Secretion of Hormones.The secretion of hormones into the blood is often a periodic activity. If a hormone is secreted on a 24-h cycle, then the rate of change of the level of the hormone in the blood may be represented by

the initial value problem\(\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = \alpha - \beta cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - kx,x(0) = }}{{\bf{x}}_{\bf{o}}}\)where x(t) is the amount of the hormone in the blood at the time t, \({\bf{\alpha }}\) is the average secretion rate, \({\bf{\beta }}\)is the amount of daily variation in the secretion, and kis a positive constant reflecting the rate at which the body removes the hormone from the blood. If \({\bf{\alpha }}\)=\({\bf{\beta }}\) = 1, k= 2, and \({{\bf{x}}_{\bf{o}}}\) = 10, solve for x(t).

Here\({\bf{\alpha }}\)=\({\bf{\beta }}\)= 1, k= 2, and \({{\bf{x}}_{\bf{o}}}\) = 10.

Now,

\(\begin{array}{c}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 1 - cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2x}}\\\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ + 2x = 1 - cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}\end{array}\)

The linear equation of the form is\(\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ + P(t)x = Q(t)}}\).

The integrating factor is\({\bf{\mu (t) = }}{{\bf{e}}^{{\bf{2t}}}}\).

Now,

\(\begin{array}{c}{{\bf{e}}^{{\bf{2t}}}}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ + 2}}{{\bf{e}}^{{\bf{2t}}}}{\bf{x = }}{{\bf{e}}^{{\bf{2t}}}}\left( {{\bf{1 - cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}} \right)\\\frac{{\bf{d}}}{{{\bf{dt}}}}{\bf{[}}{{\bf{e}}^{{\bf{2t}}}}{\bf{x] = }}\int {{{\bf{e}}^{{\bf{2t}}}}{\bf{ - }}{{\bf{e}}^{{\bf{2t}}}}} \left( {{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}} \right){\bf{dt}}\\{{\bf{e}}^{{\bf{2t}}}}{\bf{x = }}\frac{{{{\bf{e}}^{{\bf{2t}}}}}}{{\bf{2}}}{\bf{ - I + c}}\end{array}\)

\(\begin{array}{c}{\bf{I = }}\int {{{\bf{e}}^{{\bf{2t}}}}\left( {{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}} \right)} {\bf{dt}}\\{\bf{I = }}\frac{{{{\bf{e}}^{{\bf{2t}}}}}}{{\bf{2}}}\left( {{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}} \right){\bf{ - }}\frac{{\bf{\pi }}}{{{\bf{24}}}}\int {{{\bf{e}}^{{\bf{2t}}}}\left( {{\bf{sin}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}} \right)} {\bf{dt}}\\{\bf{I = }}\frac{{{{\bf{e}}^{{\bf{2t}}}}}}{{\bf{2}}}\left( {{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}} \right){\bf{ - }}\frac{{{\bf{\pi }}{{\bf{e}}^{{\bf{2t}}}}}}{{{\bf{48}}}}\left( {{\bf{sin}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}} \right){\bf{ + }}\frac{{{{\bf{\pi }}^{\bf{2}}}}}{{{\bf{576}}}}{\bf{I}}\\{\bf{I = }}\frac{{{\bf{12}}{{\bf{e}}^{{\bf{2t}}}}{\bf{(24cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ + \pi sin}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{)}}}}{{{\bf{576 + }}{{\bf{\pi }}^{\bf{2}}}}}\end{array}\)

\(\begin{array}{c}{{\bf{e}}^{{\bf{2t}}}}{\bf{x = }}\frac{{{{\bf{e}}^{{\bf{2t}}}}}}{{\bf{2}}}{\bf{ - }}\frac{{{\bf{12}}{{\bf{e}}^{{\bf{2t}}}}{\bf{(24cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ + \pi sin}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{)}}}}{{{\bf{576 + }}{{\bf{\pi }}^{\bf{2}}}}}{\bf{ + c}}\\{\bf{x = }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ - }}\frac{{{\bf{12(24cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ + \pi sin}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{)}}}}{{{\bf{576 + }}{{\bf{\pi }}^{\bf{2}}}}}{\bf{ + c}}{{\bf{e}}^{{\bf{ - 2t}}}}\\{\bf{10 = }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ - }}\frac{{{\bf{12(24)}}}}{{{\bf{576 + }}{{\bf{\pi }}^{\bf{2}}}}}{\bf{ + c}}\\{\bf{c = 10 - }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + }}\frac{{{\bf{288}}}}{{{\bf{576 + }}{{\bf{\pi }}^{\bf{2}}}}}\\{\bf{c = }}\frac{{{\bf{19}}}}{{\bf{2}}}{\bf{ + }}\frac{{{\bf{288}}}}{{{\bf{576 + }}{{\bf{\pi }}^{\bf{2}}}}}\\{\bf{x = }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ - }}\frac{{{\bf{12(24cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ + \pi sin}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{)}}}}{{{\bf{576 + }}{{\bf{\pi }}^{\bf{2}}}}}{\bf{ + }}\left( {\frac{{{\bf{19}}}}{{\bf{2}}}{\bf{ + }}\frac{{{\bf{288}}}}{{{\bf{576 + }}{{\bf{\pi }}^{\bf{2}}}}}} \right){{\bf{e}}^{{\bf{ - 2t}}}}\end{array}\)

This is the required result.