Problem 29 - Advanced Engineering Mathematics

Question

(a) The equilibrium solutions \(y(x)=2\) and \(y(x)=-2\) satisfy the initial conditions \(y(0)=2\) and \(y(0)=-2\) respectively. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(1+c e^{4 x}\right) /\left(1-c e^{4 x}\right)\) we obtain $$1=2 \frac{1+c e}{1-c e}, \quad 1-c e=2+2 c e, \quad-1=3 c e, \quad \text { and } \quad c=-\frac{1}{3 e}$$. (b) Separating variables and integrating yields \\[ \begin{aligned} \frac{1}{4} \ln |y-2|-\frac{1}{4} \ln |y+2|+\ln c_{1} &=x \\ \ln |y-2|-\ln |y+2|+\ln c &=4 x \\ \ln \left|\frac{c(y-2)}{y+2}\right| &=4 x \\ c \frac{y-2}{y+2} &=e^{4 x}.\end{aligned}\\] Solving for \(y\) we get \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right) .\) The initial condition \(y(0)=-2\) implies \(2(c+1) /(c-1)=-2\) which yields \(c=0\) and \(y(x)=-2 .\) The initial condition \(y(0)=2\) does not correspond to a value of \(c,\) and it must simply be recognized that \(y(x)=2\) is a solution of the initial-value problem. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right)\) leads to \(c=-3 e .\) Thus, a solution of the initial-value problem is $$y=2 \frac{-3 e+e^{4 x}}{-3 e-e^{4 x}}=2 \frac{3-e^{4 x-1}}{3+e^{4 x-1}}.$$

Step-by-Step Solution

Answer

Solutions are \(y(x) = 2\), \(y(x) = -2\), and \(y = 2 \frac{3-e^{4x-1}}{3+e^{4x-1}}\).

Step 1: Understand the Given Equations

We are given two equilibrium solutions, \(y(x)=2\) and \(y(x)=-2\), with initial conditions \(y(0)=2\) and \(y(0)=-2\) respectively. We also have a formula \(y=\frac{2(1+c e^{4x})}{1-c e^{4x}}\). We are tasked with verifying an implicit solution and finding a particular \(c\) such that it satisfies the initial conditions.

Step 2: Solve for c in Part (a)

We're asked to set \(x = \frac{1}{4}\) and \(y = 1\) in \(y=\frac{2(1+c e^{4x})}{1-c e^{4x}}\). Working through the algebra: \(1 = \frac{2(1+ce)}{1-ce}\), which leads to \(1-ce = 2+2ce\) and simplifying this gives \(-1 = 3ce\), so \(c = -\frac{1}{3e}\).

Step 3: Separating Variables and Integrating in Part (b)

Separate the variables by rewriting: \(\frac{1}{4} \ln |y-2| - \frac{1}{4} \ln |y+2| + \ln c_1 = x\). This implies \(\ln \left| \frac{c(y-2)}{y+2} \right| = 4x\). Exponentiating both sides gives \(c \frac{y-2}{y+2} = e^{4x}\).

Step 4: Rearranging to Solve for y

To solve for \(y\), rearrange the equation \(c \frac{y-2}{y+2} = e^{4x}\) to get \(y = \frac{2(c+e^{4x})}{c-e^{4x}}\).

Step 5: Solving with Initial Condition y(0)=-2

Using the initial condition \(y(0) = -2\), substitute into \(2(c+1)/(c-1) = -2\). Solving this equation gives \(c = 0\), which simplifies to the solution \(y(x) = -2\).

Step 6: Verifying y(x)=2 as a Solution

Since no value of \(c\) can be found to satisfy the initial condition \(y(0) = 2\) directly using the formula, it's simply acknowledged that \(y(x) = 2\) is also an equilibrium solution by the problem's setup.

Step 7: Solving with x=1/4 and y=1

Set \(x=\frac{1}{4}\) and \(y=1\) again, this time in \(y=\frac{2(c+e^{4x})}{c-e^{4x}}\) equation. Solving leads to \(c = -3e\). The solution of the initial-value problem then becomes \(y = 2\frac{3-e^{4x-1}}{3+e^{4x-1}}\).

Key Concepts

Equilibrium Solutions

Equilibrium solutions in differential equations are special kinds of solutions where the function does not change as time progresses; that is, for a function \( y(x) \), the derivative \( \frac{dy}{dx} = 0 \) at these points. In the context of our exercise, the equilibrium solutions are \( y(x) = 2 \) and \( y(x) = -2 \).

These acts like fixed points where the system can potentially settle. They satisfy the condition of no change over time or space for the specific initial conditions, \( y(0) = 2 \) and \( y(0) = -2 \). These values are central to understanding how the system behaves, as they reveal possible steady-state solutions.
When dealing with equilibrium solutions, an important step is to verify how they perform under variations of initial conditions and external changes, such as when \( x = \frac{1}{4} \) and \( y = 1 \) in the exercise, and thereby determine specific values of \( c \).

• These solutions reflect the inherent stability of the system under given conditions.
• They are not just solutions to the differential equation but also define the long-term behavior of the system.

Initial Value Problem

An initial value problem in the context of differential equations is one where we are given a differential equation along with the values of the unknown function at a specific point. Essentially, it sets the stage to solve for the function under given circumstances. In our exercise, these initial values are \( y(0) = 2 \) and \( y(0) = -2 \).

These specific values are used to find constants, like \( c \) in our equations, which make the solution unique and valid for the initial setup. Solving an initial value problem involves finding a solution that not only fits the differential equation but also matches the initial conditions provided.

• The equation must reflect the behavior described by the initial values, ensuring the solution adheres closely to the system's initial state.
• These kinds of problems highlight the significance of the starting point and how these singular points can evolve over time.

Variable Separation

Variable separation is a technique used to solve differential equations, where we separate variables \( x \) and \( y \) on opposite sides of the equation before integrating. This makes the equation more manageable and solvable by straightforward integration. In the exercise,

we begin by rearranging the differential equation into forms like \( \frac{1}{4} \ln|y-2| - \frac{1}{4} \ln|y+2| + \ln c_1 = x \). From here, each side corresponds to a distinct variable, allowing us to process integrals independently.

• Variable separation assumes that both sides of the equation can stand independently when solved, yet correspond to a valid answer in the system's context.
• The process is effective for linear and some nonlinear equations, often serving as one of the first approaches tried when solving such equations.

Once variables are separated, integrating both sides yields a solution that tends towards a simpler form, often involving exponential functions, as seen in our overall process.

Exponential Functions

Exponential functions frequently appear in the solutions of differential equations, particularly when dealing with separation of variables. They are powerful because they describe growth and decay processes and solve linear differential equations with constant coefficients effectively. In our problem, we encounter terms like \( e^{4x} \), which appear after integration.

Exponential expressions naturally arise from integration of functions that are separable with respect to variables. They can help simplify and make sense of the otherwise complex relationship between variables on either side of a differential equation.

• Exponential functions are capable of capturing both the rapid growth or decay behavior inherent in many physical systems.
• They highlight the change rate directly, often simplifying the problem by revealing underlying patterns.

Understanding their role is crucial in interpreting solutions, as seen with constant \( c \) and its interaction with terms like \( e^{4x} \), revealing distinct solution paths for differential equations based on initial values.