My question is the same as this question but more general. I am dealing with parabolic cylinder functions and misunderstand some moments. As the source, I use this. The authors state that these equations $$\frac{d^2 w}{d z^2}+(az^2+bz+c)w=0;\quad \frac{d^2 w}{dz^2}+\left(\frac{z^2}{4}-a\right)w=0;$$ $$\frac{d^2 w}{dz^2}-\left(\frac{z^2}{4}+a\right)=0;\quad \frac{d^2 w}{dz^2}+\left(\nu+\frac{1}{2}-\frac{z^2}{4}\right)w=0$$ can be easily transformed to each other. I start with the first equation and comlete square $$az^2+bz+c=a\left(z+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right),$$ then change the variable $z\rightarrow z+b/2a$ and denote $\lambda=c-b^2/4a$, it gives $$\frac{d^2 w}{d z^2}+(az^2+\lambda)w=0,$$ where I perfom the change $z\rightarrow(4a)^{-1/4}z$ and $a\rightarrow-\lambda(4a)^{-1/2}$ and obtain $$ \frac{d^2 w}{dz^2}+\left(\frac{z^2}{4}-a\right)w=0.\quad (*)$$ Ok, it's good. From the DLMF, I know that the solutions of $(*)$ equation are the functions $W(a,\,\pm z)$. Then, from this I know how to connect the solutions $W(a,\,\pm z)$ and $U$-functions (see previous link). The connection formulas between $W$ and $U$ are very complicated (for me).
Then, I know that the solutions of $$\frac{d^2 w}{dz^2}+\left(\nu+\frac{1}{2}-\frac{z^2}{4}\right)w=0\quad(**)$$ are $D_{\nu}(z)$ and easily connected with $U$-function. Also, I see that someone just write down solutions of equation with $az^2+bz+c$ in terms of $D_{\nu}(z)$. So, I misunderstand how the complicated connection formula between $W$-function and $U$-function (I need it to connect $D$-function and $W$-function) allows to easily write down the solution of equation with $az^2+bz+c$ in terms of $D$-function.
Let me try to present the answer. One can start from the following equation: $$\frac{d^2w}{dz^2}+(az^2+\lambda)=0,$$ then substitute $z\rightarrow (4a)^{-1/4}z$ and $\lambda\rightarrow(4a)^{-1/2}\lambda$ and obtain: $$\frac{d^2w}{dz^2}+\left(\frac{z^2}{4}+\lambda\right)=0.$$ In this equation we also introduce new variables as $z\rightarrow xe^{i\pi/4}$ and $\rho\rightarrow -i\lambda$. It gives $$\boxed{\frac{d^2w}{dx^2}-\left(\frac{z^2}{4}+\lambda\right)=0.}$$ This equation has the solutions $U(\rho,\,z)$ and the connection formula is $U(\rho,\,z)=D_{-\rho-1/2}(z)$.