Problem 41 - Advanced Engineering Mathematics

Question

In Problems 35-46, find the general solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=z \\ &\frac{d y}{d t}=-z \\ &\frac{d z}{d t}=y \end{aligned} $$

Step-by-Step Solution

Answer

General solutions: \( x(t) = c_1 + c_2 \cos(t) + c_3 \sin(t) \), \( y(t) = c_2 \sin(t) - c_3 \cos(t) \), \( z(t) = -c_2 \sin(t) + c_3 \cos(t) \).

Step 1: Identify the system of equations

The given system is a set of three first-order differential equations:1. \( \frac{dx}{dt} = z \)2. \( \frac{dy}{dt} = -z \)3. \( \frac{dz}{dt} = y \).

Step 2: Differentiate and substitute to eliminate variables

Differentiate the first equation \( \frac{dx}{dt} = z \) with respect to \( t \):\[ \frac{d^2x}{dt^2} = \frac{dz}{dt} \]Substitute \( \frac{dz}{dt} = y \) from the third equation:\[ \frac{d^2x}{dt^2} = y \]

Step 3: Relate second derivatives to y

Differentiate the substituted equation \( \frac{d^2x}{dt^2} = y \) with respect to \( t \) to find another relation:\[ \frac{d^3x}{dt^3} = \frac{dy}{dt} \].

Step 4: Substitute for \( \frac{dy}{dt} \) and simplify

Using \( \frac{dy}{dt} = -z \) from the original system, substitute into \( \frac{d^3x}{dt^3} \): \[ \frac{d^3x}{dt^3} = -z \]Substituting the first equation \( z = \frac{dx}{dt} \) gives:\[ \frac{d^3x}{dt^3} = -\frac{dx}{dt} \].

Step 5: Solve the characteristic equation

Consider \( \frac{d^3x}{dt^3} = -\frac{dx}{dt} \). Assume a solution of the form \( x(t) = e^{\lambda t} \). This leads to:\[ \lambda^3 e^{\lambda t} = -\lambda e^{\lambda t} \]\[ \lambda^3 + \lambda = 0 \]\[ \lambda(\lambda^2 + 1) = 0 \].

Step 6: Find the roots of the characteristic equation

Solve \( \lambda(\lambda^2 + 1) = 0 \):1. \( \lambda = 0 \)2. \( \lambda^2 = -1 \), so \( \lambda = i \) or \( \lambda = -i \).

Step 7: Write the general solution for \( x(t) \)

The general solution for \( x(t) \) given the characteristic roots is:\[ x(t) = c_1 + c_2 \cos(t) + c_3 \sin(t) \],where \( c_1, c_2, c_3 \) are constants.

Step 8: Find the solutions for \( y(t) \) and \( z(t) \)

Using the relationships \( \frac{dy}{dt} = -z \) and \( \frac{dz}{dt} = y \), and substituting the expressions for \( \cos(t) \) and \( \sin(t) \) from Step 7:\[ y(t) = c_2 \sin(t) - c_3 \cos(t) \]\[ z(t) = -c_2 \sin(t) + c_3 \cos(t) \].

Key Concepts

First-order differential equations

A first-order differential equation is one where the highest derivative of the function is the first derivative. In our example, we have three first-order differential equations:

• \( \frac{dx}{dt} = z \)
• \( \frac{dy}{dt} = -z \)
• \( \frac{dz}{dt} = y \)

The primary goal in these equations is to express the rate of change of one variable in terms of another. In practice, solving such equations allows us to understand how these variables evolve over time.
These equations form a system that can be solved simultaneously to uncover relationships between variables \( x, y, \) and \( z \) that hold true as they change.

Characteristic equation

The characteristic equation provides a crucial step in solving differential equations. It arises when you assume a solution has an exponential form, such as \( x(t) = e^{\lambda t} \). This leads to substituting and simplifying derivatives in terms of \( \lambda \).

In our exercise, the characteristic equation is derived by assuming a solution of the form \( x(t) = e^{\lambda t} \). Substituting into the original higher order equation \( \frac{d^3x}{dt^3} = -\frac{dx}{dt} \) results in the characteristic equation:
\[ \lambda^3 + \lambda = 0 \]
By factoring, \( \lambda(\lambda^2 + 1) = 0 \), we find solutions such as \( \lambda = 0, i, -i \). These roots are central in deciding the type of solution the differential equation will have, involving terms with exponential, sine, and cosine functions.

General solution

The general solution of a differential equation provides a formula that encompasses all possible solutions, using arbitrary constants. This allows the flexibility needed to meet any initial or boundary conditions the problem might have.

• For the root \( \lambda = 0 \), the solution is a constant term \( c_1 \).
• For the roots \( \lambda = i \) and \( \lambda = -i \), the solution involves oscillatory functions, specifically \( c_2 \cos(t) \) and \( c_3 \sin(t) \).

Thus, the general solution for \( x(t) \) in this problem is:\[ x(t) = c_1 + c_2 \cos(t) + c_3 \sin(t) \]This form acknowledges all potential behaviors of the system over time.
Subsequently, using the relationships found between the variables, we determine:

• \( y(t) = c_2 \sin(t) - c_3 \cos(t) \)
• \( z(t) = -c_2 \sin(t) + c_3 \cos(t) \)

These solutions fully represent the dynamic evolution of \( x, y, \) and \( z \). By understanding the general solution, one can predict the future state of the system based on initial conditions.