I'm asked to find an expression for $u(r(t))$ in the exact expression demonstrated as follows $$u(r(t))=v^2\beta r(t)^{2\beta-1}+\frac{1}{2}v^2r(t)^{2\beta}\frac{d}{dr}log(p)$$ such that $r(t)$ is modelled as follows: $$dr(t)=u(r(t))dt+vr(t)^{\beta}dW(t)$$where: $r(t)$ is our spot rate or instantaneous interest rate (hence $u(r(t)$ is the drift term), and $v$ is our volatility term (hence $vr(t)^{\beta}$ is the diffusion term). Such model has a steady state transition probability density function that satisfies the following Fokker-Planck PDE (we drop time-dependence of $r$ for simplicity and for the sake of our working out) :$$\frac{\partial p}{\partial t}=\frac{1}{2}\frac{\partial^2 }{\partial r^2}p v^2 r^{2\beta}-\frac{\partial p}{\partial r}u(r)$$ which we turn into an ODE by taking the limit as follows $$ lim_{t \to \infty}(\frac{\partial p}{\partial t}=\frac{1}{2}\frac{\partial^2 }{\partial r^2}p v^2 r^{2\beta}-\frac{\partial p}{\partial r}u(r)$$
Resulting in:
$$\frac{1}{2}\frac{d^2 }{d r^2}p v^2 r^{2\beta}-\frac{d p}{d r}u(r)=0 ⟹ \frac{1}{2}\frac{d^2 }{d r^2}p v^2 r^{2\beta}=\frac{d p}{d r}u(r)$$
since we now only differentiate with respect to the variable $r$.
We now integrate both sides as follows:
$$\frac{1}{2}\int\frac{d^2 }{d r^2}p v^2 r^{2\beta}=\int\frac{d p}{d r}u(r) ⟹\frac{1}{2}\frac{d }{d r}p v^2 r^{2\beta}=u(r)p+C$$
where $$C=0,$$ as $$r\rightarrow \infty ,\frac{dp}{dr}\rightarrow0, p \rightarrow 0 ⟹\frac{1}{2}\frac{d }{d r}p v^2 r^{2\beta}=u(r)p $$
We now do some arranging as it's clearly a first-order ODE, and integrate again. However, this is the part that I'm stuck on. Do we rearrange and integrate as it's done on $1.$? Or do we do it as it's performed on $2.$?:
$$1.$$ $$\frac{dp}{p}=2v^{-2}r^{-2\beta}u(r)dr $$ and integrate both sides as follows $$\int \frac{dp}{p}=2v^{-2}\int u(r)r^{-2\beta}dr⟹log(p)=2v^{-2}\int u(r)r^{-2\beta}dr$$ which I wouldn't know what to do on the right-hand side. $$2.$$ $$\frac{1}{u(r)}\int \frac{dp}{p}=2v^{-2}\int r^{-2\beta}dr⟹\frac{1}{u(r)}log(p)=2v^{-2} \frac{r^{-2\beta+1}}{-2\beta+1}+C$$.
As previously said, no matter what path I decide to take ($1.$ or $2.$) I still get stuck. So does someone have any idea on how we would go about this problem to reach the expression of $u(r(t))$ shown above? I can't stress how much I thank anyone in advance who tries to help me with this, as I've been trying to unravel this dead-end of a question for as long as I can remember.