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Chapter 2: Problem 31

Let \(S(t)\) represent the amount of a chemical reactant present at time \(t, t \geq 0\). Assume that \(S(t)\) can be determined by solving the initial value problem $$ S^{\prime}=-\frac{\alpha S}{K+S}, \quad S(0)=S_{0}, $$ where \(\alpha, K\), and \(S_{0}\) are positive constants. Obtain an implicit solution of the initial value problem. (The differential equation, often referred to as the Michaelis-Menten equation, arises in the study of biochemical reactions.)

Short Answer

Expert verified
Answer: The implicit solution for the given initial value problem is $S(t) = S_0 e^{-\alpha \int \frac{dt}{K+S}}$.

Step by step solution

01

Separate variables

First, let's separate the variables in the given equation: $$ S^{\prime} = -\frac{\alpha S}{K+S} $$ To separate variables, we want to write the equation in the form \(\frac{dS}{S} = f(t) dt\). We have: $$ \frac{dS}{S} = -\frac{\alpha dt}{K+S} $$
02

Integrate both sides

Now we need to integrate both sides. On the left-hand side (LHS), we have a simple integration: $$ \int \frac{dS}{S} = \int -\frac{\alpha dt}{K+S} $$ $$ \ln |S| = -\alpha \int \frac{dt}{K+S} + C $$ Here, C is the constant of integration.
03

Simplify the equation

Now we want to simplify the equation to make it more manageable. First, we can exponentiate both sides to eliminate the natural logarithm: $$ |S| = e^{-\alpha \int \frac{dt}{K+S} + C} $$ Now, we can introduce a new constant of integration \(C_1 = e^C\) to remove the absolute value sign: $$ S = C_1 e^{-\alpha \int \frac{dt}{K+S}} $$
04

Use the initial condition

We know the initial condition is \(S(0) = S_0\). Therefore, we can find our constant of integration \(C_1\): $$ S_0 = C_1 e^{-\alpha \int \frac{d(0)}{K+S(0)}} $$ Since \(\int \frac{d(0)}{K+S(0)} = 0\), \(e^0 = 1\), and we get: $$ S_0 = C_1 $$ So, our solution becomes: $$ S(t) = S_0 e^{-\alpha \int \frac{dt}{K+S}} $$
05

Implicit solution of the IVP

Since we cannot integrate the right-hand side (RHS) any further, we have arrived at the implicit solution of the initial value problem: $$ S(t) = S_0 e^{-\alpha \int \frac{dt}{K+S}} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem
The initial value problem is a fundamental concept in differential equations. It involves finding a solution to a differential equation that also satisfies an initial condition. For this exercise, we consider the Michaelis-Menten equation, which describes the rate of a biochemical reaction. This problem starts with a differential equation involving a rate of change of a chemical, denoted by \(S'(t)\).
The initial value, given as \(S(0) = S_0\), allows us to determine a unique solution. It lets us define the state of the system at the beginning, which simplifies finding solutions to differential equations.
Such problems are typical in studying reaction rates and understanding how reactants evolve over time.
Variable Separation
Variable separation is a powerful technique used to solve differential equations. It is particularly useful when the differential equation can be rearranged to have all instances of one variable on one side of the equation and the opposite variable on the other. In this exercise, the equation \(S' = -\frac{\alpha S}{K+S}\) was separated.
It involves rearranging expressions so you end up with something that looks like \(\frac{dS}{S}\) on one side, and terms involving \(t\) on the other, such as \(-\frac{\alpha dt}{K+S}\).
This separation makes it possible to individually integrate both sides of the equation, setting the stage for finding a solution.
Integration
Integration is a mathematical process that allows us to solve differential equations, especially after variables have been separated. Once the equation \(\frac{dS}{S} = -\frac{\alpha dt}{K+S}\) is set, the next step is to integrate both sides.
The left side involves a simple integration which results in \(\ln|S|\). For the right side, it turns into \(-\alpha \int \frac{dt}{K+S}\). The constant of integration, \(C\), appears as a part of the solution, recognizing that indefinite integrals involve arbitrary constants.
By integrating each side, we connect the rate of change of the chemical reactant to time, offering a richer understanding of how it diminishes over time due to the biochemical reaction.
Biochemical Reactions
Biochemical reactions are transformations within living organisms catalyzed by enzymes. The study of these reactions often leads to differential equations like the Michaelis-Menten equation used in this exercise.
It highlights how the concentration of a reactant, represented as \(S(t)\), changes over time. Here, the constant \(\alpha\) represents the reaction rate, and \(K\) denotes the Michaelis constant, offering insight into the enzyme's efficiency and affinity for substrate.
Understanding these reactions through mathematical models helps in biology to quantify and predict changes in reaction rates, facilitating advancements in fields like pharmacology and metabolic engineering. It explains how biochemical systems maintain balance within biological systems.

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Most popular questions from this chapter

Sometimes a change of variable can be used to convert a differential equation \(y^{\prime}=f(t, y)\) into a separable equation. (a) Consider a differential equation of the form \(y^{\prime}=f(\alpha t+\beta y+\gamma)\), where \(\alpha, \beta\), and \(\gamma\) are constants. Use the change of variable \(z=\alpha t+\beta y+\gamma\) to rewrite the differential equation as a separable equation of the form \(z^{\prime}=g(z)\). List the function \(g(z)\). (b) A differential equation that can be written in the form \(y^{\prime}=f(y / t)\) is called an equidimensional differential equation. Use the change of variable \(z=y / t\) to rewrite the equation as a separable equation of the form \(t z^{\prime}=g(z)\). List the function \(g(z)\).

A tank, containing 1000 gal of liquid, has a brine solution entering at a constant rate of \(2 \mathrm{gal} / \mathrm{min}\). The well-stirred solution leaves the tank at the same rate. The concentration within the tank is monitored and is found to be the function of time specified. In each exercise, determine (a) the amount of salt initially present within the tank. (b) the inflow concentration \(c_{i}(t)\), where \(c_{i}(t)\) denotes the concentration of salt in the brine solution flowing into the tank. $$c(t)=\frac{1}{20}\left(1-e^{-t / 500}\right) \mathrm{lb} / \mathrm{gal}$$

A projectile of mass \(m\) is launched vertically upward from ground level at time \(t=0\) with initial velocity \(v_{0}\) and is acted upon by gravity and air resistance. Assume the drag force is proportional to velocity, with drag coefficient \(k\). Derive an expression for the time, \(t_{m}\), when the projectile achieves its maximum height.

The motion of a body of mass \(m\), gravitationally attracted to Earth in the presence of a resisting drag force proportional to the square of its velocity, is given by $$ m \frac{d v}{d t}=-\frac{G m M_{e}}{r^{2}}+\kappa v^{2} $$ [recall equation (13)]. In this equation, \(r\) is the radial distance of the body from the center of Earth, \(G\) is the universal gravitational constant, \(M_{e}\) is the mass of Earth, and \(v=d r / d t\). Note that the drag force is positive, since it acts in the positive \(r\) direction. (a) Assume that the body is released from rest at an altitude \(h\) above the surface of Earth. Recast the differential equation so that distance \(r\) is the independent variable. State an appropriate initial condition for the new problem. (b) Show that the impact velocity can be expressed as $$ v_{\text {impact }}=-\left[2 G M_{e} \int_{0}^{h} \frac{e^{-2(\kappa / m) s}}{\left(R_{e}+s\right)^{2}} d s\right]^{1 / 2}, $$ where \(R_{e}\) represents the radius of Earth. (The minus sign reflects the fact that \(v=d r / d t<0 .)\)

A tank initially contains 400 gal of fresh water. At time \(t=0\), a brine solution with a concentration of \(0.1 \mathrm{lb}\) of salt per gallon enters the tank at a rate of \(1 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out at a rate of \(2 \mathrm{gal} / \mathrm{min}\). (a) How long does it take for the tank to become empty? (This calculation determines the time interval on which our model is valid.) (b) How much salt is present when the tank contains \(100 \mathrm{gal}\) of brine? (c) What is the maximum amount of salt present in the tank during the time interval found in part (a)? When is this maximum achieved?
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