Secretion of Hormones 1

Problem 1.

The secretion of hormones into the blood is often a periodic activity. If a hormone is secreted on a $24$-hour cycle, then the rate of change of the level of the hormone in the blood may be represented by the initial value problem $$x'(t)=\alpha-\beta \cos\left(\frac{\pi t}{12} \right)-kx(t), \qquad x(0)=x_0,$$ where $x(t)$ is the amount of the hormone in the blood at time $t$, $\alpha$ is the average secretion rate, $\beta$ is the amount of daily variation in the secretion, and $k$ is a positive constant reflecting the rate at which the body removes the hormone from the blood.
If $\alpha,\beta=1$, $k=2$, and $x_0=10$, solve for $x(t)$.

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(1) In the first step we solve the homogeneous equation, $$x'(t)+kx(t)=0.$$ Its solutions are $$x_h(t)=Ce^{-kt},$$ where $C\in\mathbf{R}$ is arbitrary constant.
(2) In the second step we solve the inhomogeneous equation, applying the method of variation of parameter(s).
$$x_p(t):=v(t)x_h(t).$$

$$\begin{align*} &\Step{1A}{x_p'(t)=v'(t)x_h(t)+v(t)x_h'(t),}\\ &\Step{2A}{x_p'(t)+kx_p(t)=v'(t)x_h(t)+v(t)[x_h'(t)+kx_h(t)],} \\ &\Step{3A}{\qquad\qquad\qquad=v'(t)x_h(t),} \\ &\Step{4A}{x_p'(t)+kx_p(t)=\alpha-\beta \cos\left(\frac{\pi t}{12} \right),}\\ &\Step{5A}{v'(t)x_h(t)=\alpha-\beta \cos\left(\frac{\pi t}{12} \right),}\\ &\Step{6A}{v'(t)=x_h^{-1}(t)\left[\alpha-\beta \cos\left(\frac{\pi t}{12} \right)\right],}\\ &\Step{7A}{v'(t)=e^{kt}\left[\alpha-\beta \cos\left(\frac{\pi t}{12} \right)\right]\qquad(C:=1).}\\ \end{align*}$$

After integration by parts

we obtain $$v(t)={\frac {{{e}^{kt}}\alpha}{k}}-12\,{\frac {\beta\,{{e}^{kt}} \left[ \pi \,\sin \left( 1/12\,\pi \,t \right) +12\,k\cos \left( 1/12\,\pi \,t \right) \right] }{{\pi }^ {2}+144\,{k}^{2}}}.$$ Hence $$x_p(t)={\frac {\alpha}{k}}-12\,{\frac {\beta\, \left[ \pi \,\sin \left( 1/12\,\pi \,t \right) +12\,k\cos \left( 1/12\,\pi \,t \right) \right] }{{\pi }^ {2}+144\,{k}^{2}}}.$$ (3) The general solution of the differential equation is

$$\begin{align*} &\Step{1B}{x(t) =x_h(t)+x_p(t),}\\ &\Step{2B}{=Ce^{-kt}+\frac {\alpha}{k}-12\,{\frac {\beta\, \left[ \pi \,\sin \left( 1/12\,\pi \,t \right) +12\,k\cos \left( 1/12\,\pi \,t \right) \right] }{\pi^ {2}+144\,{k}^{2}}}.}\\ \end{align*}$$

(4) Since $x(0)=x_0$, we obtain $$C=\frac{144 x_0 k^3+144(\beta-\alpha)k^2+\pi^2x_0 k-\pi^2\alpha}{144k^3+\pi^2 k},$$ and $$\begin{align*} x(t) &=\frac{144 x_0 k^3+144(\beta-\alpha)k^2+\pi^2x_0 k-\pi^2\alpha}{144k^3+\pi^2 k}e^{-kt}\\ &+\frac {\alpha}{k}-12\,{\frac {\beta\, \left[ \pi \,\sin \left( 1/12\,\pi \,t \right) +12\,k\cos \left( 1/12\,\pi \,t \right) \right] }{\pi^ {2}+144\,{k}^{2}}}. \end{align*}$$ Here we can transform the last term further if we use that $$\frac{a\sin(x)+b\cos(x)}{\sqrt{a^2+b^2}}=\sin(x+\delta),\qquad(a,b\gt 0),$$ where $\tan(\delta)=\frac{b}{a}$.
Hence $$\begin{align*} x(t) &=\frac{144 x_0 k^3+144(\beta-\alpha)k^2+\pi^2x_0 k-\pi^2\alpha}{144k^3+\pi^2 k}e^{-kt}\\ &+\frac {\alpha}{k}- \frac{12\beta}{\sqrt{\pi^{2}+144\,k^2}}\,\sin \left( \frac{\pi}{12}t+\delta\right), \end{align*}$$ where $\tan(\delta)=\frac{12k}{\pi}$.
If $\alpha,\beta=1$, $k=2$, and $x_0=10$ then $$\begin{align*} x(t)=&e^{-2t} \left( 10-\frac{\pi^2}{2(\pi^2+576)} \right) \\ &-\frac {24\pi\sin \left( 1/12\,\pi \,t \right) + 576\cos\left( 1/12\,\pi t \right) -\pi^2-576}{2(\pi^2+576)}. \end{align*}$$

Solution by Maple

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