$P_{n+1}(x)=P'_n(x)+2nxP_n(x)$ and looking for orthogonal relation

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Let's define

$$\begin{align*} \frac{dx}{du}&=e^{x^2}\\ \frac{d^2x}{du^2}&=2xe^{2x^2}\\ \frac{d^3x}{du^3}&=(2+8x^2)e^{3x^2}\\ \vdots\\ \frac{d^nx}{du^n}&=P_n(x)e^{nx^2}\\ \frac{d^{n+1}x}{du^{n+1}}&=(P'_n(x)+2nxP_n(x))e^{(n+1)x^2}\\ \frac{d^{n+1}x}{du^{n+1}}&=P_{n+1}(x)e^{(n+1)x^2}\\ \end{align*}$$

$$P_{n+1}(x)=P'_n(x)+2nxP_n(x)$$

I would like to find the differential equation that satisfies $P_{n}(x)$.

And then I would like to find an orthogonal relation such as if possible

if $m \neq n$ $$\int_a^{b} P_n(x)P_m(x)W(x)\;dx=0$$

if $m = n$ $$\int_a^{b} P^2_n(x)W(x)\;dx=a_n$$

Could you please help me about steps?

Thanks

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From the Rodrigues formula for the Hermite polynomials, making a few variable substitutions gives the relation

$$\exp(-z^2)\frac{\mathrm d^n}{\mathrm dz^n}\exp(z^2)=i^{-n}H_n(iz)$$

Making the appropriate substitutions into the Hermite DE yields

$$w^{\prime\prime}+2zw^\prime-2nw=0$$

I am not too sure about how one could derive an orthogonality relation from this, though; the substitutions don't really work for the usual orthogonality relation for the Hermite polynomials.