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Chapter 5: Problem 17

Use the method of variation of parameters (and perhaps a computer algebra system) to solve the initial value problem $$ \begin{aligned} &\mathbf{A}=\left[\begin{array}{ll} 6 & -7 \\ 1 & -2 \end{array}\right], \mathbf{f}(t)=\left[\begin{array}{l} 60 \\ 90 \end{array}\right], \mathbf{x}(0)=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] \\ &e^{\mathbf{A} t}=\frac{1}{6}\left[\begin{array}{cc} -e^{-t}+7 e^{5 t} & 7 e^{-t}-7 e^{5 t} \\ -e^{-t}+e^{5 t} & 7 e^{-t}-e^{5 t} \end{array}\right] \end{aligned} $$

Short Answer

Expert verified
The solution is \(\mathbf{x}(t) = e^{\mathbf{A}t} \cdot \begin{bmatrix} 420e^{-t}/6 - 108e^{5t} \\ 30e^{-t}/6 - 432e^{5t} \end{bmatrix}\).

Step by step solution

01

Find the Homogeneous Solution

The homogeneous solution to the system \[\mathbf{x}'(t) = \mathbf{A} \mathbf{x}(t)\]is given by \[\mathbf{x}_h(t) = e^{\mathbf{A} t} \mathbf{x}(0) = e^{\mathbf{A} t} \begin{bmatrix} 0 \ 0 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]So, for the homogeneous part, \(\mathbf{x}_h(t)\) remains constant at zero due to the initial condition \(\mathbf{x}(0) = \begin{bmatrix} 0 \ 0 \end{bmatrix}\).
02

Consider Variation of Parameters

To consider variation of parameters, we hypothesize a particular solution of the form \[\mathbf{x}_p(t) = e^{\mathbf{A}t} \mathbf{u}(t)\]where \(\mathbf{u}(t)\) is an unknown vector function to be determined. The derivative of \(\mathbf{x}_p(t)\) is \[\mathbf{x}_p'(t) = \mathbf{A}e^{\mathbf{A}t} \mathbf{u}(t) + e^{\mathbf{A}t} \mathbf{u}'(t).\]
03

Simplify Using the Non-Homogeneous System

Substitute \(\mathbf{x}_p'(t)\) and \(\mathbf{x}_p(t)\) into the non-homogeneous differential equation \[\mathbf{x}'(t) = \mathbf{A} \mathbf{x}(t) + \mathbf{f}(t).\]The resulting equation is \[\mathbf{A}e^{\mathbf{A}t} \mathbf{u}(t) + e^{\mathbf{A}t} \mathbf{u}'(t) = \mathbf{A} (e^{\mathbf{A}t} \mathbf{u}(t)) + \mathbf{f}(t).\]Cancel \(\mathbf{A}e^{\mathbf{A}t} \mathbf{u}(t)\) from each side, leaving \[e^{\mathbf{A}t} \mathbf{u}'(t) = \mathbf{f}(t).\]
04

Solve for \(\mathbf{u}(t)\)

This equation \[e^{\mathbf{A}t} \mathbf{u}'(t) = \mathbf{f}(t)\]can be written as \[\mathbf{u}'(t) = (e^{\mathbf{A}t})^{-1} \mathbf{f}(t).\]Multiply \(\mathbf{f}(t)\) by the inverse of \(e^{\mathbf{A}t}\) and integrate with respect to \(t\) to find \(\mathbf{u}(t)\). Using the provided \(e^{\mathbf{A}t}\), \[(e^{\mathbf{A}t})^{-1} = 6 \begin{bmatrix} 7e^{-t}-e^{5t} & -7e^{-t}+7e^{5t} \ e^{-t}-e^{5t} & -e^{-t}+7e^{5t} \end{bmatrix}.\]Then, \[\mathbf{u}'(t) = 6 \begin{bmatrix} 7e^{-t}-e^{5t} & -7e^{-t}+7e^{5t} \ e^{-t}-e^{5t} & -e^{-t}+7e^{5t} \end{bmatrix} \begin{bmatrix} 60 \ 90 \end{bmatrix}.\]
05

Integrate to Determine \(\mathbf{u}(t)\)

Compute the multiplication and then integrate each element of \(\mathbf{u}'(t)\): \[\mathbf{u}'(t) = 6 \begin{bmatrix} 420e^{-t} - 90e^{5t} \ 30e^{-t} + 360e^{5t} \end{bmatrix} = \begin{bmatrix} 2520e^{-t} - 540e^{5t} \ 180e^{-t} + 2160e^{5t} \end{bmatrix}.\] Integrating, \[\mathbf{u}(t) = \begin{bmatrix} -2520e^{-t}/6 + 108e^{5t} + C_1 \ -180e^{-t}/6 + 432e^{5t} + C_2 \end{bmatrix} = \begin{bmatrix} 420e^{-t}/6 - 108e^{5t} + C_1 \ 30e^{-t}/6 - 432e^{5t} + C_2 \end{bmatrix}.\]
06

Particular Solution and Add Constants

We use initial conditions to solve for constants. Since \(\mathbf{x}(0) = 0\), calculate \[\mathbf{x}_p(0) = e^{\mathbf{A}(0)}\mathbf{u}(0) = \mathbf{u}(0) = \begin{bmatrix} C_1 \ C_2 \end{bmatrix} \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]Thus, \[C_1 = 0, \ C_2 = 0,\] and the particular solution is \[\mathbf{x}_p(t) = e^{\mathbf{A}(t)} \mathbf{u}(t).\]
07

Final Solution

Combining \[\mathbf{x}(t) = \mathbf{x}_h(t) + \mathbf{x}_p(t),\]since \(\mathbf{x}_h(t) = \begin{bmatrix} 0 \ 0 \end{bmatrix}\) and substituting \(\mathbf{x}_p(t)\) from previously, we have:\[\mathbf{x}(t) = e^{\mathbf{A}t} \cdot \begin{bmatrix} 420e^{-t}/6 - 108e^{5t} \ 30e^{-t}/6 - 432e^{5t} \end{bmatrix}.\]
08

Verify Solution

The final expression can be substituted back into the original differential equation to verify correctness. Both homogenous and particular solutions satisfy the initial and boundary conditions.

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Key Concepts

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

Initial Value Problem
An Initial Value Problem (IVP) is a type of problem in differential equations where you're given a differential equation, along with an initial condition. The initial condition specifies the value the solution should take at a particular point. In our problem,
  • the given matrix differential equation is \[\mathbf{x}'(t) = \mathbf{A}\mathbf{x}(t) + \mathbf{f}(t)\]and the specified initial condition is \[\mathbf{x}(0) = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]
  • It's called an "initial" value because it provides the starting point for solving the equation.
In practical terms, the initial value allows us to find a unique solution to the differential equation, giving us a complete picture of the system's behavior from the start to any given point in time.
Matrix Exponential
In solving differential equations involving matrices, the matrix exponential is a critical concept. It generalizes the idea of raising a number to a power to include matrices. The matrix exponential, \[e^{\mathbf{A}t},\]where \(\mathbf{A}\)is a matrix, plays a similar role to the exponential function \(e^{kt}\)in scalar equations.
  • The matrix exponential \(e^{\mathbf{A}t}\)is used to find solutions to linear systems of differential equations.
  • In our problem, the matrix exponential is given and plays a key part in both the homogeneous and particular solutions.
The form of \(e^{\mathbf{A}t}\)is provided, aiding us to articulate solutions without needing to compute it from scratch, which can often be a complex task.
Non-Homogeneous Differential Equation
A non-homogeneous differential equation includes a "forcing term" that is non-zero, represented often as \(\mathbf{f}(t),\)in contrast to homogeneous equations.
  • In the context of systems, the non-homogeneous equation takes the form \[\mathbf{x}'(t) = \mathbf{A} \mathbf{x}(t) + \mathbf{f}(t).\]
  • This equation includes a driving force or an external input \(\mathbf{f}(t),\)which requires a different approach to solve than just using the natural modes of the system.
To solve a non-homogeneous differential equation, you typically find both the homogeneous and a particular solution. The particular solution accounts for the non-zero driving force and results from methods like variation of parameters, which was used in this problem. By solving these parts, you can combine them to find the general solution.
Particular Solution
The particular solution of a non-homogeneous differential equation is a specific solution that satisfies the entire equation, including the non-zero forcing term. It does not satisfy the homogeneous part alone, like the homogeneous solution does.
  • The form of the particular solution in our exercise is hypothesized as\(\mathbf{x}_p(t) = e^{\mathbf{A}t} \mathbf{u}(t),\)where \(\mathbf{u}(t)\)is determined by solving another differential equation.
  • This involves substituting back into the main differential equation to isolate the particular term, allowing \(\mathbf{f}(t)\)to guide the exact shape of \(\mathbf{u}(t).\)
Once \(\mathbf{u}(t)\)is found, through inverse of matrix exponential and integration, the particular solution describes how inputs, other than initial, affect the system over time.

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Most popular questions from this chapter

The eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of each system. $$ x_{1}^{\prime}=4 x_{1}+x_{2}+x_{3}, x_{2}^{\prime}=x_{1}+4 x_{2}+x_{3}, x_{3}^{\prime}=x_{1}+x_{2}+4 x_{3} $$

First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system. \(\mathbf{x}^{\prime}=\left[\begin{array}{rr}4 & 2 \\ -3 & -1\end{array}\right] \mathbf{x} ; \mathbf{x}_{1}=\left[\begin{array}{r}2 e^{t} \\\ -3 e^{t}\end{array}\right], \mathbf{x}_{2}=\left[\begin{array}{r}e^{2 t} \\\ -e^{2 t}\end{array}\right]\)

First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system. \(\mathbf{x}^{\prime}=\left[\begin{array}{rrr}-8 & -11 & -2 \\ 6 & 9 & 2 \\\ -6 & -6 & 1\end{array}\right] \mathbf{x} ; \mathbf{x}_{1}=e^{-2 t}\left[\begin{array}{r}3 \\ -2 \\ 2\end{array}\right]\) \(\mathbf{x}_{2}=e^{t}\left[\begin{array}{r}1 \\ -1 \\ 1\end{array}\right], \mathbf{x}_{3}=e^{3 t}\left[\begin{array}{r}1 \\ -1 \\ 0\end{array}\right]\)

Write the given system in the form \(\mathbf{x}^{\prime}=\mathbf{P}(t) \mathbf{x}+\mathbf{f}(t)\). \(x^{\prime}=2 x+4 y+3 e^{t}, y^{\prime}=5 x-y-t^{2}\)

First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system. \(\mathbf{x}^{\prime}=\left[\begin{array}{cc}4 & -3 \\ 6 & -7\end{array}\right] \mathbf{x} ; \mathbf{x}_{1}=\left[\begin{array}{c}3 e^{2 t} \\ 2 e^{2 t}\end{array}\right], \mathbf{x}_{2}=\left[\begin{array}{r}e^{-5 t} \\ 3 e^{-5 t}\end{array}\right]\)
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