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Chapter 7: Problem 16

Let \(\mathbf{x}=\Phi(t)\) be the general solution of \(\mathbf{x}^{\prime}=\mathbf{P}(t) \mathbf{x}+\mathbf{g}(t),\) and let \(\mathbf{x}=\mathbf{v}(t)\) be some particular solution of the same system. By considering the difference \(\boldsymbol{\phi}(t)-\mathbf{v}(t),\) show that \(\Phi(t)=\mathbf{u}(t)+\mathbf{v}(t),\) where \(\mathbf{u}(t)\) is the general solution of the homogeneous system \(\mathbf{x}^{\prime}=\mathbf{P}(t) \mathbf{x} .\)

Short Answer

Expert verified
Question: Show that the general solution of a non-homogeneous linear differential equation system can be expressed as the sum of a particular solution and the general solution of the corresponding homogeneous system. Answer: If \(\Phi(t)\) is the general solution of the non-homogeneous system and \(\mathbf{v}(t)\) is a particular solution, then their difference, \(\mathbf{u}(t) = \Phi(t) - \mathbf{v}(t),\) satisfies the homogeneous system \(\mathbf{u}^{\prime}(t) = \mathbf{P}(t) \mathbf{u}(t).\) Therefore, the general solution of the non-homogeneous system can be expressed as the sum of the general solution of the homogeneous system and a particular solution: \(\Phi(t) = \mathbf{u}(t) + \mathbf{v}(t).\)

Step by step solution

01

Define the Difference

First, let's define the difference between the general solution, \(\mathbf{x}=\Phi(t)\), and the particular solution, \(\mathbf{x}=\mathbf{v}(t)\), as \(\mathbf{u}(t) = \Phi(t) - \mathbf{v}(t).\)
02

Differentiate the Difference with Respect to t

Next, we shall find the derivative of \(\mathbf{u}(t)\) with respect to t. To do this, differentiate both \(\Phi(t)\) and \(\mathbf{v}(t)\) with respect to t and then subtract: \(\mathbf{u}^{\prime}(t) = \Phi^{\prime}(t) - \mathbf{v}^{\prime}(t).\)
03

Substituting Known Differential Equations for \(\Phi^{\prime}(t)\) and \(\mathbf{v}^{\prime}(t)\)

Now we can substitute the given differential equation system into our expression for \(\mathbf{u}^{\prime}(t)\). We know that \(\Phi^{\prime}(t) = \mathbf{P}(t) \Phi(t) + \mathbf{g}(t),\) and \(\mathbf{v}^{\prime}(t) = \mathbf{P}(t) \mathbf{v}(t) + \mathbf{g}(t).\) Substituting these expressions into the expression for \(\mathbf{u}^{\prime}(t)\), we get: \(\mathbf{u}^{\prime}(t) = (\mathbf{P}(t) \Phi(t) + \mathbf{g}(t)) - (\mathbf{P}(t) \mathbf{v}(t) + \mathbf{g}(t)).\)
04

Simplifying the Expression for \(\mathbf{u}^{\prime}(t)\)

Simplifying the expression for \(\mathbf{u}^{\prime}(t)\) gives us: \(\mathbf{u}^{\prime}(t) = \mathbf{P}(t) \Phi(t) - \mathbf{P}(t) \mathbf{v}(t).\) Next, factor out \(\mathbf{P}(t)\): \(\mathbf{u}^{\prime}(t) = \mathbf{P}(t) (\Phi(t) - \mathbf{v}(t)).\)
05

Connecting \(\mathbf{u}^{\prime}(t)\) and \(\mathbf{u}(t)\)

Recall that we previously defined \(\mathbf{u}(t) = \Phi(t) - \mathbf{v}(t)\). So we can rewrite the expression for \(\mathbf{u}^{\prime}(t)\) as: \(\mathbf{u}^{\prime}(t) = \mathbf{P}(t) \mathbf{u}(t).\) Notice that this differential equation represents the general solution to the homogeneous system.
06

Concluding the Proof

To sum up, we've shown that if \(\Phi(t)\) is the general solution of the non-homogeneous system and \(\mathbf{v}(t)\) is a particular solution, then their difference, \(\mathbf{u}(t) = \Phi(t) - \mathbf{v}(t),\) satisfies the homogeneous system \(\mathbf{u}^{\prime}(t) = \mathbf{P}(t) \mathbf{u}(t).\) Therefore, the general solution of the non-homogeneous system can be expressed as the sum of the general solution of the homogeneous system and a particular solution: \(\Phi(t) = \mathbf{u}(t) + \mathbf{v}(t).\)

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Key Concepts

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

Understanding the General Solution
The general solution is an overarching concept when dealing with differential equations, particularly non-homogeneous ones. Imagine non-homogeneous differential equations as a blend of two parts - the characteristics from the system itself and the influences from external factors. The general solution captures both these elements.
  • For a given system such as \[ \mathbf{x}' = \mathbf{P}(t) \mathbf{x} + \mathbf{g}(t), \] the general solution, denoted as \( \Phi(t) \), incorporates the full essence of this equation by including all possible particular solutions that satisfy the equation.
  • The notion of the general solution is critical because it provides a complete set of possible behaviors of the system.
  • What's unique is that \( \Phi(t) \) represents not just one solution but a family of solutions; this means we are not missing any essential characteristic of the system in question.
Ultimately, the general solution of a non-homogeneous equation consists of a particular solution that satisfies the non-homogeneous part of the system and a general solution to the corresponding homogeneous equation. This union ensures that the general solution fully captures all aspects of the system.
Unveiling the Particular Solution
Finding a particular solution for a non-homogeneous equation involves identifying one specific solution that fits the equation \[ \mathbf{x}' = \mathbf{P}(t) \mathbf{x} + \mathbf{g}(t). \]
  • Unlike the general solution, the particular solution zeros in on meeting the exact requirements without spanning a family of solutions.
  • It is essential to pinpoint because, when combined with the homogeneous solution, it constructs the most comprehensive representation of the system's behavior.
  • For example, if \( \mathbf{v}(t) \) is found as a particular solution to the non-homogeneous equation, it corresponds to one very precise trajectory or state that the system can take.
Determining this unique solution often involves various methods of substitution or applying special conditions to solve the equation directly. Once found, it serves as a crucial building block in deriving the overall solution.
Deciphering the Homogeneous System
The homogeneous system relates to a simplified model where the external forces or non-homogeneous elements disappear. It is described by the equation \[ \mathbf{x}' = \mathbf{P}(t) \mathbf{x}. \]
  • This system highlights the internal or inherent dynamics of the differential equation without external influences.
  • Understanding the homogeneous system is vital because it provides the general solution \( \mathbf{u}(t) \) that describes the behavior of the system under no external conditions.
  • The differential equation for the homogeneous part represents natural oscillations, growth, or decay typical to the system's design.
Interestingly, the insight gained from solving the homogeneous system plays a critical role in crafting the general solution \( \Phi(t) = \mathbf{u}(t) + \mathbf{v}(t) \), where \( \mathbf{u}(t) \) forms the core of all behaviors predicted by the system. Solving this part of the equation requires classical techniques such as finding eigenvalues and eigenvectors, depending on the complexity of \( \mathbf{P}(t) \). By combining this with particular solutions, we embrace the complete spectrum of system responses.

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Most popular questions from this chapter

Consider the system $$ x^{\prime}=A x=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\\ {-3} & {2} & {4}\end{array}\right) x $$ (a) Show that \(r=2\) is an eigenvalue of multiplicity 3 of the coefficient matrix \(\mathbf{A}\) and that there is only one corresponding cigenvector, namely, $$ \xi^{(1)}=\left(\begin{array}{r}{0} \\ {1} \\ {-1}\end{array}\right) $$ (b) Using the information in part (a), write down one solution \(\mathbf{x}^{(1)}(t)\) of the system (i). There is no other solution of the purely exponential form \(\mathbf{x}=\xi e^{y t}\). (c) To find a second solution assume that \(\mathbf{x}=\xi t e^{2 t}+\mathbf{\eta} e^{2 t} .\) Show that \(\xi\) and \(\mathbf{\eta}\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{I}) \xi=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{n}=\mathbf{\xi} $$ since \(\xi\) has already been found in part (a), solve the second equation for \(\eta\). Neglect the multiple of \(\xi^{(1)}\) that appears in \(\eta\), since it leads only to a multiple of the first solution \(\mathbf{x}^{(1)}\). Then write down a second solution \(\mathbf{x}^{(2)}(t)\) of the system (i). (d) To find a third solution assume that \(\mathbf{x}=\xi\left(t^{2} / 2\right) e^{2 t}+\mathbf{\eta} t e^{2 t}+\zeta e^{2 t} .\) Show that \(\xi, \eta,\) and \(\zeta\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{l}) \xi=\mathbf{0}, \quad(\mathbf{\Lambda}-2 \mathbf{I}) \mathbf{\eta}=\mathbf{\xi}, \quad(\mathbf{A}-2 \mathbf{l}) \zeta=\mathbf{\eta} $$ The first two equations are the same as in part (c), so solve the third equation for \(\zeta,\) again neglecting the multiple of \(\xi^{(1)}\) that appears. Then write down a third solution \(\mathbf{x}^{(3)}(t)\) of the system (i). (e) Write down a fundamental matrix \(\boldsymbol{\Psi}(t)\) for the system (i). (f) Form a matrix \(\mathbf{T}\) with the cigenvector \(\xi^{(1)}\) in the first column, and the generalized eigenvectors \(\eta\) and \(\zeta\) in the second and third columns. Then find \(T^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A}\).

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 \(\mathbf{x}(0)=0\).

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

In each of Problems 13 through 20 the coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues in terms of \(\alpha\). (b) Find the critical value or values of \(\alpha\) where the qualitative nature of the phase portrait for the system changes. (c) Draw a phase portrait for a value of \(\alpha\) slightly below, and for another value slightly above, each crititical value. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\alpha} & {1} \\ {-1} & {\alpha}\end{array}\right) \mathbf{x} $$

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