Question

In each exercise, an initial value problem for a first order nonlinear system is given. Rewrite the problem as an equivalent initial value problem for a higher order nonlinear scalar differential equation. $$ \frac{d}{d t}\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right]=\left[\begin{array}{c} y_{2} \\ y_{2} \tan \left(y_{1}\right)+e^{y_{2}} \end{array}\right], \quad\left[\begin{array}{l} y_{1}(0) \\ y_{2}(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \end{array}\right] $$

Short answer

Question: Rewrite the following initial value problem for a first-order nonlinear system with two equations as an equivalent initial value problem for a higher-order nonlinear scalar differential equation: $$ \frac{d}{dt}\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right]=\left[\begin{array}{c} y_{2} \\ y_{2} \tan(y_{1})+e^{y_{2}} \end{array}\right], \quad \left[\begin{array}{l} y_{1}(0) \\ y_{2}(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \end{array}\right] $$ Answer: The equivalent initial value problem for a higher-order nonlinear scalar differential equation is: $$ \frac{d^2y_1}{dt^2} = y_2 \tan(y_1) + e^{y_2}, $$ with \(y_1(0) = 0\) and \(\frac{dy_1}{dt}(0) = 1\).

Step-by-step solution

Identify the given system of equations

The given system of equations and initial conditions are: $$ \frac{d}{dt}\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right]=\left[\begin{array}{c} y_{2} \\ y_{2} \tan(y_{1})+e^{y_{2}} \end{array}\right], \quad \left[\begin{array}{l} y_{1}(0) \\ y_{2}(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \end{array}\right] $$ Breaking them down, we get: $$ \frac{dy_1}{dt} = y_2 $$ $$ \frac{dy_2}{dt} = y_2 \tan(y_1) + e^{y_2} $$ with initial conditions \(y_1(0) = 0\) and \(y_2(0) = 1\).

Convert the system of first-order equations into a single second-order equation

To do this, we can rewrite the first equation and differentiate it again with respect to \(t\): $$ \frac{d^2y_1}{dt^2} = \frac{d}{dt}\left(\frac{dy_1}{dt}\right) = \frac{d}{dt}(y_2) $$ Now, replace \(\frac{d}{dt}(y_2)\) with \(\frac{dy_2}{dt}\) from the second equation: $$ \frac{d^2y_1}{dt^2} = y_2 \tan(y_1) + e^{y_2} $$

Substitute the expressions for the initial conditions

Since \(y_1(0) = 0\) and \(y_2(0) = 1\), we can make the following substitutions: - \(y_1(0) = 0\): \((t=0) \implies y_1 = 0\) - \(y_2(0) = 1\): \((t=0) \implies y_2 = 1\) Using these substitutions, we get the initial conditions for the second-order equation: - \(\frac{dy_1}{dt}(0) = y_2(0) = 1\) for \(t = 0\)

Write the resulting single second-order equation with its initial condition

We now have a single second-order nonlinear scalar differential equation: $$ \frac{d^2y_1}{dt^2} = y_2 \tan(y_1) + e^{y_2} $$ with the initial condition \(\frac{dy_1}{dt}(0) = 1\) and \(y_1(0) = 0\). The resulting initial value problem for a higher-order nonlinear scalar differential equation is: $$ \frac{d^2y_1}{dt^2} = y_2 \tan(y_1) + e^{y_2}, $$ with \(y_1(0) = 0\) and \(\frac{dy_1}{dt}(0) = 1\).

Key concepts

First Order Systems

A first order system in the context of differential equations involves equations where the highest derivative is the first one. In this problem, we encounter a nonlinear first order system, given by:

• \( \frac{dy_1}{dt} = y_2 \)
• \( \frac{dy_2}{dt} = y_2 \tan(y_1) + e^{y_2} \)

Nonlinear Nature: Unlike linear differential equations, nonlinear equations can often describe more complex behaviors, such as chaotic systems or oscillations. The presence of terms like \( y_2 \tan(y_1) \) indicates the nonlinear nature, as these terms multiply the unknown functions together or involve nonlinear functions of them.
Representation as a System: The system is expressed in vector form, where each equation represents a component of the system's state. This enables tackling multiple interdependent quantities and is often useful in disciplines that require understanding the dynamics of systems, like engineering or physics.

Initial Value Problems

An initial value problem (IVP) in differential equations is a problem where the solution is sought subject to certain initial conditions. These initial conditions specify the values of the solution at a given point, typically when time \( t = 0 \). In our exercise, the initial conditions provided are:

• \( y_1(0) = 0 \)
• \( y_2(0) = 1 \)

Importance of Initial Conditions: Initial conditions are crucial as they determine a unique solution from a potentially infinite set of solutions. They provide the 'starting point' for solving equations, allowing us to predict how the system evolves over time.
Solving IVPs: The process involves integrating the differential equations while simultaneously considering the provided initial data. In a first-order system, it ensures that the solutions match physical constraints or specific scenarios described by these initial points.

Second Order Differential Equations

These are equations involving the second derivative of the unknown function, providing more information about the system's acceleration rather than just velocity or position. The transformation from first-order systems to a second-order differential equation in the solution steps allows simplifying multiple equations into one. In this context, we derived the second-order equation:

• \( \frac{d^2y_1}{dt^2} = y_2 \tan(y_1) + e^{y_2} \)

Why Convert? Second order equations can be simpler to analyze and solve, particularly when attempting to understand the fundamental nature of a system's behavior. They often arise in physics, especially in the study of mechanical vibrations or circuits.
Initial Conditions for Second Order: The initial conditions need adjustments to fit the new equation's terms. Because we derived this from a first-order system, conditions like \( y_1(0) = 0 \) and \( \frac{dy_1}{dt}(0) = 1 \) ensure solutions are correctly anchored and relevant to real-world applications, maintaining consistency with the original system.