda}}}+\frac{1}{2}\,{\frac { \left( \sqrt {\lambda}a-1 \right) {
{\rm e}^{-\sqrt {\lambda}x}}}{{{\rm e}^{-\sqrt {\lambda}a}}\sqrt {
\lambda}}}.$$
