Solution of System of Nonhomogeneous Differential Equations

Ordinary differential equations

Brent J. Lewis , ... Andrew A. Prudil , in Advanced Mathematics for Engineering Students, 2022

Problems

2.1

Consider the nonhomogeneous system of first-order, linear differential equations $y_1^{\prime }=y_2+\cosh t$ and $y_2^{\prime }=y_1$ with boundary conditions $y_1(0)=0$ and $y_2(0)=-\frac{1}{2}$ . Solve this initial value problem by the following three methods:

(a)

Matrix methods. Use the method of variation of parameters to determine the particular solution.

(b)

Laplace transform methods.

(c)

Convert the two first-order differential equations into a single second-order differential equation and solve this latter equation.

2.2.

Consider the following problems:

(a)

State the differential equation for the Sturm–Liouville problem where $r(x)=x$ , $p(x)=x^{-1}$ , and $q(x)=0$ .

(b)

Using the transformation $x=e^t$ , show that this differential equation reduces to $\frac{d^{2}y}{d{t}^2}+\lambda y=0$ .

(c)

Given the boundary conditions $y(1)=0$ and $y(e)=0$ for the differential equation in (a), what are the corresponding eigenvalues and eigenfunctions?

2.3.

Using the transformation $x=\cosh t$ , show that the differential equation $\frac{d^{2}y}{d{t}^2}+\coth t\frac{dy}{dt}-20y=0$ reduces to Legendre's differential equation and give the expression for the Legendre polynomial which is a solution to this transformed equation.

2.4.

Consider the hypergeometric equation $x(1-x)y^{″}+[b-(2+b)x]y^{\prime }-by=0$ , where b is a constant larger than unity. Using a Frobenius method, show that the series solution about $x=0$ is the geometric series such that $y_1(x)=1+x+x^2+\dots$ .

2.5.

Find a general solution, in terms of Bessel functions, for the differential equation $4x^2{y}^{″}+4x{y}^{\prime }+(x-n^2)y=0$ , where n is an integer, by using either the substitution or the generalized form for the solution of the Bessel equation. What is the solution that satisfies the condition $|y(0)|<\infty$ ?

2.6.

Using $J_o$ for a series expansion, show that the coefficients of the Fourier–Bessel series for $f(x)=1$ over the interval $0⩽x⩽1$ are given as $a_m=2/\left\{{\alpha }_{m0}{J}_1\left({\alpha }_{m0}\right)\right\}$ , where ${\alpha }_{m0}$ are the zeros of $J_0$ .

2.7.

Consider the function $f(x)=\{\begin{array}{cc}0\hfill & \text{if}-1⩽x⩽0,\hfill \\ 2\hfill & \text{if}0⩽x⩽1.\hfill \end{array}$

Calculate the first two terms of the Fourier–Legendre series of $f(x)$ over the interval $-1⩽x⩽1$ . What is the value of the complete Fourier–Legendre series of $f(x)$ at $x=0$ ?

2.8.

Consider the differential equation $16x^2{y}^{″}+3y=0$ .

(a)

Calculate the two basis functions using a Frobenius method and give the general solution.

(b)

Derive the second basis function from $y_1(x)$ using a method of reduction of order.

2.9.

Give the Legendre polynomial $P_n(x)$ which is a solution to the differential equation $\frac{1}{2}(1-x^2)y^{″}-x{y}^{\prime }+3y=0$ .

2.10.

Find a general solution, in terms of Bessel functions, for the differential equation $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-(1+4x^4)y=0$ by (i) using the transformation $z=x^2$ and (ii) employing the inspection method for the generalized form of Bessel's equation.

2.11.

The steady-state temperature distribution for a fin of cross-sectional area A, constant perimeter P, constant conductivity k, and length L can be determined from the following differential equation:

$\frac{d}{dx}\left(kA\frac{dT}{dx}\right)-hP(T-T_{\infty })=0,$

where h is the heat transfer coefficient for the fin surrounded by a fluid with a constant temperature $T_{\infty }$ .

(a)

Is this differential equation homogeneous or nonhomogeneous? With a change of variable $\theta =T-T_{\infty }$ , find the general solution for the variable $\theta (x)$ by putting $m^2=hP/(kA)$ .

(b)

Find a solution to the boundary value problem for the boundary conditions $\theta (0)={\theta }_0$ and ${\theta }^{\prime }(L)=0$ .

2.12.

Given the periodic function $f(x)=x$ for $-2<x<2$ and $f(x+4)=f(x)$ .

(a)

Sketch this function over the interval $-6⩽x⩽6$ .

(b)

Find the Fourier series of this function.

(c)

What is the value of $f(x)$ that one would compute from the Fourier series solution at the discontinuity of $x=2$ .

2.13.

Given is the differential equation $x{y}^{″}+(1-x)y^{\prime }+ny=0$ .

(a)

Determine one of the basis solutions for this differential equation, when $n=2$ , using a Frobenius method.

(b)

The Laguerre polynomial $L_n(x)$ is in fact a solution of this differential equation as given by the Rodrigues formula $L_n(x)=e^x\frac{d^n}{d{x}^n}\left(x^n{e}^{-x}\right)$ . Using this formula, specify the Laguerre polynomial $L_2(x)$ . What is the value of the arbitrary constant in part (a) to obtain the Laguerre polynomial solution $L_2(x)$ ?

2.14.

Consider the differential equation $\frac{d^{2}y}{d{x}^2}-xy=0$ . Using the transformations $y=u\sqrt{x}$ and $\frac{2}{3}i{x}^{3/2}=z$ , where $i=\sqrt{-1}$ , show that this differential equation reduces to the Bessel differential equation

$\frac{d^{2}u}{d{z}^2}+\frac{1}{z}\frac{du}{dz}+u(1-\frac{1}{{\left(3z\right)}^2})=0.$

What is the solution for $u(z)$ and $y(x)$ ?

2.15.

Determine the particular solution for the nonhomogeneous, second-order, ordinary differential equation $y^{″}+y^{\prime }-2y=4\sin 2x$ using the following methods:

(a)

undetermined coefficients,

(b)

variation of parameters,

(c)

Fourier transforms (see Problem 3.10).

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Elements of linear algebra

Giovanni Romeo , in Elements of Numerical Mathematical Economics with Excel, 2020

Rank of a matrix

The rank of a matrix A corresponds to the maximal number of linearly independent columns (or rows) of A. The rank can be found as the maximal order of the nonsingular minors (the determinants of the square submatrices of A) that can be extracted from the matrix A. The rank can be also seen as a measure of the "nondegenerateness" of the system of linear equations.

From the Rouché–Capelli theorem, for a nonhomogeneous system, it turns out that:

i.

if r(A)   = r(A|b)   = n, the solution is unique: we have ${\infty }^{n-r}={\infty }^0=1$ solution. Regular systems. The n vectors are all l.i.

ii.

Otherwise there are infinitely many solutions: if r(A)   = r(A|b)   < n, there exist ${\infty }^{n-r}$ solutions. If 1   ≤ r(A)   = r(A|b)   < m, there will be (m−r) redundant equations and there exist ${\infty }^{n-r}$ solutions. r vectors are l.i., while (n−r) vectors are l.d.

The regular cases where r(A)   = r(A|b)   = m  = n are called square regular systems (or Cramer's systems), where the coefficient matrix A is square and det(A)   ≠   0; the solution is unique and to find it we can apply the following Cramer's rule.

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Systems of linear differential equations

Henry J. Ricardo , in A Modern Introduction to Differential Equations (Third Edition), 2021

6.10.1 The general solution

The linear systems we have been dealing with so far are called homogeneous systems. Basically, this means that they can be expressed in the form $\dot{X}=AX$ with no "leftover" terms. If a linear system has to be written as $\dot{X}=AX+B\left(t\right)$ , where $B(t)$ is a vector of the form $\left[\begin{array}{c}\hfill b_1(t)\hfill \\ \hfill b_2(t)\hfill \end{array}\right]$ , then we say that the system is nonhomogeneous. For example, in matrix terms, the system $\frac{dx}{dt}=x+\sin t,\frac{dy}{dt}=t-y$ must be written as $\left[\begin{array}{c}\hfill \dot{x}(t)\hfill \\ \hfill \dot{y}(t)\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\\ \hfill 0\hfill & \hfill -1\end{array}\right]\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right]+\left[\begin{array}{c}\hfill \sin t\hfill \\ \hfill t\hfill \end{array}\right]$ and so is nonhomogeneous.

Don't confuse the distinction between autonomous and nonautonomous with that between homogeneous and nonhomogeneous. For example, if $b_1(t)$ and $b_2(t)$ are constant functions (not both zero), then we have a system that is both autonomous and nonhomogeneous. (See, e.g., Example 6.4.3.)

The techniques that were introduced in Section 4.2 for second-order nonhomogeneous equations generalize to systems, but the calculations are more complicated. To get a handle on solving a nonhomogeneous linear system, we need a fundamental fact about linear systems:

The general solution, $X_{\mathrm{GNH}}$ , of a linear nonhomogeneous system is obtained by finding a particular solution, $X_{\mathrm{PNH}}$ , of the nonhomogeneous system and adding it to the general solution, $X_{\mathrm{GH}}$ , of the associated homogeneous system.

You should see this as an application of the Superposition Principle and as an extension of the result we saw for single linear differential equations (Section 4.2). Symbolically, we can write $X_{\mathrm{GNH}}=X_{\mathrm{GH}}+X_{\mathrm{PNH}}$ . Using the definitions of these terms, we can see that this sum of vectors is a solution of the nonhomogeneous system:

${\dot{X}}_{\mathrm{GNH}}={\dot{X}}_{\mathrm{GH}}+{\dot{X}}_{\mathrm{PNH}}=A{X}_{\mathrm{GH}}+\{A{X}_{\mathrm{PNH}}+B(t)\}=A(X_{\mathrm{GH}}+X_{\mathrm{PNH}})+B(t)=A{X}_{\mathrm{GNH}}+B(t).$

(Be sure you follow this.) You should see that $X_{\mathrm{GH}}$ , as a general solution, must contain two arbitrary constants, so the expression for $X_{\mathrm{GNH}}$ contains two arbitrary constants.

Let's look at a simple example showing the structure of a nonhomogeneous system's solution.

Example 6.10.1 The Solution of a Nonhomogeneous System

The system

$$\begin{gathered} \dot{x}=x+y+2e^{-t} \\ \dot{y}=4x+y+4e^{-t} \end{gathered}$$

can be written in the form $\dot{X}(t)=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 1\hfill \end{array}\right]X+\left[\begin{array}{c}\hfill 2e^{-t}\hfill \\ \hfill 4e^{-t}\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 1\hfill \end{array}\right]X+2e^{-t}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$ . The system has eigenvalues ${\lambda }_1=3$ and ${\lambda }_2=-1$ , with corresponding eigenvectors $V_1=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$ and $V_2=\left[\begin{array}{c}\hfill 1\\ \hfill -2\end{array}\right]$ . (Check this.) Then the general solution of the associated homogeneous system $\dot{X}(t)=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 1\hfill \end{array}\right]X$ is

$X_{\mathrm{GH}}=c_1{e}^{3t}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]+c_2{e}^{-t}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \end{array}\right].$

We need to verify that a particular solution of the original nonhomogeneous system is given by $X_{\mathrm{PNH}}=e^{-t}\left[\begin{array}{c}\hfill 0\\ \hfill -2\end{array}\right]=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -2e^{-t}\hfill \end{array}\right]$ . Therefore, the general solution of the nonhomogeneous system is

$X_{\mathrm{GNH}}=X_{\mathrm{GH}}+X_{\mathrm{PNH}}=c_1{e}^{3t}\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]+c_2{e}^{-t}\left[\begin{array}{c}\hfill 1\\ \hfill -2\end{array}\right]+\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -2e^{-t}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill c_1{e}^{3t}+c_2{e}^{-t}\hfill \\ \hfill 2c_1{e}^{3t}-2c_2{e}^{-t}-2e^{-t}\hfill \end{array}\right].$

(Check that this is the general solution of the original nonhomogeneous system.)

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Systems of Differential Equations

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (Fifth Edition), 2018

6.3 An Introduction to Linear Systems

We first encounter systems of linear equations in elementary algebra courses. For example,

$\begin{array}{cc}\hfill 3x-5y& =-13\hfill \\ \hfill -3x+6y& =15\hfill \end{array}$

written in matrix form as

$\left(\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -3\hfill & \hfill 6\hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -13\hfill \\ \hfill 15\hfill \end{array}\right)$

is a system of two linear equations in two variables with solution $(x,y)=(-1,2)$ , which is written in matrix or vector notations using our convention as $\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 2\hfill \end{array}\right)$ . In the same manner, we can consider a system of linear differential equations.

We begin our study of systems of linear ordinary differential equations by introducing several definitions along with some convenient notation. Let

$\begin{array}{cc}\hfill \mathbf{X}(t)& =\left(\begin{array}{c}\hfill x_1(t)\hfill \\ \hfill x_2(t)\hfill \\ \hfill ⋮\hfill \\ \hfill x_n(t)\hfill \end{array}\right),\hfill \\ \hfill \mathbf{A}(t)& =\left(\begin{array}{cccc}\hfill a_{11}(t)\hfill & \hfill a_{12}(t)\hfill & \hfill \dots \hfill & \hfill a_{1n}(t)\hfill \\ \hfill a_{21}(t)\hfill & \hfill a_{22}(t)\hfill & \hfill \dots \hfill & \hfill a_{2n}(t)\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \dots \hfill & \hfill ⋮\hfill \\ \hfill a_{n1}(t)\hfill & \hfill a_{n2}(t)\hfill & \hfill \dots \hfill & \hfill a_{nn}(t)\hfill \end{array}\right),\hfill \end{array}$

and

$\mathbf{F}(t)=\left(\begin{array}{c}\hfill f_1(t)\hfill \\ \hfill f_2(t)\hfill \\ \hfill ⋮\hfill \\ \hfill f_n(t)\hfill \end{array}\right).$

Then, the homogeneous system of first order linear differential equations

(6.8) $\begin{array}{cc}\hfill x_1^{\prime }& =a_{11}(t)x_1+a_{12}(t)x_2+\cdots +a_{1n}{x}_n(t)\hfill \\ \hfill x_2^{\prime }& =a_{21}(t)x_1+a_{22}(t)x_2+\cdots +a_{2n}{x}_n(t)\hfill \\ \hfill ⋮\\ \hfill x_n^{\prime }& =a_{n1}(t)x_1+a_{n2}(t)x_2+\cdots +a_{nn}{x}_n(t)\hfill \end{array}$

is equivalent to

(6.9) ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$

and the nonhomogeneous system

(6.10) $\begin{array}{cc}\hfill x_1^{\prime }& =a_{11}(t)x_1+a_{12}(t)x_2+\cdots +a_{1n}{x}_n(t)+f_1(t)\hfill \\ \hfill x_2^{\prime }& =a_{21}(t)x_1+a_{22}(t)x_2+\cdots +a_{2n}{x}_n(t)+f_2(t)\hfill \\ \hfill ⋮\\ \hfill x_n^{\prime }& =a_{n1}(t)x_1+a_{n2}(t)x_2+\cdots +a_{nn}{x}_n(t)+f_n(t)\hfill \end{array}$

is equivalent to

(6.11) ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)+\mathbf{F}(t)$

For the nonhomogeneous system (6.11), the corresponding homogeneous system is system (6.9).

Example 6.14

(a) Write the homogeneous system $\{\begin{array}{c}{x}^{\prime }=-5x+5y\hfill \\ y^{\prime }=-5x+y\hfill \end{array}$ in matrix form. (b) Write the nonhomogeneous system $\{\begin{array}{c}{x}^{\prime }=x+2y-\sin t\hfill \\ y^{\prime }=4x-3y+t^2\hfill \end{array}$ in matrix form.

Solution: (a) The homogeneous system $\{\begin{array}{c}{x}^{\prime }=-5x+5y\hfill \\ y^{\prime }=-5x+y\hfill \end{array}$ is equivalent to the system ${\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{cc}\hfill -5\hfill & \hfill 5\hfill \\ \hfill -5\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)$ .

With our notation, ${\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{c}\hfill x^{\prime }\hfill \\ \hfill y^{\prime }\hfill \end{array}\right)$

(b) The nonhomogeneous system $\{\begin{array}{c}{x}^{\prime }=x+2y-\sin t\hfill \\ y^{\prime }=4x-3y+t^2\hfill \end{array}$ is equivalent to ${\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill -3\hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)+\left(\begin{array}{c}\hfill -\sin t\hfill \\ \hfill t^2\hfill \end{array}\right)$ .  □

The nth-order linear equation

(6.12) $\begin{array}{cc}\hfill & y^{(n)}(t)+a_{n-1}(t)y^{(n-1)}+\cdots +a_2(t)y^{″}\hfill \\ \hfill & +a_1(t)y^{\prime }+a_0(t)y=f(t),\hfill \end{array}$

The nth-order linear equation is discussed in Chapter 4.

discussed in previous chapters, can be written as a system of first-order equations as well. Let $x_1=y$ , $x_2=d{x}_1/dt=y^{\prime }$ , $x_3=d{x}_2/dt=y^{″}$ , …, $x_{n-1}=d{x}_{n-2}/dt=y^{(n-2)}$ , $x_n=d{x}_{n-1}/dt=y^{(n-1)}$ . Then, Eq. (6.12) is equivalent to the system

(6.13) $\begin{array}{cc}\hfill x_1^{\prime }& =x_2\hfill \\ \hfill x_2^{\prime }& =x_3\hfill \\ \hfill ⋮\\ \hfill x_{n-1}^{\prime }& =x_n\hfill \\ \hfill x_n^{\prime }& =-a_{n-1}{x}_n-\cdots -a_2{x}_3-a_1{x}_2-a_0{x}_1+f(t)\hfill \end{array}$

which can be written in matrix form as

(6.14) $\begin{array}{cc}\hfill & \begin{array}{c}\hfill {\left(\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill ⋮\hfill \\ \hfill x_{n-1}\hfill \\ \hfill x_n\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{ccccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \cdots \hfill & \hfill ⋮\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 1\hfill \\ \hfill -a_0\hfill & \hfill -a_1\hfill & \hfill -a_2\hfill & \hfill \cdots \hfill & \hfill -a_n\hfill \end{array}\right)\left(\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \\ \hfill ⋮\hfill \\ \hfill x_{n-1}\hfill \\ \hfill x_n\hfill \end{array}\right)\end{array}\hfill \\ \hfill & +\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \\ \hfill f(t)\hfill \end{array}\right).\hfill \end{array}$

Example 6.15

Write the equation $y^{″}+5y^{\prime }+6y=\cos t$ as a system of first order differential equations.

Solution: We let $x_1=y$ and $x_2=x_1^{\prime }=y^{\prime }$ . Then,

$x_2^{\prime }=y^{″}=\cos t-6y-5y^{\prime }=\cos t-6x_1-5x_2$

so the second-order equation $y^{″}+5y^{\prime }+6y=\cos t$ is equivalent to the system

$\begin{array}{cc}\hfill x_1^{\prime }& =x_2\hfill \\ \hfill x_2^{\prime }& =\cos t-6x_1-5x_2,\hfill \end{array}$

which can be written in matrix form as

${\left(\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -6\hfill & \hfill -5\hfill \end{array}\right)\left(\begin{array}{c}\hfill x_1\hfill \\ \hfill x_2\hfill \end{array}\right)+\left(\begin{array}{c}\hfill 0\hfill \\ \hfill \cos t\hfill \end{array}\right).\text{\hspace{1em}□}$

At this point, given a system of ordinary differential equations, our goal is to construct either an explicit, numerical, or graphical solution of the system of equations.

We now state the theorems and terminology used in establishing the fundamentals of solving systems of differential equations. All proofs are omitted but can be found in advanced differential equations textbooks. In each case, we assume that the matrix $\mathbf{A}=\mathbf{A}(t)$ in the systems ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ (Eq. (6.9)) and ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)+\mathbf{F}(t)$ (Eq. (6.11)) is an $n\times n$ matrix.

Definition 6.12 Solution Vector

A solution vector (or solution) of the system ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)+\mathbf{F}(t)$ (Eq. (6.11)) on the interval I is an $n\times 1$ matrix (or vector) of the form

$\mathbf{\Phi }(t)=\left(\begin{array}{c}\hfill {\varphi }_1(t)\hfill \\ \hfill {\varphi }_2(t)\hfill \\ \hfill ⋮\hfill \\ \hfill {\varphi }_n(t)\hfill \end{array}\right),$

where the ${\varphi }_i(t)$ are differentiable functions that satisfy ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)+\mathbf{F}(t)$ on I.

Example 6.16

Show that ${\mathbf{\Phi }}_1(t)=\left(\begin{array}{c}\hfill -2e^{2t}\hfill \\ \hfill e^{2t}\hfill \end{array}\right)$ is a solution of ${\mathbf{X}}^{\prime }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right)\mathbf{X}$ .

Solution: Notice that ${\mathbf{\Phi }}_1^{\prime }(t)=\left(\begin{array}{c}\hfill -4e^{2t}\hfill \\ \hfill 2e^{2t}\hfill \end{array}\right)$ and $\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right){\mathbf{\Phi }}_1=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right)\left(\begin{array}{c}\hfill -2e^{2t}\hfill \\ \hfill e^{2t}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -2e^{2t}-2e^{2t}\hfill \\ \hfill -4e^{2t}+6e^{2t}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -4e^{2t}\hfill \\ \hfill 2e^{2t}\hfill \end{array}\right)={\mathbf{\Phi }}_1^{\prime }$ . Then, because ${\mathbf{\Phi }}_1^{\prime }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right){\mathbf{\Phi }}_1$ , ${\mathbf{\Phi }}_1$ is a solution of the system.  □

Theorem 6.2 Principle of Superposition

Suppose that ${\mathbf{\Phi }}_1(t)$ , ${\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_m(t)$ are m solutions of the linear homogeneous system ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ (Eq. (6.9) ) on the open interval I. Then, the linear combination

$\mathbf{\Phi }(t)=c_1{\mathbf{\Phi }}_1(t)+c_2{\mathbf{\Phi }}_2(t)+\cdots +c_m{\mathbf{\Phi }}_m(t),$

where $c_1,c_2,\dots ,c_m$ are arbitrary constants, is also a solution of ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ .

Example 6.17

Show that ${\mathbf{\Phi }}_2(t)=\left(\begin{array}{c}\hfill -e^{5t}\hfill \\ \hfill 2e^{5t}\hfill \end{array}\right)$ and $\mathbf{\Phi }(t)=c_1\left(\begin{array}{c}\hfill -2e^{2t}\hfill \\ \hfill e^{2t}\hfill \end{array}\right)+c_2\left(\begin{array}{c}\hfill -e^{5t}\hfill \\ \hfill 2e^{5t}\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill -2e^{2t}\hfill & \hfill -e^{5t}\hfill \\ \hfill e^{2t}\hfill & \hfill 2e^{5t}\hfill \end{array}\right)\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \end{array}\right)$ are solutions of ${\mathbf{X}}^{\prime }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right)\mathbf{X}$ .

Solution: We let the reader follow the procedure used in Example 6.16 to show that ${\mathbf{\Phi }}_2$ satisfies ${\mathbf{X}}^{\prime }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right)\mathbf{X}$ . By the Principle of Superposition, the linear combination of the two solutions ${\mathbf{\Phi }}_1$ (from Example 6.16) and ${\mathbf{\Phi }}_2$ is also a solution. We verify this now by first writing $\mathbf{\Phi }(t)$ as $\mathbf{\Phi }(t)=\left(\begin{array}{c}\hfill -2c_1{e}^{2t}-c_2{e}^{5t}\hfill \\ \hfill c_1{e}^{2t}+2c_2{e}^{5t}\hfill \end{array}\right)$ . Then, ${\mathbf{\Phi }}^{\prime }(t)=\left(\begin{array}{c}\hfill -4c_1{e}^{2t}-5c_2{e}^{5t}\hfill \\ \hfill 2c_1{e}^{2t}+10c_2{e}^{5t}\hfill \end{array}\right)$ and

$\begin{array}{cc}\hfill & \left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right)\left(\begin{array}{c}\hfill -4c_1{e}^{2t}-5c_2{e}^{5t}\hfill \\ \hfill 2c_1{e}^{2t}+10c_2{e}^{5t}\hfill \end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{c}\hfill (-2c_1{e}^{2t}-c_2{e}^{5t})-2(c_1{e}^{2t}+2c_2{e}^{5t})\hfill \\ \hfill 2(-2c_1{e}^{2t}-c_2{e}^{5t})+6(c_1{e}^{2t}+2c_2{e}^{5t})\hfill \end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{c}\hfill -4c_1{e}^{2t}-5c_2{e}^{5t}\hfill \\ \hfill 2c_1{e}^{2t}+10c_2{e}^{5t}\hfill \end{array}\right).\hfill \end{array}$

Therefore, the linear combination of the solutions is also a solution.  □

We define linear dependence and independence of a set of vector-valued functions $S=\{{\mathbf{\Phi }}_1(t),{\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_m(t)\}$ in a manner similar as to how we defined linear dependence and independence of sets of real-valued functions. The set $S=\{{\mathbf{\Phi }}_1(t),{\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_m(t)\}$ is linearly dependent on an interval I if there is a set of constants $c_1$ , $c_2$ , …, $c_m$ not all zero such that

$c_1{\mathbf{\Phi }}_1(t)+c_2{\mathbf{\Phi }}_2(t)+\cdots +c_m{\mathbf{\Phi }}_m(t)=0;$

otherwise, the set is linearly independent. (Note that 0 is the zero vector with the same dimensions as each of the ${\mathbf{\Phi }}_i(t)$ , $i=1$ , 2, …, m.) As with two real-valued functions, two vector-valued functions are linearly dependent if they are scalar multiples of each other. Otherwise, the functions are linearly independent. For more than two vector-valued functions, we often use the Wronskian to determine if the functions are linearly independent or dependent.

Definition 6.13 Wronskian of a Set of Vector-Valued Functions

The Wronskian of $S=\{{\mathbf{\Phi }}_1(t),{\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_n(t)\}$ is the determinant of the matrix with columns ${\mathbf{\Phi }}_1(t)=\left(\begin{array}{c}\hfill {\varphi }_{11}(t)\hfill \\ \hfill {\varphi }_{21}(t)\hfill \\ \hfill ⋮\hfill \\ \hfill {\varphi }_{n1}(t)\hfill \end{array}\right)$ , ${\mathbf{\Phi }}_2(t)=\left(\begin{array}{c}\hfill {\varphi }_{12}(t)\hfill \\ \hfill {\varphi }_{22}(t)\hfill \\ \hfill ⋮\hfill \\ \hfill {\varphi }_{n2}(t)\hfill \end{array}\right)$ , …, ${\mathbf{\Phi }}_n(t)=\left(\begin{array}{c}\hfill {\varphi }_{1n}(t)\hfill \\ \hfill {\varphi }_{2n}(t)\hfill \\ \hfill ⋮\hfill \\ \hfill {\varphi }_{nn}(t)\hfill \end{array}\right)$ :

(6.15) $\begin{array}{cc}\hfill W(S)& =W\left(\{{\mathbf{\Phi }}_1(t),{\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_n(t)\}\right)\hfill \\ \hfill & =\left|\begin{array}{cccc}\hfill {\varphi }_{11}(t)\hfill & \hfill {\varphi }_{12}(t)\hfill & \hfill \cdots \hfill & \hfill {\varphi }_{1n}(t)\hfill \\ \hfill {\varphi }_{21}(t)\hfill & \hfill {\varphi }_{22}(t)\hfill & \hfill \cdots \hfill & \hfill {\varphi }_{2n}(t)\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill {\varphi }_{n1}(t)\hfill & \hfill {\varphi }_{n2}(t)\hfill & \hfill \cdots \hfill & \hfill {\varphi }_{nn}(t)\hfill \end{array}\right|.\hfill \end{array}$

Theorem 6.3 Wronskian of Solutions

Suppose that

$S=\{{\mathbf{\Phi }}_1(t),{\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_n(t)\}$

is a set of n solutions of the linear homogeneous system ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ (Eq. (6.9) ) on the open interval I, where each component of $\mathbf{A}(t)$ is continuous on I. If S is linearly dependent, then $W(S)=0$ on I. If S is linearly independent, then $W(S)\ne 0$ for all values on I.

Example 6.18

Verify that ${\mathbf{\Phi }}_1(t)=\left(\begin{array}{c}\hfill -2e^{2t}\hfill \\ \hfill e^{2t}\hfill \end{array}\right)$ and ${\mathbf{\Phi }}_2(t)=\left(\begin{array}{c}\hfill -e^{5t}\hfill \\ \hfill 2e^{5t}\hfill \end{array}\right)$ are linearly independent solutions of ${\mathbf{X}}^{\prime }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right)\mathbf{X}$ .

Solution: In Examples 6.16 and 6.17 we showed that ${\mathbf{\Phi }}_1(t)$ and ${\mathbf{\Phi }}_2(t)$ are solutions of the system ${\mathbf{X}}^{\prime }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right)\mathbf{X}$ . Therefore, we calculate $W(\{{\mathbf{\Phi }}_1,{\mathbf{\Phi }}_2\})=\left|\begin{array}{cc}\hfill -2e^{2t}\hfill & \hfill -e^{5t}\hfill \\ \hfill e^{2t}\hfill & \hfill 2e^{5t}\hfill \end{array}\right|=-3e^{7t}$ . The vector-valued functions are linearly independent because $W(\{{\mathbf{\Phi }}_1,{\mathbf{\Phi }}_2\})=-3e^{7t}\ne 0$ for all values of t.  □

Definition 6.14 Fundamental Set of Solutions

Any set

$S=\{{\mathbf{\Phi }}_1(t),{\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_n(t)\}$

of n linearly independent solutions of the linear homogeneous system ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ (Eq. (6.9)) on an open interval I is called a fundamental set of solutions on I.

Example 6.19

Which of the following is a fundamental set of solutions for

${\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{cc}\hfill -2\hfill & \hfill -8\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)?$

(a) $S_1=\{\left(\begin{array}{c}\hfill \cos 2t\hfill \\ \hfill \sin 2t\hfill \end{array}\right),\left(\begin{array}{c}\hfill \sin 2t\hfill \\ \hfill \cos 2t\hfill \end{array}\right)\}$ (b) $S_2=\{\left(\begin{array}{c}\hfill -2\sin 2t+2\cos 2t\hfill \\ \hfill \sin 2t\hfill \end{array}\right),\left(\begin{array}{c}\hfill 4\cos 2t\hfill \\ \hfill \sin 2t-\cos 2t\hfill \end{array}\right)\}$

Solution: We first remark that the equation ${\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{cc}\hfill -2\hfill & \hfill -8\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)$ is equivalent to the system $\{\begin{array}{c}{x}^{\prime }=-2x-8y\hfill \\ y^{\prime }=x+2y\hfill \end{array}$ . (a) Differentiating we see that

${\left(\begin{array}{c}\hfill \cos 2t\hfill \\ \hfill \sin 2t\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{c}\hfill -2\sin 2t\hfill \\ \hfill 2\cos 2t\hfill \end{array}\right)\ne \left(\begin{array}{c}\hfill -2\cos 2t-8\sin 2t\hfill \\ \hfill \cos 2t+2\sin 2t\hfill \end{array}\right),$

which shows us that $\left(\begin{array}{c}\hfill \cos 2t\hfill \\ \hfill \sin 2t\hfill \end{array}\right)$ is not a solution of the system. Therefore, $S_1$ is not a fundamental set of solutions. (b) You should verify that both $\left(\begin{array}{c}\hfill -2\sin 2t+2\cos 2t\hfill \\ \hfill \sin 2t\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 4\cos 2t\hfill \\ \hfill \sin 2t-\cos 2t\hfill \end{array}\right)$ are solutions of the system. Computing the Wronskian we have

$\begin{array}{cc}\hfill & \left|\begin{array}{cc}\hfill -2\sin 2t+2\cos 2t\hfill & \hfill 4\cos 2t\hfill \\ \hfill \sin 2t\hfill & \hfill \sin 2t-\cos 2t\hfill \end{array}\right|\hfill \\ \hfill & =(-2\sin 2t+2\cos 2t)(\sin 2t-\cos 2t)\hfill \\ \hfill & -(4\cos 2t)(\sin 2t)\hfill \\ \hfill & =-2{\cos }^{2}2t-2{\sin }^{2}2t=-2.\hfill \end{array}$

Thus, the set $S_2$ is a set of two linearly independent solutions of the system and, consequently, a fundamental set of solutions.  □

Show that any linear combination of $\left(\begin{array}{c}\hfill -2\sin 2t+2\cos 2t\hfill \\ \hfill \sin 2t\hfill \end{array}\right)$ and $\left(\begin{array}{c}\hfill 4\cos 2t\hfill \\ \hfill \sin 2t-\cos 2t\hfill \end{array}\right)$ is also a solution of the system.

The following theorem implies that a fundamental set of solutions cannot contain more than n vectors, because the solutions would not be linearly independent.

Theorem 6.4

Any $n+1$ nontrivial solutions of ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ are linearly dependent.

Finally, we state the following theorems, which state that a fundamental set of solutions of ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ exists and a general solution can (theoretically) be constructed.

Theorem 6.5

There is a set of n nontrivial linearly independent solutions of ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ .

Theorem 6.6 General Solution

Let $S=\{{\mathbf{\Phi }}_1(t),{\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_n(t)\}={\left\{\left(\begin{array}{c}\hfill {\mathrm{\Phi }}_{1i}\hfill \\ \hfill {\mathrm{\Phi }}_{2i}\hfill \\ \hfill ⋮\hfill \\ \hfill {\mathrm{\Phi }}_{ni}\hfill \end{array}\right)\right\}}_{i=1}^n$ be a fundamental set of solutions of ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ on the open interval I, where each component of $\mathbf{A}(t)$ is continuous on I. Then every solution of ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ is a linear combination of these solutions. Therefore a general solution of ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ is

$\mathbf{X}(t)=c_1{\mathbf{\Phi }}_1(t)+c_2{\mathbf{\Phi }}_2(t)+\cdots +c_n{\mathbf{\Phi }}_n(t).$

Thus Examples 6.16, 6.17, and 6.18 imply that $\mathbf{X}(t)=c_1\left(\begin{array}{c}\hfill -2e^{2t}\hfill \\ \hfill e^{2t}\hfill \end{array}\right)+c_2\left(\begin{array}{c}\hfill -e^{5t}\hfill \\ \hfill 2e^{5t}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -2c_1{e}^{2t}-c_2{e}^{5t}\hfill \\ \hfill c_1{e}^{2t}+2c_2{e}^{5t}\hfill \end{array}\right)=\underset{\mathbf{\Phi }\text{ "Fundamental Matrix"}}{\underbrace{\left(\begin{array}{cc}\hfill -2e^{2t}\hfill & \hfill -e^{5t}\hfill \\ \hfill e^{2t}\hfill & \hfill 2e^{5t}\hfill \end{array}\right)}}\underset{\mathbf{C}}{\underbrace{\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \end{array}\right)}}$ is a general solution of ${\mathbf{X}}^{\prime }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right)\mathbf{X}$ .

Example 6.20

Given that ${\mathbf{\Phi }}_1(t)=\left(\begin{array}{c}\hfill t^{-1}\hfill \\ \hfill t\hfill \end{array}\right)$ and ${\mathbf{\Phi }}_1(t)=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill t^{-1}\hfill \end{array}\right)$ are solutions of ${\mathbf{X}}^{\prime }=\left(\begin{array}{cc}\hfill {(t^4-t)}^{-1}\hfill & \hfill {(t^3-1)}^{-1}\hfill \\ \hfill 2t{(1-t^3)}^{-1}\hfill & \hfill (1+t^3){(t^4-t)}^{-1}\hfill \end{array}\right)\mathbf{X}$ , find a general solution of this equation.

Solution: To verify linear independence of these two solutions, we compute the Wronskian:

$\left|\begin{array}{cc}\hfill t^{-1}\hfill & \hfill 1\hfill \\ \hfill t\hfill & \hfill t^{-1}\hfill \end{array}\right|=t^{-2}-t\ne 0.$

Hence, we have two linearly independent solutions of the equation, so a general solution is given by

$\begin{array}{cc}\hfill \mathbf{X}(t)& =c_1\left(\begin{array}{c}\hfill t^{-1}\hfill \\ \hfill t\hfill \end{array}\right)+c_2\left(\begin{array}{c}\hfill 1\hfill \\ \hfill t^{-1}\hfill \end{array}\right)\hfill \\ \hfill & =\underset{\mathbf{\Phi }\text{ "Fundamental Matrix"}}{\underbrace{\left(\begin{array}{cc}\hfill t^{-1}\hfill & \hfill 1\hfill \\ \hfill t\hfill & \hfill t^{-1}\hfill \end{array}\right)}}\underset{\mathbf{C}}{\underbrace{\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \end{array}\right)}}.\text{\hspace{1em}□}\hfill \end{array}$

Definition 6.15 Fundamental Matrix

Suppose that $S=\{{\mathbf{\Phi }}_1(t),{\mathbf{\Phi }}_2(t),\dots ,{\mathbf{\Phi }}_n(t)\}={\left\{\left(\begin{array}{c}\hfill {\mathrm{\Phi }}_{1i}\hfill \\ \hfill {\mathrm{\Phi }}_{2i}\hfill \\ \hfill ⋮\hfill \\ \hfill {\mathrm{\Phi }}_{ni}\hfill \end{array}\right)\right\}}_{i=1}^n$ is a fundamental set of solutions of ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ on the open interval I, where each component of $\mathbf{A}(t)$ is continuous on I. The matrix

$\mathbf{\Phi }(t)=\left(\begin{array}{ccc}\hfill {\mathbf{\Phi }}_1(t)\hfill & \hfill {\mathbf{\Phi }}_2(t)\hfill & \hfill \cdots {\mathbf{\Phi }}_n(t)\hfill \end{array}\right)$

is called a fundamental matrix of the system ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ on I.

The theorems and definitions introduced in this section indicate that when solving an $n\times n$ homogeneous system of linear first order equations, ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ , we find n linearly independent solutions. After finding these solutions, we form a fundamental matrix that can be used to form a general solution or solve an initial value problem.

If $\mathbf{\Phi }(t)$ is a fundamental matrix of the system ${\mathbf{X}}^{\prime }(t)=\mathbf{A}(t)\mathbf{X}(t)$ , a general solution can be written as $\mathbf{X}(t)=\mathbf{\Phi }(t)\mathbf{C}$ , where $\mathbf{C}=\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill ⋮\hfill \\ \hfill c_n\hfill \end{array}\right)$ .

Thus, $\mathbf{\Phi }(t)=\left(\begin{array}{cc}\hfill e^{-2t}\hfill & \hfill -3e^{5t}\hfill \\ \hfill 2e^{-2t}\hfill & \hfill e^{5t}\hfill \end{array}\right)$ is a fundamental matrix for the system ${\mathbf{X}}^{\prime }=\left(\begin{array}{cc}\hfill 4\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill -1\hfill \end{array}\right)\mathbf{X}$ because each column vector of Φ is a solution of the system:

$\begin{array}{cc}\hfill {\mathbf{\Phi }}^{\prime }(t)& =\left(\begin{array}{cc}\hfill -2e^{-2t}\hfill & \hfill -15e^{5t}\hfill \\ \hfill -4e^{-2t}\hfill & \hfill 5e^{5t}\hfill \end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}\hfill 4\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill -1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill e^{-2t}\hfill & \hfill -3e^{5t}\hfill \\ \hfill 2e^{-2t}\hfill & \hfill e^{5t}\hfill \end{array}\right)\hfill \end{array}$

and because these two vectors are linearly independent

$\left|\begin{array}{cc}\hfill e^{-2t}\hfill & \hfill -3e^{5t}\hfill \\ \hfill 2e^{-2t}\hfill & \hfill e^{5t}\hfill \end{array}\right|=7e^{3t}\ne 0.$

Example 6.21

Show that $\mathbf{\Phi }=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill t^{-1}\hfill & \hfill 1\hfill \\ \hfill t\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill t\hfill \end{array}\right)$ , $t>0$ , is a fundamental matrix for the system ${\mathbf{X}}^{\prime }=\begin{array}{c}\hfill \left(\begin{array}{ccc}\hfill -{(t+1)}^{-1}\hfill & \hfill -t^{-2}{(t+1)}^{-1}\hfill & \hfill t^{-1}{(t+1)}^{-1}\hfill \\ \hfill -t{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \\ \hfill {(t+1)}^{-1}\hfill & \hfill -t^{-1}{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \end{array}\right)\end{array}\mathbf{X}$ . Use this matrix to find a general solution of ${\mathbf{X}}^{\prime }=\left(\begin{array}{ccc}\hfill -{(t+1)}^{-1}\hfill & \hfill -t^{-2}{(t+1)}^{-1}\hfill & \hfill t^{-1}{(t+1)}^{-1}\hfill \\ \hfill -t{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \\ \hfill {(t+1)}^{-1}\hfill & \hfill -t^{-1}{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \end{array}\right)\mathbf{X}$ .

Solution: Because

$\begin{array}{cc}\hfill & {\left(\begin{array}{c}\hfill 0\hfill \\ \hfill t\hfill \\ \hfill 1\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)\hfill \\ \hfill & =\begin{array}{c}\hfill \left(\begin{array}{ccc}\hfill -{(t+1)}^{-1}\hfill & \hfill -t^{-2}{(t+1)}^{-1}\hfill & \hfill t^{-1}{(t+1)}^{-1}\hfill \\ \hfill -t{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \\ \hfill {(t+1)}^{-1}\hfill & \hfill -t^{-1}{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \end{array}\right)\left(\begin{array}{c}\hfill 0\hfill \\ \hfill t\hfill \\ \hfill 1\hfill \end{array}\right)\end{array},\hfill \end{array}$

$\begin{array}{cc}\hfill & {\left(\begin{array}{c}\hfill t^{-1}\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{c}\hfill -t^{-2}\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)\hfill \\ \hfill & =\begin{array}{c}\hfill \left(\begin{array}{ccc}\hfill -{(t+1)}^{-1}\hfill & \hfill -t^{-2}{(t+1)}^{-1}\hfill & \hfill t^{-1}{(t+1)}^{-1}\hfill \\ \hfill -t{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \\ \hfill {(t+1)}^{-1}\hfill & \hfill -t^{-1}{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \end{array}\right)\left(\begin{array}{c}\hfill t^{-1}\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)\end{array},\hfill \end{array}$

and

$\begin{array}{cc}\hfill & {\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill t\hfill \end{array}\right)}^{\prime }=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)\hfill \\ \hfill & =\begin{array}{c}\hfill \left(\begin{array}{ccc}\hfill -{(t+1)}^{-1}\hfill & \hfill -t^{-2}{(t+1)}^{-1}\hfill & \hfill t^{-1}{(t+1)}^{-1}\hfill \\ \hfill -t{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \\ \hfill {(t+1)}^{-1}\hfill & \hfill -t^{-1}{(t+1)}^{-1}\hfill & \hfill {(t+1)}^{-1}\hfill \end{array}\right)\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill t\hfill \end{array}\right)\end{array},\hfill \end{array}$

all three columns of Φ are solutions of the system. The solutions are linearly independent because the Wronskian of these three solutions is

$\left|\begin{array}{ccc}\hfill 0\hfill & \hfill t^{-1}\hfill & \hfill 1\hfill \\ \hfill t\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill t\hfill \end{array}\right|=-1-t\ne 0\text{for}t>0.$

A general solution of the system is given by

$\begin{array}{cc}\hfill \mathbf{X}(t)& =\mathbf{\Phi }(t)\mathbf{C}=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill t^{-1}\hfill & \hfill 1\hfill \\ \hfill t\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill t\hfill \end{array}\right)\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill c_3\hfill \end{array}\right)\hfill \\ \hfill & =c_1\left(\begin{array}{c}\hfill 0\hfill \\ \hfill t\hfill \\ \hfill 1\hfill \end{array}\right)+c_2\left(\begin{array}{c}\hfill t^{-1}\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)+c_3\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill t\hfill \end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{c}\hfill c_2{t}^{-1}+c_3\hfill \\ \hfill c_{1}t+c_2\hfill \\ \hfill c_1+c_{3}t\hfill \end{array}\right).\text{\hspace{1em}□}\hfill \end{array}$

Example 6.22

Solve $\begin{array}{c}\hfill \{\begin{array}{c}{t}^2(t+1)x^{\prime }=-t^{2}x-y+tz\hfill \\ (t+1)y^{\prime }=-tx+y+z\hfill \\ t(t+1)z^{\prime }=tx-y+tz\hfill \\ x(1)=0,y(1)=1,z(1)=2.\hfill \end{array}\end{array}$

Solution: In matrix form, the system is equivalent to that in Example 6.21 so a general solution is $x=c_2{t}^{-1}+c_3$ , $y=c_{1}t+c_2$ , and $z=c_1+c_{3}t$ . Application of the initial conditions results in $x(1)=c_2+c_3=0$ , $y(1)=c_1+c_2=1$ , and $z=c_1+c_3=2$ so $c_1=3/2$ , $c_2=-1/2$ and $c_3=1/2$ and the solution to the initial value problem is $x=-\frac{1}{2}{t}^{-1}+\frac{1}{2}$ , $y=\frac{3}{2}t-\frac{1}{2}$ , and $z=\frac{3}{2}+\frac{1}{2}t$ .  □

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The One-Electron Approximation and Beyond

Giuseppe Grosso , Giuseppe Pastori Parravicini , in Solid State Physics (Second Edition), 2014

Hartree-Fock-Slater Approximation

The Slater approximation starts from the fact that the exchange integral operator in the homogeneous free-electron gas can be well approximated with a local potential (as shown in Section 4.7). Then Slater proposes to replace, also for non-homogeneous systems, the exchange integral operator with the ordinary local potential, corresponding to the free-electron gas of the same local density $n(\mathbf{r})$ ; in this way the exchange operator takes the form

$V_{\text{exch}}^{\text{(Slater)}}(\mathbf{r})=-\frac{3}{2}\frac{e^2}{\pi }{[3{\pi }^{2}n(\mathbf{r})]}^{1/3}\text{.}$

With the above approximation, the Fock operator (4.26) becomes an ordinary differential operator with a local potential to be determined self-consistently. The method (with appropriate implementations) has been of wide use mainly in solid state physics.

The Hartree-Fock-Slater approximation becomes particularly agile when applied to atoms, because of the spherical symmetry of the (average) atomic electron density $n(r)$ . The effective atomic potential in the Hartree-Fock-Slater approximation is related to $n(r)$ in the form

(4.37) $\begin{array}{cc}\hfill V^{\text{(HFS)}}(r)=& -\frac{{\mathit{Ze}}^2}{r}+\frac{e^2}{r}{\int }_0^{r}4\pi r^{\prime 2}n(r^{\prime }){\mathit{dr}}^{\prime }\hfill \\ \hfill & +e^2{\int }_r^{\infty }4\pi r^{\prime }n(r^{\prime }){\mathit{dr}}^{\prime }-\frac{3}{2}\frac{e^2}{\pi }{[3{\pi }^{2}n(r)]}^{1/3}\text{.}\hfill \end{array}$

The first term in the right-hand side of Eq. (4.37) is the nuclear potential, the second and the third are the Hartree potential of the spherically symmetric charge distribution $(-e)n(r)$ ; the last is the Slater local exchange potential [to be precise the potential (4.37) should be appropriately corrected to preserve the asymptotic behavior $-e^2/r$ , when acting on occupied wavefunctions]. Usually the Numerov method is adopted to solve the radial Schrödinger equation; self-consistency of wavefunctions and potential can be achieved with reasonable modest computational labor. Complete calculations (program code included) for all atoms is reported, for instance, by F. Herman and S. Skillman "Atomic Structure Calculations" (Prentice Hall, Englewood Cliffs, New Jersey, 1963).

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Systems of Differential Equations

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (Fourth Edition), 2014

Undetermined Coefficients

We use the method of undetermined coefficients to find a particular solution X _p to a nonhomogeneous linear system with constant coefficient matrix in much the same way as we approached nonhomogeneous higher order linear equations with constant coefficients in Chapter 4. The main difference is that the coefficients are constant vectors when we work with systems. For example, if we consider X′   = AX  + F(t) where the entries of A are constant and $\mathbf{F}\left(t\right)=\left(\begin{array}{c}\hfill {\mathrm{e}}^{-2t}\hfill \\ \hfill 4\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right){\mathrm{e}}^{-2t}+\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 4\hfill \end{array}\right)$ and none of the terms in F(t) satisfy the corresponding homogeneous system X′   = AX, we assume that a particular solution has the form X _p(t)   = ae^{−   2t}   + b where a and b are constant vectors. On the other hand if λ   =   2 is an eigenvalue of A (with multiplicity one), we assume that X _p(t)   = a te^{−   2t}   + be^{−   2t}   + c. (Note that even in this situation, a could be the zero vector.)

For constant vectors, a, b, c, …, our convention will be that $\mathbf{a}=\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \\ \hfill ⋮\hfill \\ \hfill a_n\hfill \end{array}\right),\mathbf{b}=\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \\ \hfill ⋮\hfill \\ \hfill b_n\hfill \end{array}\right),\mathbf{c}=\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill ⋮\hfill \\ \hfill c_n\hfill \end{array}\right),\dots$ .

Example 6.5.1

Solve ${\mathbf{X}}^{\mathbf{\prime }}=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right)\mathbf{X}+\left(\begin{array}{c}\hfill {\mathrm{e}}^{3t}\hfill \\ \hfill t\hfill \end{array}\right)$ .

Solution

In this case, $\mathbf{F}\left(t\right)=\left(\begin{array}{c}\hfill {\mathrm{e}}^{3t}\hfill \\ \hfill t\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right){\mathrm{e}}^{3t}+\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)t$ and a general solution of the corresponding homogeneous system is ${\mathbf{X}}_{\mathrm{h}}\left(t\right)=c_1\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \end{array}\right){\mathrm{e}}^{4t}+c_2\left(\begin{array}{c}\hfill -2\hfill \\ \hfill 1\hfill \end{array}\right){\mathrm{e}}^{4t}$ . Notice that none of the components of F(t) are in X _h(t), so we assume that there is a particular solution of the form X _p(t)   = ae^3t   + b t  + c. Then, X′_p  =   3ae^3t   + b and substitution into the nonhomogeneous system X′   = AX  + F(t), where $\mathbf{A}=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right)$ , yields

$3\mathbf{a}{\mathrm{e}}^{3t}+\mathbf{b}=\mathbf{Aa}{\mathrm{e}}^{3t}+\mathbf{Ab}t+\mathbf{Ac}+\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right){\mathrm{e}}^{3t}+\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)t.$

Collecting like terms, we obtain the system of equations

$\begin{array}{l}3\mathbf{a}=\mathbf{Aa}+\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)\left(\mathrm{Cofficient}\mathrm{of}{\mathrm{e}}^{3t}\right),\\ \mathbf{b}=\mathbf{Ac}\left(\mathrm{Constant}\mathrm{terms}\right),\\ \mathbf{Ab}+\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)=0\left(\mathrm{Cofficient}\mathrm{of}t\right)\text{.}\end{array}$

From the coefficients of e^3t , we find that $\left(\mathbf{A}-3\mathbf{I}\right)\mathbf{a}=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)$ or $\left(\begin{array}{cc}\hfill -3\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill -3\hfill \end{array}\right)\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -1\hfill \\ \hfill 0\hfill \end{array}\right)$ . This system has the unique solution $\mathbf{a}=\left(\begin{array}{c}\hfill -3/7\hfill \\ \hfill -2/7\hfill \end{array}\right)$ . Next, we solve the system $\mathbf{Ab}+\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)=\mathbf{0}$ or $\left(\begin{array}{cc}\hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{c}\hfill b_1\hfill \\ \hfill b_2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill -1\hfill \end{array}\right)$ for b. This yields the unique solution $\mathbf{b}=\left(\begin{array}{c}\hfill -1/2\hfill \\ \hfill 0\hfill \end{array}\right)$ . Finally, we solve b  = Ac or $\left(\begin{array}{cc}\hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -1/2\hfill \\ \hfill 0\hfill \end{array}\right)$ for c, which gives us $\mathbf{c}=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill -1/16\hfill \end{array}\right)$ . A particular solution to the nonhomogeneous system is then

$\begin{array}{l}{\mathbf{X}}_{\mathrm{p}}\left(t\right)=\mathbf{a}{\mathrm{e}}^{3t}+\mathbf{b}t+\mathbf{c}=\left(\begin{array}{c}\hfill -3/7\hfill \\ \hfill -2/7\hfill \end{array}\right){\mathrm{e}}^{3t}+\left(\begin{array}{c}\hfill -1/2\hfill \\ \hfill 0\hfill \end{array}\right)t\\ +\left(\begin{array}{c}\hfill 0\hfill \\ \hfill -1/16\hfill \end{array}\right)\text{,}\end{array}$

so a general solution to the nonhomogeneous system is

$\mathbf{X}\left(t\right)={\mathbf{X}}_{\mathrm{h}}\left(t\right)+{\mathbf{X}}_{\mathrm{p}}\left(t\right)=\left(\begin{array}{c}\hfill 2c_1{\mathrm{e}}^{4t}-2c_2{\mathrm{e}}^{-4t}-\frac{3}{7}{\mathrm{e}}^{3t}-\frac{1}{2}t\hfill \\ \hfill c_1{\mathrm{e}}^{4t}+c_2{\mathrm{e}}^{-4t}-\frac{2}{7}{\mathrm{e}}^{3t}-\frac{1}{16}\hfill \end{array}\right).$

To give another illustration of how the form of a particular solution is selected, suppose that $\mathbf{F}\left(t\right)=\left(\begin{array}{c}\hfill 4sin2t\hfill \\ \hfill {\mathrm{e}}^{-t}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \end{array}\right)sin2t+\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right){\mathrm{e}}^{-t}$ in Example 6.5.1. In this case, we assume that X _p(t)   = a cos 2 t  + b sin 2 t  + ce^{− t} and find the vectors a, b, and c through substitution into the nonhomogeneous system.

Find a particular solution to ${\mathbf{X}}^{\mathbf{\prime }}=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right)\mathbf{X}+\left(\begin{array}{c}\hfill 4sin2t\hfill \\ \hfill e^{-t}\hfill \end{array}\right)$ .

Example 6.5.2

Solve X′   = AX  + F(t) if $\mathbf{A}=\left(\begin{array}{cc}\hfill -3\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 3\hfill \end{array}\right)$ and (a) $\mathbf{F}\left(t\right)=\left(\begin{array}{c}\hfill -3\hfill \\ \hfill -1\hfill \end{array}\right){\mathrm{e}}^t$ and (b) $\mathbf{F}\left(t\right)=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill -2\hfill \end{array}\right){\mathrm{e}}^t$ .

Solution

A general solution of the corresponding homogeneous system ${\mathbf{X}}^{\mathbf{\prime }}=\left(\begin{array}{cc}\hfill -3\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 3\hfill \end{array}\right)\mathbf{X}$ is ${\mathbf{X}}_{\mathrm{h}}\left(t\right)=c_1\left(\begin{array}{c}\hfill 2\hfill \\ \hfill 1\hfill \end{array}\right){\mathrm{e}}^{-t}+c_2\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right){\mathrm{e}}^t$ . (a) Because an e ^t term is a solution to the corresponding homogeneous system, we search for a particular solution of the nonhomogeneous system of the form X _p(t)   = a te ^t   + be ^t . Differentiating gives us X′_p  = a te ^t   +   (a  + b)e ^t and substituting into the nonhomogeneous system results in

$\begin{array}{l}\mathbf{a}t{\mathrm{e}}^t+\left(\mathbf{a}+\mathbf{b}\right){\mathrm{e}}^t=\mathbf{A}\left(\mathbf{a}t{\mathrm{e}}^t+\mathbf{b}{\mathrm{e}}^t\right)+\left(\begin{array}{c}\hfill -3\hfill \\ \hfill -1\hfill \end{array}\right){\mathrm{e}}^t\\ \mathbf{a}t{\mathrm{e}}^t+\left(\mathbf{a}+\mathbf{b}\right){\mathrm{e}}^t=\mathbf{Aa}t{\mathrm{e}}^t+\left(\mathbf{Ab}+\left(\begin{array}{c}\hfill -3\hfill \\ \hfill -1\hfill \end{array}\right)\right){\mathrm{e}}^t.\end{array}$

Equating coefficients gives us the system

$\begin{array}{l}\mathbf{a}=\mathbf{Aa}\hfill \\ \mathbf{a}+\mathbf{b}=\mathbf{AB}+\left(\begin{array}{c}\hfill -3\hfill \\ \hfill -1\hfill \end{array}\right)\hfill \end{array}\mathrm{or}\begin{array}{l}\left(\mathbf{A}-\mathbf{I}\right)\mathbf{a}=\mathbf{0}\hfill \\ \mathbf{a}-\left(\mathbf{A}-\mathbf{I}\right)\mathbf{b}=\left(\begin{array}{c}\hfill -3\hfill \\ \hfill -1\hfill \end{array}\right).\hfill \end{array}$

The equation (A  − I)a  = 0 has solution a  = c v where v is an eigenvector corresponding to the eigenvalue λ   =   1. Choosing $\mathbf{v}=\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right)$ gives us $\mathbf{a}=c\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right)$ . The equation $\left(\mathbf{A}-\mathbf{I}\right)\mathbf{b}=\left(\begin{array}{l}3\hfill \\ 1\hfill \end{array}\right)+\mathbf{a}$ or $\left(\begin{array}{cc}\hfill -4\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 2\hfill \end{array}\right)\left(\begin{array}{l}{b}_1\hfill \\ b_2\hfill \end{array}\right)=\left(\begin{array}{l}3\hfill \\ 1\hfill \end{array}\right)+c\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right)$ has zero solutions unless 3   + c  =   2(1   + c) which yields c  =   1. With c  =   1, $\left(\begin{array}{cc}\hfill -4\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 2\hfill \end{array}\right)\left(\begin{array}{l}{b}_1\hfill \\ b_2\hfill \end{array}\right)=\left(\begin{array}{l}4\hfill \\ 2\hfill \end{array}\right)$ so   − b ₁  + b ₂  =   1. Choosing b ₁  =   0 and b ₂  =   1 gives us $\mathbf{b}=\left(\begin{array}{l}0\hfill \\ 1\hfill \end{array}\right)$ . Then a particular solution of the nonhomogeneous system is

${\mathbf{X}}_{\mathrm{p}}\left(t\right)=\mathbf{a}t{\mathrm{e}}^t+\mathbf{b}{\mathrm{e}}^t=\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right)t{\mathrm{e}}^t+\left(\begin{array}{l}0\hfill \\ 1\hfill \end{array}\right){\mathrm{e}}^t$

and a general solution is

$\begin{array}{l}\mathbf{X}\left(t\right)={\mathbf{X}}_{\mathrm{h}}\left(t\right)+{\mathbf{X}}_{\mathrm{p}}\left(t\right)=c_1\left(\begin{array}{l}2\hfill \\ 1\hfill \end{array}\right){\mathrm{e}}^{-t}+c_2\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right){\mathrm{e}}^t\\ +\left(\begin{array}{l}1\\ 1\end{array}\right)t{\mathrm{e}}^t+\left(\begin{array}{l}0\hfill \\ 1\hfill \end{array}\right){\mathrm{e}}^t\text{.}\end{array}$

(b) Proceeding in the same manner as in (a), we assume that a particular solution of the nonhomogeneous system has the form X _p(t)   = a te ^t   + be ^t . Differentiating gives us X′_p  = a te ^t   +   (a  + b)e ^t and substituting into the nonhomogeneous system results in

$\begin{array}{l}\mathbf{a}t{\mathrm{e}}^t+\left(\mathbf{a}+\mathbf{b}\right){\mathrm{e}}^t=\mathbf{A}\left(\mathbf{a}t{\mathrm{e}}^t+\mathbf{b}{\mathrm{e}}^t\right)+\left(\begin{array}{c}\hfill -4\hfill \\ \hfill -2\hfill \end{array}\right){\mathrm{e}}^t\\ \mathbf{a}t{\mathrm{e}}^t+\left(\mathbf{a}+\mathbf{b}\right){\mathrm{e}}^t=\mathbf{Aa}t{\mathrm{e}}^t+\left(\mathbf{Ab}+\left(\begin{array}{c}\hfill -4\hfill \\ \hfill -2\hfill \end{array}\right)\right){\mathrm{e}}^t.\end{array}$

Equating coefficients gives us the system

$\begin{array}{l}\mathbf{a}=\mathbf{Aa}\hfill \\ \mathbf{a}+\mathbf{b}=\mathbf{AB}+\left(\begin{array}{c}\hfill -4\hfill \\ \hfill -2\hfill \end{array}\right)\hfill \end{array}\mathrm{or}\begin{array}{l}\left(\mathbf{A}-\mathbf{I}\right)\mathbf{a}=\mathbf{0}\hfill \\ \mathbf{a}-\left(\mathbf{A}-\mathbf{I}\right)\mathbf{b}=\left(\begin{array}{c}\hfill -4\hfill \\ \hfill -2\hfill \end{array}\right).\hfill \end{array}$

The equation (A  − I)a  = 0 has solution a  = c v where v is an eigenvector corresponding to the eigenvalue λ   =   1. Choosing $\mathbf{v}=\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right)$ gives us $\mathbf{a}=c\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right)$ . The equation $\left(\mathbf{A}-\mathbf{I}\right)\mathbf{b}=\left(\begin{array}{l}3\hfill \\ 1\hfill \end{array}\right)+\mathbf{a}$ or $\left(\begin{array}{cc}\hfill -4\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 2\hfill \end{array}\right)\left(\begin{array}{l}{b}_1\hfill \\ b_2\hfill \end{array}\right)=\left(\begin{array}{l}4\hfill \\ 2\hfill \end{array}\right)+c\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right)$ has zero solutions unless 4   + c  =   2(2   + c) which yields c  =   0. With c  =   0, $\left(\begin{array}{cc}\hfill -4\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 2\hfill \end{array}\right)\left(\begin{array}{l}{b}_1\hfill \\ b_2\hfill \end{array}\right)=\left(\begin{array}{l}4\hfill \\ 2\hfill \end{array}\right)$ so   − b ₁  + b ₂  =   1 Choosing b ₁  =   0 and b ₂  =   1 gives us $\mathbf{b}=\left(\begin{array}{l}0\hfill \\ 1\hfill \end{array}\right)$ . Then a particular solution of the nonhomogeneous system is

${\mathbf{X}}_{\mathrm{p}}\left(t\right)=\mathbf{a}t{\mathrm{e}}^t+\mathbf{b}{\mathrm{e}}^t=\left(\begin{array}{l}0\hfill \\ 1\hfill \end{array}\right){\mathrm{e}}^t$

and a general solution is

$\begin{array}{l}\mathbf{X}\left(t\right)={\mathbf{X}}_{\mathrm{h}}\left(t\right)+{\mathbf{X}}_{\mathrm{p}}\left(t\right)=c_1\left(\begin{array}{l}2\hfill \\ 1\hfill \end{array}\right){\mathrm{e}}^{-t}\\ +c_2\left(\begin{array}{l}1\hfill \\ 1\hfill \end{array}\right){\mathrm{e}}^t+\left(\begin{array}{l}0\hfill \\ 1\hfill \end{array}\right){\mathrm{e}}^t\text{.}\end{array}$

Find a particular solution to X′

 

=

AX

+

F(t) if $\mathbf{A}=\left(\begin{array}{cc}\hfill -3\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 3\hfill \end{array}\right)$ and $\mathbf{F}\left(t\right)=\left(\begin{array}{l}-3\hfill \\ -3\hfill \end{array}\right){\mathrm{e}}^t$ .

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Applications related to ordinary and partial differential equations

Martha L. Abell , James P. Braselton , in Mathematica by Example (Sixth Edition), 2022

6.4.1.1 Homogeneous linear systems

The corresponding homogeneous system of Eq. (6.28) is

(6.29) ${\mathbf{X}}^{\prime }=\mathbf{AX}.$

In the same way as with the previously discussed linear equations, a general solution of Eq. (6.28) is $\mathbf{X}={\mathbf{X}}_h+{\mathbf{X}}_p$ , where ${\mathbf{X}}_h$ is a general solution of Eq. (6.29), and ${\mathbf{X}}_p$ is a particular solution of the nonhomogeneous system equation (6.28).

A particular solution to a system of ordinary differential equations is a set of functions that satisfy the system, but do not contain any arbitrary constants. That is, a particular solution to a system is a set of specific functions, containing no arbitrary constants, that satisfy the system.

If ${\mathbf{\Phi }}_1=\left(\begin{array}{c}\hfill {\varphi }_{11}\hfill \\ \hfill {\varphi }_{21}\hfill \\ \hfill ⋮\hfill \\ \hfill {\varphi }_{n1}\hfill \end{array}\right)$ , ${\mathbf{\Phi }}_2=\left(\begin{array}{c}\hfill {\varphi }_{12}\hfill \\ \hfill {\varphi }_{22}\hfill \\ \hfill ⋮\hfill \\ \hfill {\varphi }_{n2}\hfill \end{array}\right)$ , …, ${\mathbf{\Phi }}_n=\left(\begin{array}{c}\hfill {\varphi }_{1n}\hfill \\ \hfill {\varphi }_{2n}\hfill \\ \hfill ⋮\hfill \\ \hfill {\varphi }_{nn}\hfill \end{array}\right)$ are n linearly independent solutions of Eq. (6.29); a general solution of Eq. (6.29) is

$\mathbf{X}=c_1{\mathbf{\Phi }}_1+c_2{\mathbf{\Phi }}_2+\dots +c_n{\mathbf{\Phi }}_n=\left(\begin{array}{cccc}\hfill {\mathbf{\Phi }}_1\hfill & \hfill {\mathbf{\Phi }}_2\hfill & \hfill \dots \hfill & \hfill {\mathbf{\Phi }}_n\hfill \end{array}\right)\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill ⋮\hfill \\ \hfill c_n\hfill \end{array}\right)=\mathbf{\Phi }\mathbf{C},$

where

$\mathbf{\Phi }=\left(\begin{array}{cccc}\hfill {\mathbf{\Phi }}_1\hfill & \hfill {\mathbf{\Phi }}_2\hfill & \hfill \dots \hfill & \hfill {\mathbf{\Phi }}_n\hfill \end{array}\right)\text{and}\mathbf{C}=\left(\begin{array}{c}\hfill c_1\hfill \\ \hfill c_2\hfill \\ \hfill ⋮\hfill \\ \hfill c_n\hfill \end{array}\right).$

$\mathbf{\Phi }=\left(\begin{array}{cccc}\hfill {\varphi }_{11}\hfill & \hfill {\varphi }_{12}\hfill & \hfill \dots \hfill & \hfill {\varphi }_{1n}\hfill \\ \hfill {\varphi }_{21}\hfill & \hfill {\varphi }_{22}\hfill & \hfill \dots \hfill & \hfill {\varphi }_{2n}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \dots \hfill & \hfill ⋮\hfill \\ \hfill {\varphi }_{n1}\hfill & \hfill {\varphi }_{n2}\hfill & \hfill \dots \hfill & \hfill {\varphi }_{nn}\hfill \end{array}\right)$ is called a fundamental matrix for Eq. (6.29). If Φ is a fundamental matrix for Eq. (6.29), ${\mathbf{\Phi }}^{\prime }=\mathbf{A}\mathbf{\Phi }$ or ${\mathbf{\Phi }}^{\prime }-\mathbf{A}\mathbf{\Phi }=\mathbf{0}$ .

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Systems of Ordinary Differential Equations

Martha L. Abell , James P. Braselton , in Differential Equations with Mathematica (Fourth Edition), 2016

6.4.1 Undetermined Coefficients

We use the method of undetermined coefficients to find a particular solution of a nonhomogeneous system in much the same way as we approached nonhomogeneous higher-order equations in Chapter 4. The main difference is that the coefficients are constant vectors when we work with systems.

Example 6.4.1

Solve $\left\{\begin{array}{l}{x}^{\prime }=2x+y+sin3t\\ y^{\prime }=-8x-2y\\ x(0)=0,y(0)=1\end{array}\right.$ .

Solution

In matrix form, the system is equivalent to ${\mathbf{X}}^{\prime }=\left(\begin{array}{ll}\hfill 2\hfill & \hfill 1\hfill \\ \hfill -8\hfill & \hfill -2\hfill \end{array}\right)\mathbf{X}+\left(\begin{array}{l}\hfill sin3t\hfill \\ \hfill 0\hfill \end{array}\right)$ . We find a general solution of the corresponding homogeneous system ${\mathbf{X}}^{\prime }=\left(\begin{array}{ll}\hfill 2\hfill & \hfill 1\hfill \\ \hfill -8\hfill & \hfill -2\hfill \end{array}\right)\mathbf{X}$ with DSolve.

Clear[x, y, xh, xp] homsol=DSolve[{x′[t]==2x[t]+y[t], y′[t]==−8x[t]−2y[t]}, {x[t], y[t]}, t]

$\{\{x[t]\to \frac{1}{2}C[2]\text{Sin}[2t]+C[1](\text{Cos}[2t]+\text{Sin}[2t]),y[t]\to C[2](\text{Cos}[2t]-\text{Sin}[2t])-4C[1]\text{Sin}[2t]\}\}$

These results indicate that a general solution of the corresponding homogeneous system is

$\begin{array}{l}\hfill {\mathbf{X}}_h=\left(\begin{array}{ll}\hfill cos2t+sin2t\hfill & \hfill \frac{1}{2}sin2t\hfill \\ \hfill 4sin2t\hfill & \hfill cos2t-sin2t\hfill \end{array}\right)\left(\begin{array}{l}\hfill c_1\hfill \\ \hfill c_2\hfill \end{array}\right).\end{array}$

xh[t_]={{x[t]}, {y[t]}}/.homsol[[1]];

Thus, we search for a particular solution of the nonhomogeneous system of the form ${\mathbf{X}}_p=\mathbf{a}sin3t+\mathbf{b}cos3t$ , where $\mathbf{a}=\left(\begin{array}{l}\hfill a_1\hfill \\ \hfill a_2\hfill \end{array}\right)$ and $\mathbf{b}=\left(\begin{array}{l}\hfill b_1\hfill \\ \hfill b_2\hfill \end{array}\right)$ . After defining $\mathbf{A}=\left(\begin{array}{ll}\hfill 2\hfill & \hfill 1\hfill \\ \hfill -8\hfill & \hfill -2\hfill \end{array}\right)$ and ${\mathbf{X}}_p=\mathbf{a}sin3t+\mathbf{b}cos3t$ , we substitute X _p into the nonhomogeneous system.

capa={{2, 1}, {−8, −2}}; MatrixForm[capa]

$\left(\begin{array}{ll}2& 1\\ -8& -2\\ \end{array}\right)$

xp[t_]={{a1}, {a2}}Sin[3t]+{{b1}, {b2}}Cos[3t]; MatrixForm[xp[t]]

$\left(\begin{array}{l}\text{b1}\text{Cos}[3t]+\text{a1}\text{Sin}[3t]\\ \text{b2}\text{Cos}[3t]+\text{a2}\text{Sin}[3t]\\ \end{array}\right)$

step1= xp ′[t]==capa.xp[t]+{{Sin[3t]}, {0}}

{{3a1Cos[3t]−3b1Sin[3t]}, {3a2Cos[3t]−3b2Sin[3t]}} == {{b2Cos[3t]+Sin[3t]+a2Sin[3t]+2(b1Cos[3t]+a1Sin[3t])}, {−8(b1Cos[3t] + a1Sin[3t]) − 2(b2Cos[3t] + a2Sin[3t])}}

The result represents a system of equations that is true for all values of t. In particular, substituting t = 0 yields

eq1=step1/.t→0

{{3a1}, {3a2}} == {{2b1 + b2}, {−8b1 − 2b2}}

which is equivalent to the system of equations

$\begin{array}{l}\hfill \left\{\begin{array}{l}3a_1=2b_1+b_2\\ 3a_2=-2\left(4b_1+b_2\right)\end{array}\right..\end{array}$

Similarly, substituting t = π/2 results in

eq2=step1/.t→Pi/2

{{3b1}, {3b2}} == {{−1 − 2a1 −a2}, {8a1 + 2a2}}

which is equivalent to the system of equations

$\begin{array}{l}\hfill \left\{\begin{array}{l}3b_1=-1-2a_1-a_2\\ 3b_2-2\left(-4a_1-a_2\right)\end{array}\right..\end{array}$

We now use Solve to solve these four equations for a ₁, a ₂, b ₁, and b ₂

coeffs=Solve[{eq1, eq2}]

$\{\{\text{a1}\to -\frac{2}{5},\text{a2}\to \frac{8}{5},\text{b1}\to -\frac{3}{5},\text{b2}\to 0\}\}$

and then substitute these values into X _p to obtain a particular solution to the nonhomogeneous system.

xp[t_]=xp[t]/.coeffs[[1]]

$\{\{-\frac{3}{5}\text{Cos}[3t]-\frac{2}{5}\text{Sin}[3t]\},\{\frac{8}{5}\text{Sin}[3t]\}\}$

A general solution to the nonhomogeneous system is then given by X =X _h +X _p .

xh[t]

$\{\{\frac{1}{2}C[2]\text{Sin}[2t]+C[1](\text{Cos}[2t]+\text{Sin}[2t])\},\{C[2](\text{Cos}[2t]-\text{Sin}[2t])-4C[1]\text{Sin}[2t]\}\}$

x[t_]=xh[t]+xp[t]

$\{\{-\frac{3}{5}\text{Cos}[3t]+\frac{1}{2}C[2]\text{Sin}[2t]+C[1](\text{Cos}[2t]+\text{Sin}[2t])-\frac{2}{5}\text{Sin}[3t]\},\{C[2](\text{Cos}[2t]-\text{Sin}[2t])-4C[1]\text{Sin}[2t]+\frac{8}{5}\text{Sin}[3t]\}\}$

To solve the initial value problem, we apply the initial condition and solve for the unknown constants.

x[0]

$\{\{-\frac{3}{5}+C[1]\},\{C[2]\}\}$

cvals=Solve[x[0]=={{0}, {1}}]

$\{\{C[1]\to \frac{3}{5},C[2]\to 1\}\}$

We obtain the solution to the initial value problem by substituting these values back into the general solution.

x[t_]=x[t]/.cvals[[1]]//Flatten//Simplify

$\{\frac{1}{10}(6\text{Cos}[2t]-6\text{Cos}[3t]+11\text{Sin}[2t]-4\text{Sin}[3t]),\text{Cos}[2t]-\frac{17}{5}\text{Sin}[2t]+\frac{8}{5}\text{Sin}[3t]\}$

We confirm this result by graphing x(t) and y(t) together in Figure 6-19 (a) as well as parametrically in B.

Figure 6-19. (a) x(t) and y(t). (b) Parametric plot of x(t) versus y(t)

p1=Plot[Evaluate[x[t]], {t, 0, 4Pi}, PlotRange→{−2Pi, 2Pi}, AspectRatio→1, PlotLabel→"(a)"]

p2=ParametricPlot[x[t], {t, 0, 4Pi}, PlotRange→{{−6, 5}, {−5, 6}}, AspectRatio→1, PlotLabel→"(b)"]Show[GraphicsRow[{p1, p2}]]

Finally, we note that DSolve is able to find a general solution of the nonhomogeneous system

Clear[x, y, t]Clear[x, y, xh, xp]homsol=DSolve[{x′[t]==2x[t]+y[t], y′[t]==−8x[t]−2y[t]}, {x[t], y[t]}, t]

$\{\{x[t]\to \frac{1}{2}C[2]\text{Sin}[2t]+C[1](\text{Cos}[2t]+\text{Sin}[2t]),y[t]\to C[2](\text{Cos}[2t]-\text{Sin}[2t])-4C[1]\text{Sin}[2t]\}\}$

as well as solve the initial value problem.

Clear[x, y, xh, xp]homsol=DSolve[{x′[t]==2x[t]+y[t], y′[t]==−8x[t]−2y[t], x[0]==0, y[0]==1}, {x[t], y[t]}, t]

$\{\{x[t]\to \frac{1}{2}\text{Sin}[2t],y[t]\to \text{Cos}[2t]-\text{Sin}[2t]\}\}$

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The Algebraic Eigenvalue Problem

William Ford , in Numerical Linear Algebra with Applications, 2015

Applications of the Eigenvalue Problem

Resonance is the tendency of a system to oscillate at a greater amplitude at some frequencies than at others. With little or no damping, if a periodic driving force matches a natural frequency of vibration, oscillations of large amplitude can occur. The are many types of resonance, including mechanical resonance, acoustic resonance, electromagnetic resonance, and nuclear magnetic resonance. The phenomenon is illustrated by a mass-spring system that is modeled by a system of second-order ordinary differential equations. The solution to the system with no driving force (homogeneous system) depends on eigenvalue and eigenvector computations, that give rise to the natural frequencies of vibration. After adding a driving force and solving the nonhomogeneous system, resonance is illustrated by choosing the frequency of the driving force to be close to a natural frequency.

The model of Leslie is a heavily used tool in population ecology. It models an age-structured population which predicts how distinct populations change over time. The heart of the model is the Leslie matrix, which is irreducible and has an eigenvector with strictly positive entries. Its characteristic function has exactly one real positive root, λ₁, the largest eigenvalue of the Leslie matrix in magnitude. All the other eigenvalues are negative or imaginary. By studying powers of the Leslie matrix, we find that at equilibrium the proportion of individuals belonging to each age class will remain constant, and the number of individuals will increase by λ₁ times each period. The eigenvector with all positive entries can be used to determine the percentage of females in each age class after an extended period of time.

The buckling of a elastic column is determined by solving a boundary value problem. The column will not buckle unless subjected to a critical load, in which case the deflection curve is of the form $k\text{sin}\left(\frac{n\pi x}{L}\right)$ , where L is the length of the column and k is a constant. The functions are termed eigenfunctions, and the corresponding eigenvalues are ${\text{λ}}_n=\frac{n^2{\pi }^2}{L^2}$ . Associated with the eigenvalues are the critical loads $P_n=\frac{EI{\pi }^2{n}^2}{L^2}$ , the only forces that will cause buckling.

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Applications of Systems of Ordinary Differential Equations

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (Fourth Edition), 2014

Mixture Problems

Consider the interconnected tanks that are shown in Figure 7.7 in which a solution with a given concentration of some substance (like salt) is allowed to flow according to the given information. Let x(t) and y(t) represent the amount of the substance in Tank 1 and Tank 2, respectively. Using this information, we set up two differential equations to describe the rate at which x and y change with respect to time. Notice that the rate at which liquid flows into each tank equals the rate at which it flows out, so the volume of liquid in each tank remains constant. If we consider Tank 1, we can determine a first-order differential equation for dx/dt with

Figure 7.7. Illustrating a mixture problem for two interconnected tanks.

$\begin{array}{c}\frac{\mathrm{d}x}{\mathrm{d}y}=\left(\mathrm{Rate}\mathrm{at}\mathrm{which}\mathrm{substance}\mathrm{enters}\mathrm{Tank}1\right)\\ -\left(\mathrm{Rate}\mathrm{at}\mathrm{which}\mathrm{substance}\mathrm{leaves}\mathrm{Tank}1\right),\end{array}$

where the rate at which the substance enters Tank I is R gal/min   × C lb/gal   = RC lb/min, and the rate at which it leaves is R gal/min   × x/V ₁  lb/gal =   Rx/V ₁  lb/min, where x/V ₁ is the substance concentration in Tank 1. Therefore, dx/dt   =   RC – (R/V ₁)x. Similarly, we find dy/dt to be $\frac{\mathrm{d}y}{\mathrm{d}t}=\left(R/V_1\right)x-\left(R/V_2\right)y.$ We use the initial conditions x(0)   = x ₀ and y(0)   = y ₀ to solve the nonhomogeneous system

(7.6) $\begin{array}{l}\frac{\mathrm{d}x}{\mathrm{d}t}=\mathit{RC}-\frac{R}{V_1}x,\\ \frac{\mathrm{d}y}{\mathrm{d}t}=\frac{R}{V_1}x-\frac{R}{V_2}y,\end{array}$

for x(t) and y(t).

In deriving this system of equations, we used rates in gal/min and concentrations in lb/gal. In general, rates are given by (volume of liquid)/(time) and (concentration of substance)/(volume of liquid).

Example 7.2.2

Determine the amount of salt in each tank in Figure 7.7 if V ₁  = V ₂  =   500   gal, R   =  5   gal/min, C   =  3   lb /gal, x ₀  =   50   lb and y ₀  =   100   lb.

Solution

In this case, the IVP is

$\begin{array}{l}\frac{\mathrm{d}x}{\mathrm{d}t}=5\times 3-\frac{5}{500}x=15-\frac{1}{100}x\\ \frac{\mathrm{d}y}{\mathrm{d}t}=\frac{5}{500}x-\frac{5}{500}y=\frac{1}{100}x-\frac{1}{100}y,\\ x\left(0\right)=50,y\left(0\right)=100,\end{array}$

which in matrix form is. $\mathbf{X}\text{'}=\left(\begin{array}{cc}\hfill -1/100\hfill & \hfill 0\hfill \\ \hfill 1/100\hfill & \hfill -1/100\hfill \end{array}\right)\mathbf{X}\mathbf{+}\left(\begin{array}{c}\hfill 15\hfill \\ \hfill 0\hfill \end{array}\right)=\mathbf{AX}\mathbf{+}\mathbf{F}\left(t\right),\mathbf{X}\left(0\right)=\left(\begin{array}{c}\hfill 50\hfill \\ \hfill 100\hfill \end{array}\right).$ The matrix A has the repeated eigenvalue λ_1,2  =   –1/100, for which we can find only one (linearly independent) eigenvector, ${\mathrm{v}}_1=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)$ . Therefore, one solution of the corresponding homogenous system is ${\mathrm{X}}_1\left(t\right)=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right){\mathrm{e}}^{-t/100},$ and after some work a second linearly independent solution is found to be ${\mathrm{X}}_2\left(t\right)=\left[\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)t+\left(\begin{array}{c}\hfill 100\hfill \\ \hfill 0\hfill \end{array}\right)\right]{\mathrm{e}}^{-t/100},$ so a general solution of the corresponding homogeneous system is

$\begin{array}{c}{\mathbf{X}}_{\mathrm{h}}\left(t\right)=c_1\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right){\mathrm{e}}^{-t/100}+c_2\left[\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)t+\left(\begin{array}{c}\hfill 100\hfill \\ \hfill 0\hfill \end{array}\right)\right]{\mathrm{e}}^{-t/100}\\ =\left(\begin{array}{c}\hfill {100}_{c2{\mathrm{e}}^{\mathit{‐}t/100}}\hfill \\ \hfill c_{1^{\mathrm{e}‐t/100}}{c}_{2^{t{\mathrm{e}}^{-t/100}}}\hfill \end{array}\right).\end{array}$

Notice that $\mathrm{F}\left(t\right)=\left(\begin{array}{c}\hfill 15\hfill \\ \hfill 0\hfill \end{array}\right)$ is not a solution to the corresponding homogeneous system, so with the method of undetermined coefficients we assume a particular solution of the nonhomogeneous system has the form ${\mathbf{X}}_{\mathrm{p}}\left(t\right)=\mathrm{a}=\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \end{array}\right)$ and substitute this vector-valued function into the nonhomogeneous equation X′   =   AX   +   F(t). This yields

$\begin{array}{c}\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill -1/100\hfill & \hfill 0\hfill \\ \hfill 1/100\hfill & \hfill -1/100\hfill \end{array}\right)\left(\begin{array}{c}\hfill a_1\hfill \\ \hfill a_2\hfill \end{array}\right)+\left(\begin{array}{c}\hfill 15\hfill \\ \hfill 0\hfill \end{array}\right)\\ =\left(\begin{array}{c}\hfill -a_1/100+15\hfill \\ \hfill a_1/100-a_2/100\hfill \end{array}\right)\end{array}$

with solution a ₁  =   1500 and a ₂  =   1500. Therefore, ${\mathbf{X}}_{\mathrm{p}}\left(t\right)=\left(\begin{array}{c}\hfill 1500\hfill \\ \hfill 1500\hfill \end{array}\right)\mathrm{and}$

$\begin{array}{c}\mathrm{X}\left(t\right)={\mathrm{X}}_{\mathrm{h}}\left(t\right)+{\mathrm{X}}_{\mathrm{p}}\left(t\right)\\ =\left(\begin{array}{c}\hfill 100c_2{e}^{-t/100}+1500\hfill \\ \hfill c_1{e}^{-t/100}+c_{2}t{e}^{-t/100}+1500\hfill \end{array}\right).\end{array}$

Application of the initial conditions then gives us $\mathbf{X}\left(0\right)=\left(\begin{array}{c}\hfill 100c_2+1500\hfill \\ \hfill c_1+1500\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 50\hfill \\ \hfill 100\hfill \end{array}\right),\mathrm{so}{c}_1=-1400$ and c ₂  =   – 1450/100   =   – 29/2. The solution to the IVP is

$\begin{array}{c}\mathrm{X}\left(t\right)=\left(\begin{array}{c}\hfill x\left(t\right)\hfill \\ \hfill y\left(t\right)\hfill \end{array}\right)\\ =\left(\begin{array}{c}\hfill -1450e^{-t/100}+1500\hfill \\ \hfill -1400e^{-t/100}-\frac{29}{2}t{e}^{-t/100}+1500\hfill \end{array}\right).\end{array}$

Notice that lim _{t  →   ∞} x(t)   =   lim _{t  →   ∞} y(t)   =   1500, which means that the amount of salt in each tank tends toward a value of 1500   lb.

In Example 7.2.2 , is there a value of t for which x(t)   =   y(t)? If so, what is this value? Which function increases most rapidly for smaller values of t?

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Solution of System of Nonhomogeneous Differential Equations

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