Problem 16 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Exercise 18.1 .16 Find the particular solution, \(y_{p}=A \cos 0.3 t+B \sin 0.3 t,\) and \(y_{h},\) the solution to the homogeneous equation, for the systems below. a. $$ y^{\prime \prime}+3 y^{\prime}+2 y=\cos 0.3 t \quad \text { b. } \quad y^{\prime \prime}+0.1 y^{\prime}+0.12 y=\cos 0.3 t $$ $$ \text { c. } y^{\prime \prime}+0.05 y^{\prime}+0.0004 y=\cos 0.3 t $$ d. \(y^{\prime \prime}+0.13 y^{\prime}+0.0036 y=\cos 0.3 t\) The algebra to solve these problems is extensive and you will find the MATLAB code below helpful at least to read and at best to run. It is written for the case of distinct real roots, and the system is initially at rest: \(y(0)=0, y^{\prime}(0)=0 .\) NOTE: \(\omega \neq 0.3\) in the code. Solutions to \(y^{\prime \prime}+0.1 y^{\prime}+0.12 y=\cos \omega t, y(0)=0, y^{\prime}(0)=0,\) for \(\omega=0.1\) and \(0.2 .\) The blue curves are the solution; the green curves are the solutions to the homogeneous equation, \(y^{\prime \prime}+0.1 y^{\prime}+0.125 y=0,\) and the red curves are the particular solutions. The solution curves (blue) are asymptotic to the particular solutions (red).

Step-by-Step Solution

Answer

Find the homogeneous and particular solution for each part and combine them.

Step 1: Identify the Differential Equation

For each part, the differential equation is of the form \( y'' + ext{coefficient}_1 y' + ext{coefficient}_2 y = \text{cos} 0.3 t \). We will use this form to determine the particular solution \( y_p \) and the homogeneous solution \( y_h \).

Step 2: Solve the Homogeneous Equation for each Case

The homogeneous equation is \( y'' + ext{coefficient}_1 y' + ext{coefficient}_2 y = 0 \). Find the roots of the characteristic equation to determine \( y_h \). For example, in part (a), the characteristic equation is \( r^2 + 3r + 2 = 0 \), which has roots \( r_1 = -1 \), \( r_2 = -2 \), leading to \( y_h = C_1 e^{-2t} + C_2 e^{-t} \). Repeat similarly for parts b, c, and d.

Step 3: Identify the Form of the Particular Solution

The form of the particular solution is given as \( y_p = A \cos 0.3 t + B \sin 0.3 t \) or similar, based on the forcing function \( \text{cos} 0.3 t \). This form is used because the forcing function involves a trigonometric term.

Step 4: Substitute into the Differential Equation

Substitute \( y_p = A \cos 0.3 t + B \sin 0.3 t \) and its derivatives into the non-homogeneous differential equation to find the coefficients \( A \) and \( B \). For instance, set \( y' = -0.3A \sin 0.3t + 0.3B \cos 0.3t \) and \( y'' = -0.09A \cos 0.3t - 0.09B \sin 0.3t \), then plug these into the equation and equate the coefficients with the right side, \( \cos 0.3 t \).

Step 5: Solve for Coefficients A and B

Match the coefficients from both sides of the equation to solve for \( A \) and \( B \). For example, balancing the coefficients of \( \cos 0.3 t \) and \( \sin 0.3 t \) will give you a system of equations to solve for \( A \) and \( B \). Repeat for each part (a, b, c, d) of the exercise to find the specific \( A \) and \( B \) values.

Step 6: Combine the Solutions

The complete solution for each equation is a sum of the homogeneous and particular solutions, \( y = y_h + y_p \). This combines both the natural response of the system and the response to the forcing function.

Key Concepts

Particular Solutions

In the realm of differential equations, particular solutions are crucial when dealing with non-homogeneous equations. Such equations present a "forcing function," or an external influence, commonly denoted by terms like \( \cos 0.3t \) as seen in the original exercise. When we talk about finding a particular solution, we mean identifying a specific solution that fits the entire non-homogeneous equation.

The particular solution takes on a form inspired by the inhomogeneous part of the equation, which is often a trigonometric, exponential, or polynomial function. In our exercise, the forcing function is cosine, prompting a particular solution of the form \(y_p = A \cos 0.3t + B \sin 0.3t\). This form is particularly effective for trigonometric inputs, and involves using derivatives to balance the equation all the way through.

To determine the coefficients \( A \) and \( B \), you need to substitute \( y_p \) and its derivatives into the original differential equation and equate coefficients. This method allows us to match the left and the right side of the equation perfectly, ensuring the solution is valid.

Homogeneous Equations

Homogeneous equations form the basis when solving both homogeneous and non-homogeneous differential equations. They are characterized by having zero on the right side of the equation, for example, \( y'' + 3y' + 2y = 0 \). Solving these equations leads to what is known as the "complementary solution."

To solve the homogeneous equation, you must first find its characteristic equation. This is done by assuming a solution of the form \( y = e^{rt} \). Substituting this into the homogeneous equation results in a polynomial equation in terms of \( r \), known as the characteristic equation.

For example, a characteristic equation might look like \( r^2 + 3r + 2 = 0 \), which can be solved using standard algebraic techniques to find the roots \( r_1 \) and \( r_2 \). These roots indicate the form of the complementary solution. If the roots are distinct and real, the solution will be a linear combination of exponentials, such as \( y_h = C_1 e^{-2t} + C_2 e^{-t} \). This approach showcases the natural behaviour of the system without external forces.

Characteristic Equation

The characteristic equation is a key concept in solving differential equations, particularly homogeneous ones. It arises when you substitute the assumed solution form, \( y = e^{rt} \), into the homogeneous differential equation. For instance, in the exercise given, this approach transforms the equation \( y'' + ay' + by = 0 \) into an algebraic equation, \( r^2 + ar + b = 0 \).

By solving this quadratic characteristic equation, you extract the roots, \( r_1 \) and \( r_2 \). These roots provide invaluable information.

• If the roots are real and distinct, the solution takes the form \( y_h = C_1 e^{r_1t} + C_2 e^{r_2t} \).
• If they are real and repeated, such as \( r_1 = r_2 \), the solution becomes \( y_h = (C_1 + C_2t)e^{r_1t} \).
• For complex roots, the solution is \( y_h = e^{\lambda t}(C_1 \cos \beta t + C_2 \sin \beta t) \), where \( \lambda + i\beta \) are the complex roots.

This diversity in solutions due to the nature of roots makes the characteristic equation a foundational element in analyzing the behaviour of differential systems.