If $K(x, t) \neq 0$ at the upper and lower limit of the integral equation, then $R(x, t ; \lambda)=K(x, t) \cdot e^{\lambda(x-t)}.$

We have, $K(x, t)=e^{x^{2}-t^{2}}$

Let $K_{m}(x, t)$ be the $m^{t h}$ iterated kernel. Then we have,

$K_{1}(x, t)=K(x, t)$

$K_{m}(x, t)=\int_{t}^{x} K(x, z) K_{m-1}(z, t) d z$

Now, $K_{1}(x, t)=K(x, t)=e^{x^{2}-t^{2}}$

Putting m = 2, we have

$\begin{aligned} K_{2}(x, t) &=\int_{t}^{x} K(x, z) K_{1}(z, t) d z \\ &=\int_{t}^{x} e^{x^{2}-z^{2}} e^{z^{2}-t^{2}} d z \\ &=e^{x^{2}-t^{2}} \int_{t}^{x} d z \\ &=e^{x^{2}-t^{2}}(x-t) \end{aligned}$

Next putting m = 3, we have

$\begin{aligned} K_{3}(x, t) &=\int_{t}^{x} K(x, z) K_{2}(z, t) d z \\ &=\int_{t}^{x} e^{x^{2}-z^{2}} \cdot e^{z^{2}-t^{2}}(z-t) d z \\ &=e^{x^{2}-t^{2}} \int_{t}^{x}(z-t) d z \\ &=e^{x^{2}-t^{2}}\left[\frac{(z-t)^{2}}{2}\right]_{t}^{x} \\ &=e^{x^{2}-t^{2}} \frac{(x-t)^{2}}{2 !} \end{aligned}$

Now putting m = 4, we have

$\begin{aligned} K_{4}(x, t) &=\int_{t}^{x} K(x, z) K_{3}(z, t) d z \\ &=\int_{t}^{x} e^{x^{2}-z^{2}} e^{z^{2}-t^{2}} \frac{(z-t)^{2}}{2 !} d z \\ &=\frac{e^{x^{2}-t^{2}}}{2 !}\left[\frac{(z-t)^{3}}{3}\right]_{t}^{x} \\ &=e^{x^{2}-t^{2}} \frac{(x-t)^{3}}{3 !} \end{aligned}$

And so on. By Mathematical Induction, we have

$K_{m}(x, t)=e^{x^{2}-t^{2}} \frac{(x-t)^{m-1}}{(m-1) !}, m=1,2,3, \ldots$

Now, by definition of the resolvent kernel, we have

$\begin{aligned} R(x, t ; \lambda) &=\sum_{m=1}^{\infty} \lambda^{m-1} K_{m}(x, t) \\ &=K_{1}(x, t)+\lambda K_{2}(x, t)+\lambda^{2} K_{3}(x, t)+\cdots \\ &=e^{x^{2}-t^{2}}+e^{x^{2}-t^{2}} \lambda \frac{(x-t)}{1 !}+e^{x^{2}-t^{2}} \lambda^{2} \frac{(x-t)^{2}}{2 !}+\cdots \\ &=e^{x^{2}-t^{2}}\left[1+\lambda \frac{(x-t)}{1 !}+\lambda^{2} \frac{(x-t)^{2}}{2 !}+\cdots\right] \end{aligned}$

Therefore, $R(x, t ; \lambda)=e^{x^{2}-t^{2}} e^{\lambda(x-t)}$

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Fredholm and Volterra Type

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