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Chapter 9: Problem 13

Find the general solution of the system of equations. \(x^{\prime}=5 x-y-5, y^{\prime}=2 x+2 y+2\)

Short Answer

Expert verified
The general solution is \[ \begin{pmatrix} x(t) \ y(t) \end{pmatrix} = \begin{pmatrix} -1 \ 2 \end{pmatrix} + C_1 e^{3t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + C_2 e^{4t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \].

Step by step solution

01

Write the system in matrix form

Express the system of differential equations in matrix form. The given equations are \[ x' = 5x - y - 5 \] and \[ y' = 2x + 2y + 2 \].This can be written as \[ \begin{pmatrix} x' \ y' \ \ \end{pmatrix} = \begin{pmatrix} 5 & -1 \ 2 & 2 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} + \begin{pmatrix} -5 \ 2 \end{pmatrix} \ \ \end{pmatrix} \]
02

Find the eigenvalues of the coefficient matrix

The coefficient matrix is \[ A = \begin{pmatrix} 5 & -1 \ 2 & 2 \end{pmatrix} \]. Find its eigenvalues by solving the characteristic equation \[ \text{det}(A - \theta I) = 0 \], where \[ I \] is the identity matrix. Calculate the determinant: \[ \text{det}\begin{pmatrix} 5 - \theta & -1 \ 2 & 2 - \theta \ \end{pmatrix} = (5 - \theta)(2 - \theta) - (-1)(2) = \theta^2 - 7\theta + 12 = 0 \]. Solving for \[ \theta \], we get \[ \theta = 3 \] and \[ \theta = 4 \].
03

Find the eigenvectors

For each eigenvalue, find the corresponding eigenvector. For \[ \theta = 3 \]: Solve \[ (A - 3I)v = 0 \]. This simplifies to \[ \begin{pmatrix} 2 & -1 \ 2 & -1 \ \ \end{pmatrix} \begin{pmatrix} x_1 \ y_1 \end{pmatrix} = 0 \]. This system has solution \[ x_1 = y_1 \]. Thus, an eigenvector is \[ v_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \].For \[ \theta = 4 \]: Solve \[ (A - 4I)v = 0 \]. This simplifies to \[ \begin{pmatrix} 1 & -1 \ 2 & -2 \ \ \end{pmatrix} \begin{pmatrix} x_2 \ y_2 \end{pmatrix} = 0 \]. This system has solution \[ x_2 = y_2 \]. Thus, an eigenvector is \[ v_2 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \].
04

Formulate the general solution of the homogeneous system

The general solution to the homogeneous system \[ x' = 5x - y \] and \[ y' = 2x + 2y \] is a linear combination of the eigenvectors multiplied by exponential functions of the eigenvalues:\[ \begin{pmatrix} x_h(t) \ y_h(t) \ \end{pmatrix} = C_1 e^{3t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + C_2 e^{4t} \begin{pmatrix} 1 \ 1 \end{pmatrix} \].
05

Find the particular solution

For the non-homogeneous system, \[ \begin{pmatrix} -5 \ 2 \end{pmatrix} \]is a constant vector. Assume a particular solution of the form: \[ \begin{pmatrix} x_p(t) \ y_p(t) \ \ \end{pmatrix} = \begin{pmatrix} x_p \ y_p \end{pmatrix} \].Substitute into the differential equations: \[ \begin{pmatrix} 0 \ 0 \end{pmatrix} = \begin{pmatrix} 5 & -1 \ 2 & 2 \ \ \end{pmatrix} \begin{pmatrix} x_p \ y_p \end{pmatrix} + \begin{pmatrix} -5 \ 2 \end{pmatrix} \].Solve this matrix equation to find \[ x_p \ \ \ y_p \]. This yields \[ x_p = -1 \ y_p = 2 \].
06

Combine the homogeneous and particular solutions

The total solution is the sum of the homogeneous and particular solutions: \[ \begin{pmatrix} x(t) \ y(t) \ \ \end{pmatrix} = \begin{pmatrix} -1 \ 2 \end{pmatrix} + C_1 e^{3t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + C_2 e^{4t} \begin{pmatrix} 1 \ 1 \end{pmatrix}\].

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Key Concepts

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

Eigenvalues
Eigenvalues are special numbers associated with a matrix. They play a key role in solving differential equations, especially when those equations can be written in matrix form. To find the eigenvalues of a matrix, you calculate the roots of its characteristic equation. For a given matrix A, the characteristic equation is found by solving \(\text{det}(A - \theta I) = 0\), where I is the identity matrix.

When solving the problem, we used the coefficient matrix \(\begin{pmatrix} 5 & -1 \ 2 & 2 \ \end{pmatrix}\). We then formed and solved the characteristic equation \(\theta^2 - 7\theta + 12 = 0\) to get the eigenvalues: \(\theta = 3\) and \(\theta = 4\).
Eigenvectors
Once you have the eigenvalues of a matrix, you can find the eigenvectors associated with each eigenvalue. The eigenvectors provide a basis in which the system's behavior can be expressed.

For the eigenvalue \(\theta = 3\), solving \( (A - 3I)v = 0\) gives us the eigenvector \(\begin{pmatrix} 1 \ 1 \ \end{pmatrix}\). For the eigenvalue \(\theta = 4\), solving \( (A - 4I)v = 0\) also results in the eigenvector \(\begin{pmatrix} 1 \ 1 \ \end{pmatrix}\).

Each eigenvector provides a direction in which the differential equations' solutions evolve over time.
Homogeneous Differential Equations
A homogeneous differential equation is one in which there is no term that is simply a function of the independent variable (usually time, t). In such systems, all terms involve the dependent variable or its derivatives.

Our system of equations \( x' = 5x - y \) and \( y' = 2x + 2y \) is homogeneous when we set the non-homogeneous vector to zero. The general solution for the homogeneous system involves finding the exponential combination of the eigenvectors. Therefore, \(\begin{pmatrix} x_h(t) \ y_h(t) \ \end{pmatrix} = C_1 e^{3t} \begin{pmatrix} 1 \ 1 \ \end{pmatrix} + C_2 e^{4t} \begin{pmatrix} 1 \ 1 \ \end{pmatrix}\).
Non-Homogeneous Differential Equations
Non-homogeneous differential equations contain terms that are functions of the independent variable. These equations require an additional particular solution to solve.

In our original system, the term \(\begin{pmatrix} -5 \ 2 \ \end{pmatrix}\) makes it non-homogeneous. To find the particular solution, we assume a constant vector such that \(\begin{pmatrix} 0 \ 0 \ \end{pmatrix} = A \begin{pmatrix} x_p \ y_p \ \end{pmatrix} + \begin{pmatrix} -5 \ 2 \ \end{pmatrix}\). Solving this equation gives us the particular solution: \(\begin{pmatrix} -1 \ 2 \ \end{pmatrix}\).
Matrix Form of Differential Equations
Writing differential equations in matrix form makes it easier to handle systems of equations. This compact representation helps in using matrix algebra to find solutions and understand the system's behavior.

In our exercise, we converted the given system into matrix form: \(\begin{pmatrix} x' \ y' \ \end{pmatrix} = \begin{pmatrix} 5 & -1 \ 2 & 2 \ \end{pmatrix} \begin{pmatrix} x \ y \ \end{pmatrix} + \begin{pmatrix} -5 \ 2 \ \end{pmatrix}\).

This form encapsulates the system in a single equation, making further analysis, such as finding eigenvalues and eigenvectors, straightforward.

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Most popular questions from this chapter

Rewrite the system of differential equations into matrix form. \(x^{\prime}=x-y, y^{\prime}=3 x-2 y\)

The displacement \(x(t)\) of a spring from its rest position after \(t\) seconds follows the differential equation $$ m x^{\prime \prime}+\gamma x^{\prime}+k x=q(t) $$ where \(m\) is the mass of the object attached to the spring, \(q(t)\) is the forcing function, and \(\gamma\) and \(k\) are the stiffness and damping coefficients, respectively. Suppose that the spring starts at rest, so that \(x(0)=0\) and \(x^{\prime}(0)=0 .\) Solve for \(x(t)\) given the following conditions. \(m=2, k=50, \gamma=20, q(t)=25 t\)

When our vocal folds vibrate under normal speech, their initial movement is governed by a second-order differential equation. Let \(x(t)\) denote the lateral displacement of the lower vocal folds at time \(t\). Then the equation of motion for the lower vocal folds is $$ m x^{\prime \prime}+\frac{1}{c}\left(x-x_{0}\right)=p d $$ In this equation, \(m\) is the mass of the lower vocal folds per unit length, \(c\) is the mechanical compliance per unit length, \(x_{0}\), is the resting position, \(p\) is the subglottal pressure, and \(d\) is the length of the lower vocal folds. a) Find the general solution of this differential equation. b) Assume that the vocal folds are at rest before the outward displacement begins, so that \(x(0)=0\) and \(x^{\prime}(0)=0 .\) Solve this initial-value problem.

Zinc Depletion. After intake of zinc into the body, the zinc may be found in either the plasma or a portion of the liver. Let \(P(t)\) and \(L(t)\) be the amount of zinc in the plasma and liver after \(t\) days, respectively. The transfer rate of zinc from the plasma to the liver is 3 per day. The transfer rate from the liver to the plasma is \(0.6\) per day. Finally, the zinc is removed from the body via plasma at a rate of \(2.24\) per day. a) Draw a two-compartment model for \(L\) and \(P\). b) Find a system of differential equations satisfied by \(L\) and \(P\). c) Solve for \(P(t)\) and \(L(t)\) given that \(P(0)=0\) and \(L(0)=241 / 15\)

Verify by substitution that the given functions solve the system of differential equations. \(\left[\begin{array}{l}x \\\ y\end{array}\right]^{\prime}=\left[\begin{array}{rr}-5 & -4 \\ 10 & 7\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]\); \(x=-2 e^{t} \sin 2 t, y=3 e^{t} \sin 2 t+e^{t} \cos 2 t\)
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