Solve the inhomogeneous Volterra integral equation and hence find its resolvent kernel : $$Y(t)=F(t)+\int_0^t (t^2-x^2)Y(x)dx$$
We know that if the kernel $K(t,x)$ is of the form $K(t-x)$, then it can be solved easily by using Laplace transform. But the given kernel is of the form $K(t^2-x^2)$. So we use the transformations $t^2=u$ and $x^2=v$. After transformation, the equation takes up the form $$Y(\sqrt{u})=F(\sqrt{u})+\int_0^u (u-v)Y(\sqrt{v})d(\sqrt{v}) \\ \implies Y(\sqrt{u})=F(\sqrt{u})+\int_0^u (u-v)\frac{Y(\sqrt{v})}{2\sqrt{v}}dv$$ Now if we use $Y(\sqrt{u})=y(u)$ and $F(\sqrt{u})=f(u)$ then we get $$y(u)=f(u)+\int_0^u \frac{u-v}{2\sqrt{v}}y(v)dv$$ and that's where I'm stuck. Because instead of the "nice" kernel $u-v$, we now have a somewhat "ugly" kernel $\displaystyle{\frac{u-v}{2\sqrt{v}}}$ which is not Laplace transformable. Can someone take a look what's happening here? Is the trasnformation I started with is wrong? If so, what's the general transformation for this type of kernel? Thanks in advance.
Taking the derivative w.r.t. $t$ on both sides, we get
$$Y'(t) = F'(t) + (t^2-t^2)Y(t) + \int_0^t 2tY(x)dx = F'(t) + 2t\int_0^t Y(x) dx$$
Then taking Laplace Transforms on both sides,
$$sy(s) - Y(0) = sf(s) - F(0) + 2\frac{y(s)-sy'(s)}{s^2}$$
$$ \implies y'(s) + \left(\frac{1}{2}s^2-\frac{1}{s}\right)y(s) = \frac{1}{2}s^2f(s) + \frac{1}{2}s\left(Y(0)-F(0)\right) = \frac{1}{2}s^2f(s)$$
This can then be solved with the integrating factor method in the frequency domain.