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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

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.

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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.

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Most popular questions from this chapter

The method of successive approximations (see Section \(2.8)\) can also be applied to systems of equations. For example, consider the initial value problem $$ \mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}, \quad \mathbf{x}(0)=\mathbf{x}^{0} $$ where \(\mathbf{A}\) is a constant matrix and \(\mathbf{x}^{0}\) a prescribed vector. (a) Assuming that a solution \(\mathbf{x}=\Phi(t)\) exists, show that it must satisfy the integral equation $$ \Phi(t)=\mathbf{x}^{0}+\int_{0}^{t} \mathbf{A} \phi(s) d s $$ (b) Start with the initial approximation \(\Phi^{(0)}(t)=\mathbf{x}^{0} .\) Substitute this expression for \(\Phi(s)\) in the right side of Eq. (ii) and obtain a new approximation \(\Phi^{(1)}(t) .\) Show that $$ \phi^{(1)}(t)=(1+\mathbf{A} t) \mathbf{x}^{0} $$ (c) Reppeat this process and thereby obtain a sequence of approximations \(\phi^{(0)}, \phi^{(1)}\), \(\phi^{(2)}, \ldots, \phi^{(n)}, \ldots\) Use an inductive argument to show that $$ \phi^{(n)}(t)=\left(1+A t+A^{2} \frac{2}{2 !}+\cdots+A^{x} \frac{r^{2}}{n !}\right) x^{0} $$ (d) Let \(n \rightarrow \infty\) and show that the solution of the initial value problem (i) is $$ \phi(t)=\exp (\mathbf{A} t) \mathbf{x}^{0} $$

Consider the initial value problem $$ x^{\prime}=A x+g(t), \quad x(0)=x^{0} $$ (a) By referring to Problem \(15(c)\) in Section \(7.7,\) show that $$ x=\Phi(t) x^{0}+\int_{0}^{t} \Phi(t-s) g(s) d s $$ (b) Show also that $$ x=\exp (A t) x^{0}+\int_{0}^{t} \exp [\mathbf{A}(t-s)] \mathbf{g}(s) d s $$ Compare these results with those of Problem 27 in Section \(3.7 .\)

In each of Problems 1 through 8 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 \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{3} & {-2} \\ {4} & {-1}\end{array}\right) \mathbf{x} $$

Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{-1} & {-1} \\ {-\alpha} & {-1}\end{array}\right) \mathbf{x} $$ $$ \begin{array}{l}{\text { (a) Solve the system for } \alpha=0.5 \text { . What are the eigennalues of the coefficient mattix? }} \\ {\text { Classifith the equilitrium point a the natare the cigemalues of the coefficient matrix? Classify }} \\ {\text { the equilithessm for } \alpha \text { . What as the cigemalluce of the coefficient matrix Classify }} \\ {\text { the equilibrium poin at the oigin as to the styse. ematitue different types of behwior. }} \\\ {\text { (c) the parts (a) and (b) solutions of thesystem exhibit two quite different ypes of behwior. }}\end{array} $$ $$ \begin{array}{l}{\text { Find the eigenvalues of the coefficient matrix in terms of } \alpha \text { and determine the value of } \alpha} \\ {\text { between } 0.5 \text { and } 2 \text { where the transition from one type of behavior to the other occurs. This }} \\ {\text { critical value of } \alpha \text { is called a bifurcation point. }}\end{array} $$ $$ \begin{array}{l}{\text { Electric Circuits. Problems } 32 \text { and } 33 \text { are concerned with the clectric circuit described by the }} \\ {\text { system of differential equations in Problem } 20 \text { of Section } 7.1 \text { : }}\end{array} $$ $$ \frac{d}{d t}\left(\begin{array}{l}{l} \\\ {V}\end{array}\right)=\left(\begin{array}{cc}{-\frac{R_{1}}{L}} & {-\frac{1}{L}} \\ {\frac{1}{C}} & {-\frac{1}{C R_{2}}}\end{array}\right)\left(\begin{array}{l}{I} \\ {V}\end{array}\right) $$

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{2} & {-1} \\ {3} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{1} \\\ {-1}\end{array}\right) e^{t} $$

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