Volterra integral equation with variable boundaries

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

$$ \phi \left( x \right) = \frac{1}{2}\,{\frac { \left( \sqrt {\lambda}a+1 \right) {{\rm e}^{\sqrt {\lambda}x}}}{{{\rm e}^{\sqrt {\lambda}a}} \sqrt {\lambda}}}+\frac{1}{2}\,{\frac { \left( \sqrt {\lambda}a-1 \right) { {\rm e}^{-\sqrt {\lambda}x}}}{{{\rm e}^{-\sqrt {\lambda}a}}\sqrt { \lambda}}}.$$