Difference distribution of iid noncentral chi2 variables Given are two identically distributed independent noncentral chi-square variables $$Z_1\sim \chi^2(k,\lambda)\\Z_2\sim\chi^2(k,\lambda)$$ with $k=2$ degrees of freedom and location parameter $\lambda$.
What is the pdf of $Z_1-Z_2$?
A related post deals with the difference of such variables. However the solution is not given there. I hope the solution for the difference of identically distributed variables is easier to show.

If you are accepting an Integral representation for this one, then you could do the following:
$$
\begin{aligned}
Z &= X - Y\\
f_Z(z)&=\frac{\partial}{\partial z}P(Z\le z) = \frac{\partial}{\partial z}P(X-Y\le z)\\
&=\frac{\partial}{\partial z}P(X\le Y + z)\\
&=\frac{\partial}{\partial z}\int_0^\infty P(X\le Y+z | Y=y) f_Y(y) ~dy\quad\text{law of Total probability}\\
&=\frac{\partial}{\partial z}\int_0^\infty F_X(y+z) f_Y(y) ~dy\\
&=\int_0^\infty \frac{\partial}{\partial z}F_X(y+z) f_Y(y) ~dy\quad\text{Leibniz Rule}\\
&=\boxed{\int_0^\infty f_X(y+z)f_Y(y) ~dy,\quad -\infty<z<\infty}
\end{aligned}
$$
We can confirm the above by using a programing Language:
R
## Our pdf:
fz_pdf <- Vectorize(function(z, df, ncp){
  g<- function(y)dchisq(y+z, df, ncp)*dchisq(y, df,ncp)
  integrate(g, 0, Inf)$value
})

## Sample random Data And see whether the function above fits the data.
n <- 100000
k <- 2
lambda <- 5

X <- rchisq(n, k, lambda)
Y <- rchisq(n, k, lambda)
plot(density(X-Y), lwd = 3)## This is the density we want to match.

# Superimpose the pdf from our line
curve(fz_pdf(z, k, lambda), 
      from = -20, 
      xname = 'z', add = TRUE, col='red', lwd = 2, n=200)


The resulting image shows that the pdf matches the density of Z
