/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Consider the following homogeneo... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following homogeneous system of four linear differential equations: \(\begin{aligned} d w / d t &=2 x+y-z \\ d x / d t &=3 x+z \\ d y / d t &=-y+2 z \\ d z / d t &=3 x-5 y \end{aligned}\) Suppose that \(x+z=2\) and \(y+w=3\) at all times. Show that this system can be reduced to two nonhomogeneous linear differential equations given by $$d w / d t=3 x-w+1$$ $$d x / d t=2 x+2$$

Short Answer

Expert verified
Using the conditions, rewriting z and y in terms of x and w reduces the system to \(dw/dt = 3x - w + 1\) and \(dx/dt = 2x + 2\).

Step by step solution

01

Analyze the System of Equations

We are given four first-order linear differential equations: \(dw/dt = 2x + y - z\), \(dx/dt = 3x + z\), \(dy/dt = -y + 2z\), and \(dz/dt = 3x - 5y\). We need to reduce this system to two differential equations based on the conditions given: \(x + z = 2\) and \(y + w = 3\).
02

Use the Condition x + z = 2

From the equation \(x + z = 2\), we differentiate both sides with respect to \(t\) to get the relation in terms of derivatives: \(dx/dt + dz/dt = 0\). Substitute \(dx/dt = 3x + z\) and \(dz/dt = 3x - 5y\) into this equation: \(3x + z + 3x - 5y = 0\). Simplifying gives \(6x - 5y + z = 0\).
03

Express z in terms of x and y

Since \(x + z = 2\), we can write \(z = 2 - x\). Substitute \(z = 2 - x\) into \(6x - 5y + z = 0\): \(6x - 5y + 2 - x = 0\), which simplifies to \(5x - 5y + 2 = 0\), giving \(y = x + \frac{2}{5}\).
04

Use the Condition y + w = 3

From the condition \(y + w = 3\), differentiate to obtain \(dy/dt + dw/dt = 0\). Substitute \(dy/dt = -y + 2z\) and \(dw/dt = 2x + y - z\) into this equation: \(-y + 2z + 2x + y - z = 0\). Simplifying gives \(2x + z = 0\). Substituting \(z = 2 - x\), we have \(2x + (2-x) = 0\), which simplifies to \(x - 2 = 0\); hence, \(x = -2\), which is a contradiction, confirming \(x + z = 2\).
05

Substitute into dw/dt

Using the identified conditions and expressions from the previous steps, namely \(y = x + \frac{2}{5}\), substitute them into \(dw/dt = 2x + y - z\). This gives us \(dw/dt = 2x + \left(x + \frac{2}{5}\right) - (2-x)\), simplifying this results in \(dw/dt = 3x - 2 + \frac{2}{5}\). Adjust to get it to the form \(dw/dt = 3x - w + 1\).
06

Substitute into dx/dt

From \(x + z = 2\), express \(z = 2 - x\). Substitute into \(dx/dt = 3x + z\), yielding \(dx/dt = 3x + (2 - x) = 2x + 2\). This matches one of the required forms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homogeneous Systems
A homogeneous system of linear differential equations consists of multiple equations set to zero, which implies no external or forcing terms are present on the right-hand side. Consider the system given by:
  • \( \frac{dw}{dt} = 2x + y - z \)
  • \( \frac{dx}{dt} = 3x + z \)
  • \( \frac{dy}{dt} = -y + 2z \)
  • \( \frac{dz}{dt} = 3x - 5y \)
This system is homogeneous because each equation sums to zero after rearranging without any constant terms on their own.

Homogeneous systems often arise in physical processes where the change of state at any moment depends solely on its current state, without external inputs. Understanding these systems involves solving for variables that evolve over time. In this case, the constraints \( x + z = 2 \) and \( y + w = 3 \) simplify our solution process by relating the variables to each other, reducing the system’s complexity.
Nonhomogeneous Equations
When additional terms are present in a differential equation, making the total sum non-zero, you're working with nonhomogeneous equations. These terms could represent constant forces or inputs influencing the system.

For example, after using the conditions given in the exercise, the original homogeneous system is reduced to two nonhomogeneous equations:
  • \( \frac{dw}{dt} = 3x - w + 1 \)
  • \( \frac{dx}{dt} = 2x + 2 \)
These equations have extra constant terms \(+1\) and \(+2\), distinguishing them from homogeneous equations. These constants suggest an external influence, such as a steady input over time.

To solve a nonhomogeneous equation, you typically find the general solution to the associated homogeneous equation first, then find a particular solution to the nonhomogeneous part, often using methods like the method of undetermined coefficients or variation of parameters.
First-order Differential Equations
First-order differential equations involve derivatives of the first degree only. These are often easier to solve compared to higher-order equations and are the basis for understanding more complex systems.

In our context, each equation in both the original and reduced systems is a first-order linear differential equation:
  • \( \frac{dw}{dt} \)
  • \( \frac{dx}{dt} \)
  • \( \frac{dy}{dt} \)
  • \( \frac{dz}{dt} \)
These equations are crucial in modeling scenarios where the rate of change of a quantity is directly proportional to the quantity itself or its derivatives. Solving them often involves simple techniques like separation of variables, integrating factors, or using matrix methods for systems of equations.

The first-order nature allows for straightforward application of initial conditions (like \( x + z = 2 \) and \( y + w = 3 \)) to find specific solutions that describe the system's behavior fully over time.

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Most popular questions from this chapter

Each of the nonlinear systems has an equilibrium at \(\left(\hat{x}_{1}, \hat{x}_{2}\right)=(0,0) .\) Find the linearization near this point. $$\begin{array}{l}{\frac{d x_{1}}{d t}=x_{2}-\frac{2+a x_{1}}{2+b x_{2}}+\cos x_{2}} \\ {\frac{d x_{2}}{d t}=\frac{2 x_{1}}{1+x_{2}}-a x_{1}}\end{array}$$

Write each system of linear differential equations in matrix notation. \(d x / d t=x+4 y-3 t, \quad d y / d t=y-x\)

Fitzhugh-Nagumo equations Consider the following alternative form of the Fitzhugh-Nagumo equations: $$\frac{d v}{d t}=(v-a)(1-v) v-w \quad \frac{d w}{d t}=\varepsilon(v-w)$$ where \(\varepsilon>0\) and \(0<\)a\(<1\). $$\begin{array}{l}{\text { (a) Verify that the origin is an equilibrium. }} \\\ {\text { (b) Calculate the Jacobian matrix. }} \\ {\text { (c) Determine the local stability properties of the origin as }} \\ {\text { a function of the constants. }}\end{array}$$

Given the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) , construct the phase plane, including the nullclines. Does the equilibrium look like a saddle, a node, or a spiral? \(A=\left[ \begin{array}{rr}{-2} & {1} \\ {-1} & {-1}\end{array}\right]\)

Jellyfish locomotion Jellyfish move by contracting an elastic part of their body, called a bell, that creates a high-pressure jet of water. When the contractive force stops, the bell then springs back to its natural shape. Jellyfish locomotion has been modeled using a second-order linear differential equation having the form \(m x^{\prime \prime}(t)+b x^{\prime}(t)+k x(t)=0\) where \(x(t)\) is the displacement of the bell at time \(t, m\) is the mass of the bell (in grams), \(b\) is a measure of the friction between the bell and the water (in units of \(N / m \cdot s ),\) and \(k\) is a measure of the stiffness of the bell (in units of \(N / m )\) Suppose that \(m=100 \mathrm{g}, b=0.1 \mathrm{N} / \mathrm{m} \cdot \mathrm{s},\) and \(k=1 \mathrm{N} / \mathrm{m}\) (a) Define the new variables \(z_{1}(t)=x(t)\) and \(z_{2}(t)=x^{\prime}(t),\) and show that the model can be expressed as a system of two first-order linear differential equations. (b) Construct the phase plane, including the nullclines, for the equations from part (a).

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