= Solution
Substitute the nonzero product $V=f(S)g(t)$ into the <Black-Scholes equation with continuous stock dividends> and divide by $fg$ on intervals where it is nonzero. The result is
$$
-\frac{\dot g}{g}+r
=\frac{\tfrac12\sigma^2S^2f''+(r-q)Sf'}{f}.
$$
The left side depends only on time and the right side only on spot, so both equal a constant $\lambda$. Thus
$$
\boxed{\dot g=(r-\lambda)g,\qquad
\tfrac12\sigma^2S^2f''+(r-q)Sf'-\lambda f=0.}
$$
These are the <separated solutions of the Black-Scholes equation>. The zero solution is included separately, so division loses no relevant case.
For the unheaded parity request, the terminal difference of the power-call and power-put payoffs is $S_T^\beta-K^\beta$. Under the <risk-neutral measure>, $S_T=S\exp[(r-q-\sigma^2/2)\tau+\sigma(W_T-W_t)]$, where $\tau=T-t$. The <Gaussian> exponential moment gives $\mathbb E_Q[S_T^\beta\mid S_t=S]=S^\beta e^{\lambda_\beta\tau}$ with $\lambda_\beta=\beta(r-q)+\sigma^2\beta(\beta-1)/2$. Discounting the difference proves <power put-call parity>:
$$
\boxed{C_\beta(S,t)-P_\beta(S,t)=S^\beta e^{(\lambda_\beta-r)(T-t)}-K^\beta e^{-r(T-t)}.}
$$
For $\beta=1$ this is the familiar dividend-adjusted <put-call parity>, $Se^{-q\tau}-Ke^{-r\tau}$.
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