Q37E Secretion of Hormones.聽The secr... [FREE SOLUTION]

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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 2: Q37E (page 57)

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 $\frac{dx}{dt} = 伪 - 尾肠辞蝉 \frac{蟺迟}{12} - kx, x (0) = x_{o}$ , where $x_{1} t_{2}$ is the amount of the hormone in the blood at timet, a is the average secretion rate, b is the amount of daily variation in the secretion, andkis a positive constant reflecting the rate at which the body removes the hormone from the blood. If $伪 = 尾 = 1$ , $k = 2$ , and $x_{o} = 10$ , solve for $x (t)$ .

Short Answer

Expert verified

$x (t) = \frac{1}{2} - \frac{2 \cos \frac{蟺 t}{12}}{{(\frac{蟺}{12})}^{2} + 4} - \frac{[\frac{蟺}{12} \sin (\frac{蟺 t}{12})]}{{(\frac{蟺}{12})}^{2} + 4} + e^{- 2 t} (\frac{19}{2} + \frac{2}{{(\frac{蟺}{12})}^{2} + 4})$

Step by step solution

Important formula

The required formula is $\frac{d}{dx} (x^{n}) = n x^{n - 1}$ .

Find the value ofxt

Given $\frac{dx}{dt} = 伪 - 尾 \cos \frac{蟺 t}{12} - kx, x (0) = x_{o}$

\frac{dx}{dt} = 伪 - 尾 \cos \frac{蟺 t}{12} - kx \frac{dx}{dt} + kx = 伪 - 尾 \cos \frac{蟺 t}{12}

Integrating factor is $e^{鈭� kdt} = e^{kt}$ .

$x (t) = e^{- kt} 鈭� e^{kt} (伪 - 尾 \cos \frac{蟺 t}{12}) dt + e^{- kt} C x (t) = \frac{伪}{k} + e^{- kt} C - 尾 e^{- kt} I$

Where $I = 鈭� \cos \frac{蟺 t}{12} dt$

Solving by integration by parts

$I = e^{kt} \frac{12}{蟺} \sin (\frac{蟺 t}{12}) - \frac{12 k}{蟺} 鈭� \sin (\frac{蟺 t}{12}) e^{kt} dt$

Again solving by integration by parts

$I = \frac{e^{kt} [\frac{蟺}{12} \sin (\frac{蟺 t}{12}) + k \cos \frac{蟺 t}{12}]}{{(\frac{蟺}{12})}^{2} + k^{2}}$

Therefore . $x (t) = \frac{伪}{k} + e^{- kt} C - 尾 \frac{[\frac{蟺}{12} \sin (\frac{蟺 t}{12}) + k \cos \frac{蟺 t}{12}]}{{(\frac{蟺}{12})}^{2} + k^{2}}$

Put $k = 2 and 伪 = 尾 = 1$ then

$x (t) = \frac{1}{2} + e^{- 2 t} C - \frac{[\frac{蟺}{12} \sin (\frac{蟺 t}{12}) + 2 \cos \frac{蟺 t}{12}]}{{(\frac{蟺}{12})}^{2} + 4}$

Evaluate the value C

For $t = 0 and x_{o} = 10$ then

$C = 9.5 + \frac{2}{{(\frac{蟺}{12})}^{2} + 4}$

Therefore,.

role="math" localid="1664162247706" $x (t) = \frac{1}{2} - \frac{2 \cos \frac{蟺迟}{12}}{{(\frac{蟺}{12})}^{2} + 4} - \frac{[\frac{蟺}{12} \sin (\frac{蟺迟}{12})]}{{(\frac{蟺}{12})}^{2} + 4} + e^{- 2 t} (\frac{19}{2} + \frac{2}{{(\frac{蟺}{12})}^{2} + 4})$

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91影视

Short Answer

Step by step solution

Important formula

Find the value ofxt

Evaluate the value C

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