Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 7: Problem 5

Show that the general solution of \(\mathbf{x}^{\prime}=\mathbf{P}(t) \mathbf{x}+\mathbf{g}(t)\) is the sum of any particular solution \(\mathbf{x}^{(p)}\) of this equation and the general solution \(\mathbf{x}^{(i)}\) of the corresponding homogeneous equation.

Short Answer

Expert verified
Answer: The general solution of a nonhomogeneous system of linear differential equations is the sum of any particular solution and the general solution of the corresponding homogeneous system.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards
Start your free trial Skip

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Write down the nonhomogeneous and homogeneous systems

We are given the nonhomogeneous system of linear differential equations: \(\mathbf{x}' = \mathbf{P}(t) \mathbf{x} + \mathbf{g}(t)\) The corresponding homogeneous system is obtained by setting \(\mathbf{g}(t) = 0\): \(\mathbf{x}' = \mathbf{P}(t) \mathbf{x}\)
02

Represent the general solution of the nonhomogeneous system

Let \(\mathbf{x}^{(p)}(t)\) be any particular solution of the nonhomogeneous system, and let \(\mathbf{x}(t)\) represent the general solution of the nonhomogeneous system. Then, we can write the general solution as: \(\mathbf{x}(t) = \mathbf{x}^{(p)}(t) + \mathbf{x}^{(i)}(t)\) where \(\mathbf{x}^{(i)}(t)\) is the general solution of the corresponding homogeneous system.
03

Write down the differential equations for \(\mathbf{x}^{(p)}\) and \(\mathbf{x}^{(i)}\)

Next, we need to find the differential equations satisfied by \(\mathbf{x}^{(p)}(t)\) and \(\mathbf{x}^{(i)}(t)\). Since \(\mathbf{x}^{(p)}(t)\) is a particular solution of the nonhomogeneous system, it satisfies: \(\mathbf{x}^{(p)'} = \mathbf{P}(t) \mathbf{x}^{(p)} + \mathbf{g}(t)\) Since \(\mathbf{x}^{(i)}(t)\) is the general solution of the homogeneous system, it satisfies: \(\mathbf{x}^{(i)'} = \mathbf{P}(t) \mathbf{x}^{(i)}\)
04

Show that the sum of \(\mathbf{x}^{(p)}\) and \(\mathbf{x}^{(i)}\) satisfies the nonhomogeneous equation

Now, we need to show that the sum of any particular solution \(\mathbf{x}^{(p)}\) and the general solution \(\mathbf{x}^{(i)}\) of the homogeneous equation satisfies the nonhomogeneous equation. To do this, let's differentiate the sum \(\mathbf{x}^{(p)} + \mathbf{x}^{(i)}\) with respect to \(t\): \((\mathbf{x}^{(p)} + \mathbf{x}^{(i)})' = \mathbf{x}^{(p)'} + \mathbf{x}^{(i)'}\) From Steps 3, we have the differential equations satisfied by \(\mathbf{x}^{(p)}\) and \(\mathbf{x}^{(i)}\). Substituting these into the above expression, we get: \((\mathbf{x}^{(p)'} + \mathbf{x}^{(i)'}) = (\mathbf{P}(t) \mathbf{x}^{(p)} + \mathbf{g}(t)) + (\mathbf{P}(t) \mathbf{x}^{(i)})\) Combining the terms gives: \((\mathbf{x}^{(p)'} + \mathbf{x}^{(i)'}) = \mathbf{P}(t)(\mathbf{x}^{(p)} + \mathbf{x}^{(i)}) + \mathbf{g}(t)\) Since by the definition above, \(\mathbf{x}(t) = \mathbf{x}^{(p)}(t) + \mathbf{x}^{(i)}(t)\), we can write the above expression as: \(\mathbf{x}' = \mathbf{P}(t) \mathbf{x} + \mathbf{g}(t)\) Therefore, we have shown that the general solution of the nonhomogeneous system is the sum of a particular solution of the nonhomogeneous system and the general solution of the homogeneous system.

Key Concepts

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

Differential Equations
A differential equation is an equation that describes a relationship between a function and its derivatives. It serves as a cornerstone in mathematics, being used to model various phenomena such as growth rates, movements, and changes across disciplines like physics, engineering, biology, and economics.

Differential equations can be divided into several types. One primary categorization is based on linearity: linear differential equations maintain a linear relationship between the function and its derivatives, while nonlinear equations do not. Linearity makes the equations more manageable, as they satisfy superposition principles, allowing us to combine solutions in specific ways.
Homogeneous Linear Differential Equations
Homogeneous linear differential equations are a specific kind of differential equation. They are characterized by the absence of a stand-alone term, that is, the term without the dependent variable or its derivatives (often represented as g(t) in nonhomogeneous equations).

In formal terms, a homogeneous linear differential equation has a form where the zero function is a solution. This brings us to an important property of homogeneous equations - any linear combination of solutions to a homogeneous linear differential equation is also a solution. This fact underpins why we can talk about the general solution to these equations.
Particular Solution
In contrast, a particular solution to a differential equation pertains to one specific instance of the infinite set of solutions. For nonhomogeneous linear differential equations, which include an extra non-zero function (like g(t)), the particular solution satisfies the entire nonhomogeneous equation, not just the homogeneous part.

This particular solution is determined by the specific initial conditions or boundary conditions of the problem. It's important to note that while there are infinitely many solutions to the homogeneous equation, there's typically only one particular solution for a given set of initial conditions in a nonhomogeneous equation.
General Solution
The general solution of a differential equation is the most comprehensive expression that contains all possible particular solutions of the equation. For linear differential equations, the general solution can be expressed as the sum of any particular solution to the nonhomogeneous equation and the general solution to the corresponding homogeneous equation.

This duality of the general solution — incorporating both the homogeneous and nonhomogeneous parts — is instrumental for understanding how to approach problems involving differential equations. It illustrates how we can use the structure and properties of these equations to determine a broad class of solutions before honing in on a single, particular solution that meets specific criteria.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The clectric circuit shown in Figure 7.9 .1 is described by the system of differential equations $$ \frac{d \mathbf{x}}{d t}=\left(\begin{array}{cc}{-\frac{1}{2}} & {-\frac{1}{8}} \\ {2} & {-\frac{1}{2}}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{\frac{1}{2}} \\ {0}\end{array}\right) I(t) $$ where \(x_{1}\) is the current through the inductor, \(x_{2}\) is the voltage drop across the capacitor, and \(I(t)\) is the current supplied by the external source. (a) Determine a fundamental matrix \(\Psi(t)\) for the homogeneous system corresponding to Eq. (i). Refer to Problem 25 of Section \(7.6 .\) (b) If \(I(t)=e^{-t / 2}\), determine the solution of the system (i) that also satisfies the initial conditions \(\mathbf{x}(0)=0\).

Let \(\Phi(t)\) denote the fundamental matrix satisfying \(\Phi^{\prime}=A \Phi, \Phi(0)=L\) In the text we also denoted this matrix by \(\exp (A t)\), In this problem we show that \(\Phi\) does indeed have the principal algebraic properties associated with the exponential function. (a) Show that \(\Phi(t) \Phi(s)=\Phi(t+s) ;\) that is, \(\exp (\hat{\mathbf{A}} t) \exp (\mathbf{A} s)=\exp [\mathbf{A}(t+s)]\) Hint: Show that if \(s\) is fixed and \(t\) is variable, then both \(\Phi(t) \Phi(s)\) and \(\Phi(t+s)\) satisfy the initial value problem \(\mathbf{Z}^{\prime}=\mathbf{A} \mathbf{Z}, \mathbf{Z}(0)=\mathbf{\Phi}(s)\) (b) Show that \(\Phi(t) \Phi(-t)=\mathbf{I}\); that is, exp(At) \(\exp [\mathbf{A}(-t)]=\mathbf{1}\). Then show that \(\Phi(-t)=\) \(\mathbf{\Phi}^{-1}(t) .\) (c) Show that \(\mathbf{\Phi}(t-s)=\mathbf{\Phi}(t) \mathbf{\Phi}^{-1}(s)\)

Verify that the given vector is the general solution of the corresponding homogeneous system, and then solve the non-homogeneous system. Assume that \(t>0 .\) $$ t \mathrm{x}^{\prime}=\left(\begin{array}{cc}{3} & {-2} \\ {2} & {-2}\end{array}\right) \mathrm{x}+\left(\begin{array}{c}{-2 t} \\\ {t^{4}-1}\end{array}\right), \quad \mathbf{x}^{(c)}=c_{1}\left(\begin{array}{c}{1} \\ {2}\end{array}\right) t^{-1}+c_{2}\left(\begin{array}{c}{2} \\ {1}\end{array}\right) t^{2} $$

Consider the system $$ \mathbf{x}^{\prime}=\mathbf{A x}=\left(\begin{array}{rrr}{5} & {-3} & {-2} \\\ {8} & {-5} & {-2} \\ {-4} & {-5} & {-4} \\ {-4} & {3} & {3}\end{array}\right) \mathbf{x} $$ (a) Show that \(r=1\) is a triple eigenvalue of the coefficient matrix \(\mathbf{A},\) and that there are only two linearly independent eigenvectors, which we may take as $$ \xi^{(1)}=\left(\begin{array}{l}{1} \\ {0} \\ {2}\end{array}\right), \quad \xi^{(2)}=\left(\begin{array}{r}{0} \\ {2} \\ {-3}\end{array}\right) $$ Find two linearly independent solutions \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) of Eq. (i). (b) To find a third solution assume that \(\mathbf{x}=\xi t e^{t}+\mathbf{\eta} e^{\lambda} ;\) thow that \(\xi\) and \(\eta\) must satisfy $$ (\mathbf{A}-\mathbf{1}) \xi=0 $$ \((\mathbf{A}-\mathbf{I}) \mathbf{\eta}=\mathbf{\xi}\) (c) Show that \(\xi=c_{1} \xi^{(1)}+c_{2} \mathbf{\xi}^{(2)},\) where \(c_{1}\) and \(c_{2}\) are arbitrary constants, is the most general solution of Eq. (iii). Show that in order to solve Eq. (iv) it is necessary that \(c_{1}=c_{2}\) (d) It is convenient to choose \(c_{1}=c_{2}=2 .\) For this choice show that $$ \xi=\left(\begin{array}{r}{2} \\ {4} \\ {-2}\end{array}\right), \quad \mathbf{\eta}=\left(\begin{array}{r}{0} \\ {0} \\ {-1}\end{array}\right) $$ where we have dropped the multiples of \(\xi^{(1)}\) and \(\xi^{(2)}\) that appear in \(\eta\). Use the results given in Eqs. (v) to find a third linearly independent solution \(\mathbf{x}^{(3)}\) of Eq. (i). (e) Write down a fundamental matrix \(\Psi(t)\) for the system (i). (f) Form a matrix T with the cigenvector \(\xi^{(1)}\) in the first column and with the eigenvector \(\xi\) and the generalized eigenvector \(\eta\) from Eqs. (v) in the other two columns. Find \(\mathbf{T}^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A} .\)

Express the general solution of the given system of equations in terms of real-valued functions. In each of Problems 1 through 6 also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-\frac{3}{2}} \\\ {\frac{9}{5}} & {-1}\end{array}\right) \mathbf{x} $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free