The nonlinear Michaelis-Menton model equation ẏ=-a y /(b+y), with a, b positive constants, descr

The nonlinear Michaelis-Menton model equation ẏ=-a y /(b+y),...

Key Concepts

Plotting Solutions

Plotting solutions involves graphically representing the response of the system over a specific interval. By visualizing the analytic solution of an ODE, one can directly observe the dynamic behavior of the substrate concentration over time, validate the correctness of the solution, and provide insights into the system’s transient and steady-state characteristics.

Analytic Solution

An analytic solution refers to a solution of a differential equation expressed in a closed-form expression, as opposed to a numerical or approximate solution. Finding an analytic solution means deriving an explicit formula for the function, which is crucial for understanding the behavior and properties of the system over time.

Michaelis-Menten Kinetics

Michaelis-Menten kinetics is a widely used model in enzyme kinetics that describes the rate of enzymatic reactions. The model is based on the assumption that the formation and dissociation of the enzyme-substrate complex reach a quasi-steady state. Although the example given simplifies the model, interpreting and solving the resulting non-linear ODE is essential in understanding enzyme behavior.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve functions of one independent variable and their derivatives. Solving ODEs involves finding a function that satisfies the given relationship between the function itself and its derivatives. In contexts such as enzyme kinetics, ODEs model the rates of changes in the concentrations of substances over time.

Separation of Variables

Separation of variables is a method used to solve differential equations by rearranging the terms so that each variable and its differential appear on opposite sides of the equation. This technique simplifies the process by reducing the problem to integrating both sides independently, which is particularly effective for first-order ODEs.

Initial Value Problem

An initial value problem (IVP) is a differential equation together with a specified value of the unknown function at a given point, commonly referred to as the initial condition. This specification determines a unique solution for many types of ODEs, ensuring that the evolution of the system is well-defined from a starting state.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic technique used to break down rational functions into simpler fractions. When an ODE yields an integrand expressed as a rational function, this method allows the integration to be performed term-by-term, thereby facilitating the solution of the integral.

Transcript

0:00 Hello everyone.

00:01 Today we're going to determine and sketch the equation of motion for the undamped system at resonance governed by the differential equation d2y dt squared plus y equals 5 cosine t with initial conditions y of 0 equals 0 and y prime of 0 equals 1.

00:17 So all we're doing is finding a solution to this initial value problem.

00:22 So the first thing that we're going to do is going to determine the homogeneous solution.

00:27 So first when we write d2y dt squared plus y equals equals 5 cosine t, we make this equal to zero, and then we write the characteristic polynomial of this equation, which is r squared plus 1.

00:41 Now we know that r squared plus 1 equals 0 whenever r is equal to plus or minus i.

00:48 This tells us that a homogeneous solution y h of t is equal to some constant c1 times cosine t plus another constant c2 times sine of t, since all roots are complex.

01:06 Next thing that we're going to do is we're going to obtain the particular solution based on the method of undetermined coefficients.

01:15 So we write yp of t, which equals t raised to the power of 1 since our roots are simple times a times cosine t plus b sine of t.

01:30 And now we start taking derivatives.

01:33 Y prime subp of t equals a cosine t plus b sine t plus t plus t times negative a sine t plus b cosine t and then we take derivatives again y cp double prime of t equals negative 2a sine of t which comes from the a cosine t and the negative a sine of t from the product rule plus 2b cosine t minus t times a cosine t plus a cosine t plus b sine of t since all our terms here are negative so we can pull the negative sign out and now we plug this y sub p into our differential equation and so we get y double prime sub p of t plus y sub p of t equals 5 cosine t which then becomes negative 2a sine a t plus 2b cosine t minus t times a cosine t plus b cc plus b sine t plus t equals 5 cosine t.

03:18 Now notice that these two terms cancel and we're left with negative 2a sine t plus 2b cosine t equals 5 cosine t.

03:31 And so we match the coefficients here and here.

03:39 So this means that a is equal to 0, and b is equal to 5 over 2.

03:45 So we have that our particular solution, y sub p of t, is equal to 5 over 2, t, sine of t...

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