I want to prove that $$r_t = \theta + (r_0 -\theta)e^{-kt}$$ satisfies $$dr_t = k(\theta-r_t)dt, \ r(0) = r_0$$ I have \begin{split}\frac{1}{\theta - r_t} dr_t = kdt \Rightarrow & \int_0^t \frac{1}{\theta - r_s} dr_s = \int_0^t kds\\ \Rightarrow &- \ln|\theta - r_s|\big\lvert_0^t = ks\big\lvert_0^t\\ \Rightarrow &-\ln|\theta - r_t| + \ln|\theta - r_0| = kt\\ \Rightarrow &\ln|\theta - r_t| - \ln|\theta - r_0| = -kt\\ \Rightarrow &\ln\left|\frac{\theta - r_t}{\theta - r_0}\right|= -kt\\ \Rightarrow &\frac{r_t - \theta}{r_0-\theta} = e^{-kt}\\ \Rightarrow & r_t = \theta + (r_0 -\theta)e^{-kt} \end{split} Is this the way to proceed? Thanks!
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$\begingroup$ Why don’t you just differentiate your first equation? $\endgroup$– Daneel OlivawNov 25, 2021 at 16:59
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1$\begingroup$ I would say you proved that the solution of the differential equation is $r_t$ (which is quite harder), rather than proved that $r$ satisfies the differential equation (which is easier: just compute the derivatives and check) $\endgroup$– siou0107Nov 25, 2021 at 17:10
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You have: $$r(t):=\theta+(r_0-\theta)e^{-kt}\tag{1}$$ Then: $$r^\prime(t)=-k(r_0-\theta)e^{-kt}\tag{2}$$ which is clearly equal to: $$k(\theta-r(t))\tag{3}$$ based on $(1)$.
answered Nov 25, 2021 at 17:05