Suppose that the concentration of medication in the bloodstream follows M? +4 3- М Mt+1 M - %3D where M is the medication in the bloodstream given in mg/L. (Here the 4 is the supplement and the fraction term deals with the absorption rate of the mediation.) (a) Find all biological equilibria of this system.
Suppose that the concentration of medication in the bloodstream follows M? +4 3- М Mt+1 M - %3D where M is the medication in the bloodstream given in mg/L. (Here the 4 is the supplement and the fraction term deals with the absorption rate of the mediation.) (a) Find all biological equilibria of this system.
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Need help with part B. Tried to do it but am now stuck. Thanks!
![**Problem 10: Medication Concentration in the Bloodstream**
Suppose that the concentration of medication in the bloodstream follows the formula:
\[ M_{t+1} = M_t - \frac{M_t^2}{3 - M_t} + 4 \]
where \( M_t \) is the medication in the bloodstream given in mg/L. (Here the 4 is the supplement and the fraction term deals with the absorption rate of the medication.)
**(a) Find all biological equilibria of this system.**
**Solution:**
Let \( x \) be the equilibrium concentration.
\[ x = x - \frac{x^2}{3-x} + 4 \]
\[ x^2 = \frac{x^2}{3-x} \cdot (3-x) \]
\[ x^2 = 4(3 - x) \]
\[ x^2 = 12 - 4x \]
\[ x^2 + 4x - 12 = 0 \]
**Solving the quadratic equation:**
\[ (x + 6)(x - 2) = 0 \]
The solutions are:
\[ x = -6 \quad \text{and} \quad x = 2 \]
Reject \( x = -6 \) because concentration cannot be negative.
Thus, the biological equilibrium is \( x = 2 \).
**(b) For each biological equilibrium, decide if it is stable and also if it oscillates.**
**Analysis:**
Calculate the derivative \( f'(X) \):
\[ f'(X) = 1 - \frac{(3-X)(2X) - X^2 (1)}{(3-X)^2} \]
\[ f'(2) = 1 - \frac{(3-2)(2 \cdot 2) - (2)^2}{(3-2)^2} \]
\[ = \left| 1 - \frac{(1)(4) - 4}{1^2} \right| \]
\[ = |1 - 0| = |1| = 1 \]
The magnitude is 1, indicating a neutral stability with no oscillation at the equilibrium \( x = 2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fda8257-32e2-4685-ad4f-320e3e6c6af2%2F2c0d5a4d-95cf-46cc-b08c-5b17c3e6913f%2F4pqptjy.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 10: Medication Concentration in the Bloodstream**
Suppose that the concentration of medication in the bloodstream follows the formula:
\[ M_{t+1} = M_t - \frac{M_t^2}{3 - M_t} + 4 \]
where \( M_t \) is the medication in the bloodstream given in mg/L. (Here the 4 is the supplement and the fraction term deals with the absorption rate of the medication.)
**(a) Find all biological equilibria of this system.**
**Solution:**
Let \( x \) be the equilibrium concentration.
\[ x = x - \frac{x^2}{3-x} + 4 \]
\[ x^2 = \frac{x^2}{3-x} \cdot (3-x) \]
\[ x^2 = 4(3 - x) \]
\[ x^2 = 12 - 4x \]
\[ x^2 + 4x - 12 = 0 \]
**Solving the quadratic equation:**
\[ (x + 6)(x - 2) = 0 \]
The solutions are:
\[ x = -6 \quad \text{and} \quad x = 2 \]
Reject \( x = -6 \) because concentration cannot be negative.
Thus, the biological equilibrium is \( x = 2 \).
**(b) For each biological equilibrium, decide if it is stable and also if it oscillates.**
**Analysis:**
Calculate the derivative \( f'(X) \):
\[ f'(X) = 1 - \frac{(3-X)(2X) - X^2 (1)}{(3-X)^2} \]
\[ f'(2) = 1 - \frac{(3-2)(2 \cdot 2) - (2)^2}{(3-2)^2} \]
\[ = \left| 1 - \frac{(1)(4) - 4}{1^2} \right| \]
\[ = |1 - 0| = |1| = 1 \]
The magnitude is 1, indicating a neutral stability with no oscillation at the equilibrium \( x = 2 \).
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