Find a Fundamental Solution to an Ivp

Higher-Order Differential Equations

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

4.5.1 Second-Order Equations

Let $S=\left\{y_1,y_2\right\}$ be a fundamental set of solutions for equation y″ + p(t)y′q(t)y = 0. To solve the nonhomogeneous equation y″ + p(t)y′ + q(t)y = f(t), we need to find a particular solution, y _p of equation y″ + p(t)y′ + q(t)y = f(t). We search for a particular solution of the form

(4.13) $\begin{array}{l}{y}_p=u_1(t)y_1(t)+u_2(t)y_2(t),\hfill \end{array}$

A particular solution, y _p , is a solution that does not contain any arbitrary constants.

where u ₁ and u ₂ are functions of t. Differentiating equation (4.13) gives us

$\begin{array}{l}{y_p}^{\prime }={u_1}^{\prime }{y}_1+u_1{y_1}^{\prime }+{u_2}^{\prime }{y}_2+u_2{y_2}^{\prime }.\hfill \end{array}$

Observe that it is pointless to search for solutions of the form y _p = c ₁ y ₁ + c ₂ y ₂ where c ₁ and c ₂ are constants because for every choice of c ₁ and c ₂, c ₁ y ₁ + c ₂ y ₂ is a solution to the corresponding homogeneous equation.

Assuming that

(4.14) $\begin{array}{l}{y}_1{u_1}^{\prime }+y_2{u_2}^{\prime }=0\hfill \end{array}$

results in y _p ′ = u ₁ y ₁′ + u ₂ y ₂′. Computing the second derivative then yields

$\begin{array}{l}{y_p}^{″}={u_1}^{\prime }{y_1}^{\prime }+u_1{y_1}^{″}+{u_2}^{\prime }{y_2}^{\prime }+u_2{y_2}^{″}.\hfill \end{array}$

Substituting y _p , y _p ′, and y _p ″ into the equation y″ + p(t)y′ + q(t)y = f(t) and using the facts that

$\begin{array}{l}{u}_1\left({y_1}^{″}+p{y_1}^{\prime }+q{y}_1\right)=0\text{and}{u}_2\left({y_2}^{″}+p{y_2}^{\prime }+q{y}_2\right)=0\hfill \end{array}$

(because y ₁ and y ₂ are solutions to the corresponding homogeneous equation, y″ + p(t)y′ + q(t)y = 0) results in

(4.15) $\begin{array}{l}\begin{array}{cc}\frac{d^2{y}_p}{d{t}^2}+p(t)\frac{d{y}_p}{dt}+q(t)y_p\hfill & ={u_1}^{\prime }{y_1}^{\prime }+u_1{y_1}^{″}\hfill \\ & +{u_2}^{\prime }{y_2}^{\prime }+u_2{y_2}^{″}+p(t)\left(u_1{y_1}^{\prime }+u_2{y_2}^{\prime }\right)\hfill \\ & +q(t)\left(u_1{y}_1+u_2{y}_2\right)\hfill \\ & ={y_1}^{\prime }{u_1}^{\prime }+{y_2}^{\prime }{u_2}^{\prime }=f(t).\hfill \end{array}\hfill \end{array}$

Observe that equations (4.14) and (4.15) form a system of two linear equations in the unknowns u ₁′ and u ₂′:

(4.16) $\begin{array}{l}\begin{array}{cc}{y}_1{u_1}^{\prime }+y_2{u_2}^{\prime }& =0\\ {y_1}^{\prime }{u_1}^{\prime }+{y_2}^{\prime }{u_2}^{\prime }& =f(t).\end{array}\hfill \end{array}$

Applying Cramer's Rule gives us

(4.17) $\begin{array}{l}{u_1}^{\prime }=\frac{\left|\begin{array}{ll}\hfill 0\hfill & \hfill y_2\hfill \\ \hfill f(t)\hfill & \hfill {y_2}^{\prime }\hfill \end{array}\right|}{\left|\begin{array}{ll}\hfill y_1\hfill & \hfill y_2\hfill \\ \hfill {y_1}^{\prime }\hfill & \hfill {y_2}^{\prime }\hfill \end{array}\right|}=-\frac{y_2(t)f(t)}{W(S)}\text{and}{u_2}^{\prime }=\frac{\left|\begin{array}{ll}\hfill y_1\hfill & \hfill 0\hfill \\ \hfill {y_1}^{\prime }\hfill & \hfill f(t)\hfill \end{array}\right|}{\left|\begin{array}{ll}\hfill y_1\hfill & \hfill y_2\hfill \\ \hfill {y_1}^{\prime }\hfill & \hfill {y_2}^{\prime }\hfill \end{array}\right|}=\frac{y_1(t)f(t)}{W(S)},\hfill \end{array}$

where W(S) is the Wronskian, $W(S)=\left|\begin{array}{ll}\hfill y_1\hfill & \hfill y_2\hfill \\ \hfill {y_1}^{\prime }\hfill & \hfill {y_2}^{\prime }\hfill \end{array}\right|$ . After integrating to obtain u ₁ and u ₂, we form y _p and then a general solution, y = y _h + y _p .

Summary of Variation of Parameters for Second-Order Equations

Given the second-order equation a ₂(t)y″ + a ₂(t)y′ + a ₀(t)y = g(t).

1.

Divide by a ₂(t) to rewrite the equation in standard form, y″ + p(t)y′ + q(t)y = f(t).

2.

Find a general solution, y _h = c ₁ y ₁ + c ₂ y ₂, where S = {y ₁, y ₂ } is a fundamental set of solutions for the corresponding homogeneous equation, of the corresponding homogeneous equation y″ + p(t)y′ + q(t)y = 0.

3.

Let $W(S)=\left|\begin{array}{ll}\hfill y_1\hfill & \hfill y_2\hfill \\ \hfill y_1^{\prime }\hfill & \hfill y_2^{\prime }\hfill \end{array}\right|$ .

4.

Let ${u_1}^{\prime }=-\frac{y_{2}f(t)}{W}$ and ${u_2}^{\prime }=\frac{y_{1}f(t)}{W}$ .

5.

Integrate to obtain u ₁ and u ₂.

6.

A particular solution of a ₂(t)y″ + a ₂(t)y′ + a ₀(t)y = g(t) is given by y _p = u ₁ y ₁ + u ₂ y ₂.

7.

A general solution of a ₂(t)y″ + a ₂(t)y′ + a ₀(t)y = g(t) is given by y = y _h + y _p .

Example 4.5.1

Solve $y^{″}+9y=sec3t$ , y(0) = 0, y′(0) = 0, 0 ≤ t < π/6.

Solution

The corresponding homogeneous equation is y″ + 9y = 0 with general solution $y_h=c_{1}cos3t+c_{2}sin3t$ . Then, a fundamental set of solutions is $S=\left\{cos3t,sin3t\right\}$ and W(S) = 3, as we see using Wronskian.

fs={Cos[3t], Sin[3t]};wm={fs, D[fs, t]};wm//MatrixForm

$\left(\begin{array}{ll}\text{Cos}[3t]& \text{Sin}[3t]\\ -3\text{Sin}[3t]& 3\text{Cos}[3t]\\ \end{array}\right)$

Wronskian[fs, t]

3

We use equation (4.17) to find $u_1=\frac{1}{9}lncos3t$ and $u_2=\frac{1}{3}t$ .

u1=Integrate[−Sin[3t]Sec[3t]/3, t]u2=Integrate[Cos[3t]Sec[3t]/3, t]

$\frac{1}{9}\text{Log}[\text{Cos}[3t]]$

$\frac{t}{3}$

It follows that a particular solution of the nonhomogeneous equation is

$\begin{array}{l}{y}_p=\frac{1}{9}cos3tlncos3t+\frac{1}{3}tsin3t\hfill \end{array}$

and a general solution is

$\begin{array}{l}y=y_h+y_p=c_{1}cos3t+c_{2}sin3t+\frac{1}{9}cos3tlncos3t+\frac{1}{3}tsin3t.\hfill \end{array}$

yp=u1Cos[3t]+u2Sin[3t]

$\frac{1}{9}\text{Cos}[3t]\text{Log}[\text{Cos}[3t]]+\frac{1}{3}t\text{Sin}[3t]$

Identical results are obtained using DSolve.

The negative sign in the output does not affect the result because C[1] is arbitrary.

DSolve[y"[t]+9y[t]==Sec[3t], y[t], t]

$\{\{y[t]\to C[1]\text{Cos}[3t]+C[2]\text{Sin}[3t]+\frac{1}{9}(\text{Cos}[3t]\text{Log}[\text{Cos}[3t]]+3t\text{Sin}[3t])\}\}$

Applying the initial conditions gives us c ₁ = c ₂ = 0 so we conclude that the solution to the initial value problem is

$\begin{array}{l}y=\frac{1}{9}cos3tlncos3t+\frac{1}{3}tsin3t.\hfill \end{array}$

sol=DSolve[{y"[t]+9y[t]==Sec[3t], y[0]==0, y′[0]==0}, y[t], t]

$\{\{y[t]\to \frac{1}{9}(\text{Cos}[3t]\text{Log}[\text{Cos}[3t]]+3t\text{Sin}[3t])\}\}$

We graph the solution with Plot in Figure 4-25.

Figure 4-25. The domain of the solution is − π/6 &lt; t &lt; π/6

Plot[y[t]/.sol, {t, 0, Pi/6}, AxesLabel→{t, y}, PlotStyle->CMYKColor[0, 0.89, 0.94, 0.28]]

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Fluid Dynamics

Elaine S. Oran , Jay P. Boris , in Encyclopedia of Physical Science and Technology (Third Edition), 2002

II.E Nonequilibrium Gas Flows

The interactions of atoms, ions, and molecules in a gas or liquid may be described by a hierarchy of mathematical models, ranging from fundamental solutions of sets of elementary interactions of particles (such as molecular dynamics methods) to approximations of systems in which the individual particles are replaced by continuum fluid elements (such as the Navier-Stokes equations). In between these extremes, there are various levels of statistical and particle-based theories that account for the nonequilibrium or particulate nature of matter.

Gases may be characterized by the Knudsen number of the flow, defined as Kn   =   λ/L, the ratio of λ, the mean free path between molecular collisions, and L, the characteristic length of the system. When the value of Kn is typically less than 0.05 or even 0.1, that is, there are sufficient collisions between the molecules, we can assume that the gas behaves like a continuum fluid. In these cases, the Navier-Stokes equations are an excellent model of the physics, given the required input, initial, and boundary conditions. However, as Kn increases, averaging assumptions that produce the fluid approximation are not valid, and representing the gas as a continuum fluid becomes a poor approximation. Some account must then be taken of the nonequilibrium, or particle nature of the material.

There are now many applications of high Kn flows that are of practical scientific and engineering importance. Spacecraft reentry into planetary atmospheres, the function of thrusters used on spacecraft to adjust orbits, and the behavior of out-gased plumes are all space-related problems involving high Kn gases. A variety of vapor-phase processing methods are used to produce thin films, and plasma-etching techniques are used to produce semiconductor components. These material processing applications also involve high Kn gases. In both space dynamics and materials processing, the densities of the gases are quite low (a few torr or less). However, in the relatively new development of microsystems, such as microelectro-mechanical systems (MEMS), gases that flow in micron-sized channels at relatively high densities (atmospheric pressure or higher) are still characterized by high values of Kn because of the very small size.

Statistically based Monte Carlo methods offer alternate approaches to the direct solution of such nonequilibrium flow problems. These approaches have been extremely successful for predicting the behavior of a wide range of nonequilibrium gas flows. Because of their computational simplicity, they are now being used extensively for atmospheric and space predictions, as well as for materials processing.

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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

6.2.4 Variation of parameters

Let $S=\{y_1,y_2\}$ be a fundamental set of solutions for Eq. (6.9). To solve the nonhomogeneous equation (6.8), we need to find a particular solution, $y_p$ of Eq. (6.8). We search for a particular solution of the form

A particular solution, $y_p$ , is a solution that does not contain any arbitrary constants.

Key concept: A general solution of the nonhomogeneous linear equation is $y=y_h+y_p$ , where $y_h$ is a general solution of the corresponding homogeneous equation, and $y_p$ is a particular solution of the nonhomogeneous equation.

(6.14) $y_p=u_1(t)y_1(t)+u_2(t)y_2(t),$

where $u_1$ and $u_2$ are functions of t. Differentiating Eq. (6.14) gives us

Observe that it is pointless to search for solutions of the form $y_p=c_1{y}_1+c_2{y}_2$ where $c_1$ and $c_2$ are constants because for every choice of $c_1$ and $c_2$ , $c_1{y}_1+c_2{y}_2$ is a solution to the corresponding homogeneous equation.

${y_p}^{\prime }={u_1}^{\prime }{y}_1+u_1{y_1}^{\prime }+{u_2}^{\prime }{y}_2+u_2{y_2}^{\prime }.$

Assuming that

(6.15) $y_1{u_1}^{\prime }+y_2{u_2}^{\prime }=0$

results in ${y_p}^{\prime }=u_1{y_1}^{\prime }+u_2{y_2}^{\prime }$ . Computing the second derivative then yields

${y_p}^{″}={u_1}^{\prime }{y_1}^{\prime }+u_1{y_1}^{″}+{u_2}^{\prime }{y_2}^{\prime }+u_2{y_2}^{″}.$

Substituting $y_p$ , ${y_p}^{\prime }$ , and ${y_p}^{″}$ into Eq. (6.8) and using the facts that

$u_1({y_1}^{″}+p{y_1}^{\prime }+q{y}_1)=0\text{and}{u}_2({y_2}^{″}+p{y_2}^{\prime }+q{y}_2)=0$

(because $y_1$ and $y_2$ are solutions to the corresponding homogeneous equation) results in

(6.16) $\frac{d^2{y}_p}{d{t}^2}+p(t)\frac{d{y}_p}{dt}+q(t)y_p={u_1}^{\prime }{y_1}^{\prime }+u_1{y_1}^{″}+{u_2}^{\prime }{y_2}^{\prime }+u_2{y_2}^{″}+p(t)(u_1{y_1}^{\prime }+u_2{y_2}^{\prime })+q(t)(u_1{y}_1+u_2{y}_2)={y_1}^{\prime }{u_1}^{\prime }+{y_2}^{\prime }{u_2}^{\prime }=f(t).$

Observe that Eq. (6.15) and Eq. (6.16) form a system of two linear equations in the unknowns ${u_1}^{\prime }$ and ${u_2}^{\prime }$ :

(6.17) $$\begin{gathered} y_1{u_1}^{\prime }+y_2{u_2}^{\prime }=0 \\ {y_1}^{\prime }{u_1}^{\prime }+{y_2}^{\prime }{u_2}^{\prime }=f(t). \end{gathered}$$

Applying Cramer's Rule gives us

(6.18) ${u_1}^{\prime }=\frac{\left|\begin{array}{cc}\hfill 0\hfill & \hfill y_2\hfill \\ \hfill f(t)\hfill & \hfill {y_2}^{\prime }\hfill \end{array}\right|}{\left|\begin{array}{cc}\hfill y_1\hfill & \hfill y_2\hfill \\ \hfill {y_1}^{\prime }\hfill & \hfill {y_2}^{\prime }\hfill \end{array}\right|}=-\frac{y_2(t)f(t)}{W(S)}\text{and}{u_2}^{\prime }=\frac{\left|\begin{array}{cc}\hfill y_1\hfill & \hfill 0\hfill \\ \hfill {y_1}^{\prime }\hfill & \hfill f(t)\hfill \end{array}\right|}{\left|\begin{array}{cc}\hfill y_1\hfill & \hfill y_2\hfill \\ \hfill {y_1}^{\prime }\hfill & \hfill {y_2}^{\prime }\hfill \end{array}\right|}=\frac{y_1(t)f(t)}{W(S)},$

where $W(S)$ is the Wronskian, $W(S)=\left|\begin{array}{cc}\hfill y_1\hfill & \hfill y_2\hfill \\ \hfill {y_1}^{\prime }\hfill & \hfill {y_2}^{\prime }\hfill \end{array}\right|$ . After integrating to obtain $u_1$ and $u_2$ , we form $y_p$ , and then a general solution, $y=y_h+y_p$ .

Example 6.15

Solve $y^{″}+9y=\sec 3t$ , $y(0)=0$ , $y^{\prime }(0)=0$ , $0⩽t<\pi /6$ .

Solution

The corresponding homogeneous equation is $y^{″}+9y=0$ with general solution $y_h=c_1\cos 3t+c_2\sin 3t$ . Then, a fundamental set of solutions is $S=\{\cos 3t,\sin 3t\}$ and $W(S)=3$ , as we see using Det, and Simplify.

$\mathbf{\text{fs}}\mathbf{=}\mathbf{\{}\mathbf{\text{Cos}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{,}\mathbf{\text{Sin}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{\}}\mathbf{;}$

$\mathbf{\text{wm}}\mathbf{=}\mathbf{\{}\mathbf{\text{fs}}\mathbf{,}\mathit{D}\mathbf{[}\mathbf{\text{fs}}\mathbf{,}\mathit{t}\mathbf{]}\mathbf{\}}\mathbf{;}$

$\mathbf{\text{wm}}\mathbf{\text{//}}\mathbf{\text{MatrixForm}}$

$\mathbf{\text{wd}}\mathbf{=}\mathbf{\text{Det}}\mathbf{[}\mathbf{\text{wm}}\mathbf{]}\mathbf{\text{//}}\mathbf{\text{Simplify}}$

$\left(\begin{array}{cc}\hfill \text{Cos}[3t]\hfill & \hfill \text{Sin}[3t]\hfill \\ \hfill -3\text{Sin}[3t]\hfill & \hfill 3\text{Cos}[3t]\hfill \end{array}\right)$

3

We use Eq. (6.18) to find $u_1=\frac{1}{9}\ln \cos 3t$ and $u_2=\frac{1}{3}t$ .

$\mathbf{\text{u1}}\mathbf{=}\mathbf{\text{Integrate}}\mathbf{[}\mathbf{-}\mathbf{\text{Sin}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{\text{Sec}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{/}\mathbf{3}\mathbf{,}\mathit{t}\mathbf{]}$

$\mathbf{\text{u2}}\mathbf{=}\mathbf{\text{Integrate}}\mathbf{[}\mathbf{\text{Cos}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{\text{Sec}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{/}\mathbf{3}\mathbf{,}\mathit{t}\mathbf{]}$

$\frac{1}{9}\text{Log}[\text{Cos}[3t]]$

$\frac{t}{3}$

It follows that a particular solution of the nonhomogeneous equation is $y_p=\frac{1}{9}\cos 3t\ln \cos 3t+\frac{1}{3}t\sin 3t$ , and a general solution is $y=y_h+y_p=c_1\cos 3t+c_2\sin 3t+\frac{1}{9}\cos 3t\ln \cos 3t+\frac{1}{3}t\sin 3t$ .

Absolute value is not needed in the antiderivatives, because we are restricting the domain to $0⩽t\&lt;\pi /6$ and $\cos t\&gt;0$ on this interval.

$\mathbf{\text{yp}}\mathbf{=}\mathbf{\text{u1}}\mathbf{\text{Cos}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{+}\mathbf{\text{u2}}\mathbf{\text{Sin}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}$

$\frac{1}{9}\text{Cos}[3t]\text{Log}[\text{Cos}[3t]]+\frac{1}{3}t\text{Sin}[3t]$

Identical results are obtained using DSolve.

The negative sign in the output does not affect the result, because C[1] is arbitrary.

$\mathbf{\text{DSolve}}\mathbf{[}\mathit{y}"\mathbf{[}\mathit{t}\mathbf{]}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{=}\mathbf{=}\mathbf{\text{Sec}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{,}\mathit{y}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathit{t}\mathbf{]}$

$\{\{y[t]\to C[1]\text{Cos}[3t]+C[2]\text{Sin}[3t]+\frac{1}{9}(\text{Cos}[3t]\text{Log}[\text{Cos}[3t]]+3t\text{Sin}[3t])\}\}$

Applying the initial conditions gives us $c_1=c_2=0$ , so we conclude that the solution to the initial-value problem is $y=\frac{1}{9}\cos 3t\ln \cos 3t+\frac{1}{3}t\sin 3t$ .

$\mathbf{\text{sol}}\mathbf{=}\mathbf{\text{DSolve}}\mathbf{[}\mathbf{\{}\mathit{y}"\mathbf{[}\mathit{t}\mathbf{]}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{=}\mathbf{=}\mathbf{\text{Sec}}\mathbf{[}\mathbf{3}\mathit{t}\mathbf{]}\mathbf{,}\mathit{y}\mathbf{[}\mathbf{0}\mathbf{]}\mathbf{=}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\prime }}\mathbf{[}\mathbf{0}\mathbf{]}\mathbf{=}\mathbf{=}\mathbf{0}\mathbf{\}}\mathbf{,}\mathit{y}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathit{t}\mathbf{]}$

$\{\{y[t]\to \frac{1}{9}(\text{Cos}[3t]\text{Log}[\text{Cos}[3t]]+3t\text{Sin}[3t])\}\}$

We graph the solution with Plot in Fig. 6.18.

Figure 6.18. The domain of the solution is 0 ⩽t &lt;π/6 (University of Kentucky colors).

$\mathbf{\text{Plot}}\mathbf{[}\mathit{y}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{\text{/.}}\mathbf{\text{sol}}\mathbf{,}\mathbf{\{}\mathit{t}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{\text{Pi}}\mathbf{/}\mathbf{6}\mathbf{\}}\mathbf{,}$

$\mathbf{\text{PlotStyle}}\mathbf{\to }\mathbf{\{}\mathbf{\{}\mathbf{\text{Thickness}}\mathbf{[}\mathbf{.}\mathbf{01}\mathbf{]}\mathbf{,}\mathbf{\text{CMYKColor}}\mathbf{[}\mathbf{1}\mathbf{,}\mathbf{.}\mathbf{75}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{]}\mathbf{\}}\mathbf{\}}\mathbf{]}$   □

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Answers to Selected Exercises

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

Exercises 4.2

1.

y ^″ = 0 has characteristic equation r ² = 0 so r = 0 has multiplicity two. Two linearly independent solutions to the equation are y ₁ = 1 and y ₂ = t ; a fundamental set of solutions is S = {1,t}; and a general solution is y = c ₁ + c ₂ t.

3.

y ^″ + y′ = 0 has characteristic equation r ² + r = 0, which has solutions r ₁ = 0 and r ₂ = −1. Two linearly independent solutions to the equation are y ₁ = 1 and y ₂ = e^−t ; a fundamental set of solutions is S = {1,e^−t }; and a general solution is y = c ₁ + c ₂e^−t .

5.

y = c ₁e^−6t + c ₂e^−2t .

6.

6r ² + 5r + 1 = (3r + 1)(2r + 1) = 0 so r ₁ = −1/3 and r ₂ = −1/2. Thus, two linearly independent solutions to the equation are y ₁ = e^−t/3 and y ₂ = e^−t/2; a fundamental set of solutions is S = {e^−t/3,e^−t/2}; and a general solution is y = c ₁e^−t/3 + c ₂e^−t/2.

7.

y = c ₁e^−t/4 + c ₂e^−t/2.

9.

$y=c_{1}cos4t+c_{2}sin4t$ .

11.

$y=c_{1}cos\left(\sqrt{7}t\right)+c_{2}sin\left(\sqrt{7}t\right)$ .

13.

y = c ₁e^−t + c2e^3t/7.

15.

y = c ₁e^3t + c ₂ te^3t .

17.

General: y = c ₁ + c ₂e ^t/3; IVP: y = −21(1 −e ^t/3).

19.

General: y = c ₁e^3t + c ₂e^4t ; IVP: y = 14e^3t − 11e^4t .

21.

General: y = c ₁e^5t + c ₂e^2t ; IVP: y = e^5t .

23.

General: $y=c_{1}cos10t+c_{2}sin10t$ ; IVP: $y=cos10t+sin10t$ .

25.

General: y = e^−2t (c ₁ + c ₂ t); IVP: y = e^−2t (1 + 5t).

27.

General: $y=e^{-2t}(c_{1}cos4t+c_{2}sin4t)$ ; IVP: $y=e^{-2t}(2cos4t+sin4t)$ .

29.

$y=e^{-t∕2}cos\left(\frac{\sqrt{3}}{2}t\right)+\frac{1}{\sqrt{3}}{e}^{-t∕2}sin\left(\frac{\sqrt{3}}{2}t\right)$ .

31.

$y=-\frac{e^{\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)t}-e^{\left(\frac{1}{2}+\frac{\sqrt{5}}{2}\right)t}}{\sqrt{5}}$ .

33.

y = 9e^−t/3 − 8e^−t/2.

35.

$y=\frac{3}{4}{e}^{-2t}sin\left(4t\right)+e^{-2t}cos\left(4t\right)$ .

37.

(a) y = c ₁ t ^2/3 + c ₂ t; (b) $y=t(c_1+c_{2}lnt)$ .

41.

$y=C{e}^{-t}sin2t$ , y = 0, $y=e^{-t}cos2t$ .

43.

$y=\left(\frac{3}{2}a+\frac{1}{2}b\right)e^{-t}+\left(-\frac{1}{2}a-\frac{1}{2}b\right)e^{-3t}$ so $y^{\prime }=-\left(\frac{3}{2}a+\frac{1}{2}b\right)e^{-t}-3\left(-\frac{1}{2}a-\frac{1}{2}b\right)e^{-3t}$ ; y′ = 0 if $t=-ln\left(\pm \frac{1}{3}\frac{\sqrt{3}\sqrt{\left(b+a\right)\left(3a+b\right)}}{b+a}\right)$ ; For none, (b + a)(3a + b) ≤ 0 while for one, (b + a)(3a + b) > 0.

45.

(a) No (b) To be a general solution, a fundamental set for the equation is $S=\left\{tcost,tsint\right\}$ . Now substitute each of these functions into the differential equation and set the result equal to zero. Solve the resulting system for p(t) and q(t) to obtain p(t) = −2/t and q(t) = (t ² + 2)/t ².

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Higher Order Equations

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

Solving Second-Order Equations with Constant Coefficients

Let ay ^″ + by ^′ + cy = 0 be a linear homogeneous second-order equation with constant real coefficients and let r ₁ and r ₂ be the solutions of the characteristic equation ar ² + br + c = 0.

1.

If r ₁≠r ₂ and both r ₁ and r ₂ are real, a general solution of ay ^″ + by ^′ + cy = 0 is $y=c_1{e}^{r_{1}t}+c_2{e}^{r_{2}t}$ ; a fundamental set of solutions for the equation is $S=\{e^{r_{1}t},e^{r_{2}t}\}$ .

2.

If r ₁ = r ₂, where r ₁ is real, a general solution of ay ^″ + by ^′ + cy = 0 is $y=c_1{e}^{r_{1}t}+c_{2}t{e}^{r_{1}t}$ ; a fundamental set of solutions for the equation is $S=\{e^{r_{1}t},t{e}^{r_{1}t}\}$ .

3.

If r ₁ = α + iβ, β > 0, and $m_2=\overline{m_1}=\alpha -\mathit{i\beta }$ , a general solution of ay ^″ + by ^′ + cy = 0 is $y=e^{\mathit{\alpha t}}(c_{1}cos\mathit{\beta t}+c_{2}sin\mathit{\beta t})$ ; a fundamental set of solutions for the equation is $S=\{e^{\mathit{\alpha t}}cos\mathit{\beta t},e^{\mathit{\alpha t}}sin\mathit{\beta t}\}$ .

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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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Polysplines on annuli in ℝ2

Ognyan Kounchev , in Multivariate Polysplines, 2001

8.6 The fundamental set of solutions for the operator M_k,p (d/dv)

The main point here is that the operator M_k,p (d/dv) has constant coefficients and its set of fundamental solutions is simpler to describe. Indeed, for every λ _j there are only two possibilities.

•

The first is that λ _j be a simple root of the equation

$M_{k,p}\left(z\right)=0.$

This is equivalent to the fact that (z – λ _j ) divides M_k,p (z) but (z – λ _j )² does not divide M_k,p (z). Then the function e ^{λ _j v} belongs to the set of fundamental solutions of equation (8.9).

•

The second possibility is that the root λ _j is a double root, i.e.

$M_{k,p}\left({\lambda }_j\right)=M_{k,p}^{\prime }\left({\lambda }_j\right)=0.$

This is equivalent to the fact that (z – λ _j )² divides M_k,p (z). Then the function e ^{λ _j v} belongs to the set of fundamental solutions of the equation (8.9) as well as the function ve ^{λ _j v} .

Evidently, in order to obtain, for some k ≥ 0, the multiple root λ _j of the polynomial M_k,p (z), the two sets of λ _j in (8.8) have to overlap, which happens precisely if we have the inequality

$-k+2p-2\ge k,$

i.e. if

$k\le p-1.$

By definition every fundamental set of solutions of equation

$M_{k,p}\left(d/dv\right)w\left(v\right)=0$

has exactly 2p linearly independent functions. We will be interested in those having the form e ^{λ _j v} and ve ^{λ _j v} . The above elementary arguments prove the following proposition.

Proposition 8.12

The function e ^{λ _j v} belongs to ${\tilde{\mathcal{U}}}_{k,p}$ if M_k,p (λ _j ) = 0. The function ve ^{λ _j v} belongs to ${\tilde{\mathcal{U}}}_{k,p}$ if $M_{k,p}\left({\lambda }_j\right)=M_{k,p}^{\prime }\left({\lambda }_j\right)=0$ . So we have the following rigorous description of a basis of the set ${\tilde{\mathcal{U}}}_{k,p}$ of solutions of the equation M_k,p (d/dv)w(v) = 0, namely

$\begin{array}{ll}{\tilde{\mathcal{U}}}_{k,p}\hfill & =\text{lin}\{{\displaystyle \underset{j=1}{\overset{2p}{\cup }}\left\{e^{{\lambda }_{j}v}:\mathrm{for}{M}_{k,p}\left({\lambda }_j\right)=0,M_{k,p}^{\prime }\left({\lambda }_j\right)\ne 0\right\}\cup }\hfill \\ \hfill & \cup {\displaystyle \underset{j=1}{\overset{2p}{\cup }}\left\{e^{{\lambda }_{j}v},v{e}^{{\lambda }_{j}v}:\mathrm{for}{M}_{k,p}\left({\lambda }_j\right)=0,M_{k,p}^{\prime }\left({\lambda }_j\right)=0\right\}\}.}\hfill \end{array}$

(see Figure 8.2, p. 112.)

Now let us return to the space of the solutions of the operator $L_{\left(k\right)}^p$ . By the discussion in the previous section based on formula (8.6), p. 111, we saw that by the inverse change the function e ^{λ _j v} is mapped into r ^{λ _j} and the function ve ^{λ _j v} is mapped into log r · r ^{λ _j} . Now Proposition 8.12 provides a full description of the space of solutions of equation L^p _(k) u(r) = 0, as a linear hull of such terms.

In particular, for k = 0 we have $L_{\left(0\right)}^{p}u\left(r\right)={\Delta }_r^{p}u\left(r\right)$ , where Δ _r is the radial part of the Laplace operator defined in (7.1), p. 78. We can easily find the basis for the set U_k,p .

Exercise 8.13

1.

Prove that all radially symmetric polyharmonic functions in the annulus in ℝ² are linear combinations of the system of functions

$U_{0,p}:=\text{lin}\left\{1,r^2,\dots ,r^{2p-2},\log r,r^2\log r,\dots ,r^{2p-2}\log r\right\}.$

Hint: By the definition of the λ _j s in (8.8), p. 111, see that M _0,p (z) = z ²(z − 2)² · · · (z − 2p + 2)², find a basis of the set ${\tilde{\mathcal{U}}}_{0,p}$ and apply the variable change r = e^v to obtain the basis of ${\mathcal{u}}_{0,p}$ .

2.

Prove that every radially symmetric polyharmonic function h(x) of order p in the ball B(0; R) admits the representation

$h\left(x\right)={\displaystyle \sum _{j=0}^{p-1}{C}_j{r}^{2j}}$

for some constants C_j . Hint: The only continuous solutions of the ordinary differential equation ${\Delta }_r^{p}f\left(r\right)=0$ are r ^2j , for j = 0, 2, …, p − 1. Those having log terms are not continuous.

3.

Prove that every radially symmetric polyharmonic function h(x) of order p in the annulus A_a,b admits the representation

$h\left(x\right)={\displaystyle \sum _{j=0}^{p-1}\left(C_j+D_j\log r\right)r^{2j}}$

for some constants C_j , D_j .

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Differential Equations, Ordinary

Anthony N. Michel , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

IV.D Linear nth-Order Ordinary Differential Equations

We conclude this section by considering some of the more important aspects of linear nth-order ordinary differential equations. We shall consider equations of the form,

(70) ${\text{y}}^{\left(\text{n}\right)}+{\text{a}}_{\text{n}-1}\left(\text{t}\right){\text{y}}^{\left(\text{n}-1\right)}+\cdots +{\text{a}}_1\left(\text{t}\right){\text{y}}^{\left(1\right)}+{\text{a}}_0\left(\text{t}\right)\text{y}=\text{b}\left(\text{t}\right)$

(71) ${\text{y}}^{\left(\text{n}\right)}+{\text{a}}_{\text{n}-1}\left(\text{t}\right){\text{y}}^{\left(\text{n}-1\right)}+\cdots +{\text{a}}_1\left(\text{t}\right){\text{y}}^{\left(1\right)}+{\text{a}}_0\left(\text{t}\right)\text{y}=0$

and

(72) ${\text{y}}^{\left(\text{n}\right)}+{\text{a}}_{\text{n}-1}{\text{y}}^{\left(\text{n}-1\right)}+\cdots +{\text{a}}_1{\text{y}}^{\left(1\right)}+{\text{a}}_0\text{y}=0$

In Eqs. (70) and (71) the functions a _k (t) and b(t), k  =   1,   …, n −1, are continuous on some appropriate time interval J. If we define the differential operator L _n by:

(73) ${\text{L}}_{\text{n}}=\frac{{\text{d}}^{\text{n}}}{\text{d}{\text{t}}^{\text{n}}}+{\text{a}}_{\text{n}-1}(\text{t})\frac{{\text{d}}^{\text{n}-1}}{\text{d}{\text{t}}^{\text{n}-1}}+\cdots +{\text{a}}_1(\text{t})\frac{\text{d}}{\text{d}\text{t}}+{\text{a}}_0(\text{t})$

then we can rewrite Eqs. (70) and (71) more compactly as

(74) ${\text{L}}_{\text{n}}\text{y}=\text{b}\left(\text{t}\right)$

and

(75) ${\text{L}}_{\text{n}}\text{y}=0$

respectively. We can rewrite Eq. (72) similarly by defining a differential operator L in the obvious way.

Following the procedure in Section II, we can reduce the study of Eq. (71) to the study of the system of n first-order ordinary differential equations,

(LH) ${\text{x}}^{\prime }=\text{A}\left(\text{t}\right)\text{x}$

where A(t) is the companion matrix given by:

(76) $\text{A}(\text{t})=\left[\begin{array}{lllll}0\hfill & 1\hfill & 0\hfill & \cdots \hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill & \cdots \hfill & 0\hfill \\ ⋮\hfill & ⋮\hfill & ⋮\hfill & \hfill & \hfill \\ 0\hfill & 0\hfill & 0\hfill & \cdots \hfill & 1\hfill \\ -{\text{a}}_0(\text{t})\hfill & -{\text{a}}_1(\text{t})\hfill & -{\text{a}}_2(\text{t})\hfill & \cdots \hfill & -{\text{a}}_{\text{n}-1}(\text{t})\hfill \end{array}\right]$

Since A(t) is continuous on J, we know from Section III that there exists a unique solution ϕ(t), for all t  ∈ J, to the initial-value problem,

$\begin{array}{ccc}\hfill {\text{x}}^{\prime }=\text{A}\left(\text{t}\right)\text{x},& \hfill \text{x}(\tau )=\mathit{\xi },\hfill & \tau \in \text{J}\hfill \end{array}$

where ξ   =   (ξ₁,   …,ξ _n )^T  ∈ R ⁿ . The first component of this solution is a solution of L _n y  =   0 satisfying y(τ)   =   ξ₁, y′(τ)   =   ξ₂,   …, y ⁽ⁿ⁻¹⁾(τ)   =   ξ _n .

Now let ϕ₁,   …,ϕ _n be n solutions of Eq. (75). Then we can easily show that the matrix,

$\Phi (\text{t})=\left[\begin{array}{llll}{\varphi }_1\hfill & {\varphi }_2\hfill & \cdots \hfill & {\varphi }_{\text{n}}\hfill \\ {{\varphi }^{\prime }}_1\hfill & {{\varphi }^{\prime }}_2\hfill & \cdots \hfill & {{\varphi }^{\prime }}_{\text{n}}\hfill \\ ⋮\hfill & ⋮\hfill & \hfill & ⋮\hfill \\ {\varphi }_1^{(\text{n}-1)}\hfill & {\varphi }_2^{(\text{n}-1)}\hfill & \cdots \hfill & {\varphi }_{\text{n}}^{(\text{n}-1)}\hfill \end{array}\right]$

is a solution of the matrix equation,

(77) ${\text{X}}^{\prime }=\text{A}\left(\text{t}\right)\text{X}$

where A(t) is defined by Eq. (76). We call the determinant of Φ the Wronskian for Eq. (75) with respect to the solutions ϕ₁,   …,ϕ _n and we denote it by:

$\text{W}({\varphi }_1,\dots ,{\varphi }_{\text{n}})=det\Phi (\text{t})$

Note that W(ϕ₁,   …,ϕ _n )(t) depends on t  ∈ J. Since Φ is a solution of matrix equation (77), then by Abel's formula it follows that for any τ   ∈ J and for any t  ∈ J,

(78) $\begin{array}{ll}\text{W}({\varphi }_1,\dots ,{\varphi }_{\text{n}})(\text{t})\hfill & =det\Phi (\tau )exp\left[{\int }_{\tau }^{\text{t}}\text{t}\text{r}\text{A}(\text{s})\text{d}\text{s}\right]\hfill \\ \hfill & =\text{W}({\varphi }_1,\dots ,{\varphi }_{\text{n}})(\tau )\hfill \\ \hfill & \times exp\left\{{\int }_{\tau }^{\text{t}}-{\text{a}}_{\text{n}-1}(\text{s})\text{d}\text{s}\right\}\hfill \end{array}$

As an example, consider the second-order differential equation

$\begin{array}{cc}\hfill {\text{t}}^2\text{y}″+\text{t}\text{y}\prime -\text{y}=0,& \hfill 0<\text{t}<\mathrm{\infty }\hfill \end{array}$

which can be written equivalently as:

(79) $\begin{array}{cc}\hfill \text{y}″+(1/\text{t})\text{y}\prime -(1/{\text{t}}^2)\text{y}=0,& \hfill 0<\text{t}<\mathrm{\infty }\hfill \end{array}$

The functions ϕ₁(t  = t and ϕ₂(t)   =   1/t are clearly solutions of Eq. (79). We now form the matrix,

$\Phi (\text{t})=\left[\begin{array}{cc}{\varphi }_1& {\varphi }_2\\ {{\varphi }^{\prime }}_1& {{\varphi }^{\prime }}_2\end{array}\right]=\left[\begin{array}{cc}\text{t}& 1/\text{t}\\ 1& 1/{\text{t}}^2\end{array}\right]$

which yields the Wronskian

$\begin{array}{cc}\hfill \text{W}\left(\varphi 1,\varphi 2\right)\left(\text{t}\right)=\text{d}\text{e}\text{t}\Phi \left(\text{t}\right)=-2/\text{t},& \hfill \text{t}>0\hfill \end{array}$

In the notation of Eq. (76) we have a ₁(t)   =   1/t, a ₀(t)   =   − 1/t ², and thus a ₁(s)   =   1/s. In view of Eq. (78), we have for any τ   >   0,

$\begin{array}{c}\text{W}({\varphi }_1,{\varphi }_2)(\text{t})\\ \\ \end{array}\begin{array}{c}=det\Phi (\text{t})\\ =\text{W}({\varphi }_1,{\varphi }_2)(\tau )exp\left\{{\int }_{\tau }^{\text{t}}-{\text{a}}_1(\text{s})\text{d}\text{s}\right\}\\ =-(2/\tau ){\text{e}}^{\text{In}(\tau /\text{t})}=-2/\text{t},\text{t}>0\end{array}$

as expected.

Similarly as in the case of systems of equations, we can prove the following result for nth-order differential equations:

A set of n solutions of Eq. (75), ϕ₁,   …,ϕ _n , is linearly independent on J if and only if W(ϕ₁,   …,ϕ _n )(t) ≠ 0 for all t  ∈ J. Moreover, every solution of Eq. (75) is a linear combination of any set of n linearly independent solutions.

The above result enables us to make the following definition:

A set of n linearly independent solutions of Eq. (75) on j,ϕ₁,   …,ϕ _n , is called a fundamental set of solutions for Eq. (75).

Next, we turn our attention to nonhomogeneous linear nth-order ordinary differential equations of the form (70). As shown in Section II, the study of Eq. (70) reduces to the study of the system of n first-order ordinary differential equations,

(80) ${\text{x}}^{\prime }=\text{A}\left(\text{t}\right)\text{x}+\text{g}\left(\text{t}\right)$

where A(t) is given by Eq. (76) and g(t)   =   [0,   …, 0, b(t)]^T. Recall that for given τ   ∈ J and given x(τ)   =   ξ   ∈ R ⁿ , Eq. (80) has a unique solution given by ϕ   =   ϕ_h  +   ϕ_p, where ϕ _h (t)   =   Φ(t, τ)ξ is a solution of (LH), ϕ(t, τ) denotes the state transition matrix of A(t), and ϕ_p is a particular solution of Eq. (80), given by:

$\begin{array}{cc}{\varphi }_{\text{p}}\left(\text{t}\right)\hfill & ={\int }_{\tau }^{\text{t}}\Phi \left(\text{t},\text{s}\right)\text{g}\left(\text{s}\right)\text{d}\text{s}\hfill \\ \hfill & =\Phi \left(\text{t}\right){\int }_{\tau }^{\text{t}}{\Phi }^{-1}\left(\text{s}\right)\text{g}\left(\text{s}\right)\text{d}\text{s}\hfill \end{array}$

We now specialize this result from the n-dimensional system (80) to the corresponding nth-order equation (70) to obtain the following result:

If {ϕ₁,   …,ϕ _n } is a fundamental set for the equation L _n y   =   0, then the unique solution ψ of the equation L _n y   = b(t) satisfying ψ(τ)   =   ξ₁,   …,ψ⁽ⁿ⁻¹⁾(τ)   =   ξ _n is given by:

(81) $\begin{array}{cc}\psi (\text{t})=\hfill & {\psi }_{\text{h}}(\text{t})+{\psi }_{\text{P}}(\text{t})={\psi }_{\text{h}}(\text{t})+\sum _{\text{k}-1}^{\text{n}}{\varphi }_{\text{k}}(\text{t})\hfill \\ \hfill & \times {\int }_{\tau }^{\text{t}}\frac{{\text{W}}_{\text{k}}({\varphi }_1,\dots ,{\varphi }_{\text{n}})(\text{s})}{\text{W}({\varphi }_1,\dots ,{\varphi }_{\text{n}})(\text{s})}\text{b}(\text{s})\text{d}\text{s}\hfill \end{array}$

Here, ψ_h is the solution of L _n y  =   0 such that ψ(τ)   =   ξ₁, ψ′(τ)   =   ξ₂,   …,ψ⁽ⁿ⁻¹⁾(τ)   =   ξ _n , and W _k (ϕ₁,   …,ϕ _n )(t) is obtained from W(ϕ₁,   …,ϕ _n )(t) by replacing the kth column in W(ϕ₁,   …,ϕ _n )(t) by (0,   …,0,1)^T.

We apply the above example to the second-order differential equation,

$\begin{array}{cc}\hfill \text{y}″+(1/\text{t})\text{y}\prime -(\text{y}/{\text{t}}^2)=\text{b}(\text{t}),& \hfill 0<\text{t}<\mathrm{\infty }\hfill \end{array}$

where b(t) is a real continuous function for all t  >   0. From the example involving Eq. (79) we have ϕ₁(t)   = t, ϕ₂(t)   =   1/t, and W(ϕ₁,ϕ₂)(t)   =   −2/t, t  >   0. Also,

$\begin{array}{cc}{\text{W}}_1({\varphi }_1,{\varphi }_2)(\text{t})\hfill & =\left|\begin{array}{cc}0\hfill & 1/\text{t}\hfill \\ 1\hfill & -1/{\text{t}}^2\hfill \end{array}\right|=-\frac{1}{\text{t}},\hfill \\ {\text{W}}_1({\varphi }_1,{\varphi }_2)(\text{t})\hfill & =\left|\begin{array}{cc}\text{t}\hfill & 0\hfill \\ 1\hfill & 1\hfill \end{array}\right|=\text{t}\hfill \end{array}$

From Eq. (81) we now have

$\begin{array}{cc}\psi \left(\text{t}\right)\hfill & ={\psi }_{\text{h}}\left(\text{t}\right)+{\psi }_{\text{p}}\left(\text{t}\right)\hfill \\ \hfill & ={\psi }_{\text{h}}\left(\text{t}\right)+\frac{\text{t}}{2}{\int }_{\tau }^{\text{t}}\text{b}\left(\text{s}\right)\text{d}\text{s}-\frac{1}{2\text{t}}{\int }_{\tau }^{\text{t}}{\text{s}}^2\text{b}\left(\text{s}\right)\text{d}\text{s}\hfill \end{array}$

Next, we consider nth-order ordinary differential equations with constant coefficients given by Eq. (72) which can equivalently be written as L _n y  =   0, where

${\text{L}}_{\text{n}}=\frac{{\text{d}}^{\text{n}}}{\text{d}{\text{t}}^{\text{n}}}+{\text{a}}_{\text{n}-1}\frac{{\text{d}}^{\text{n}-1}}{\text{d}{\text{t}}^{\text{n}-1}}+\cdots +{\text{a}}_1\frac{\text{d}}{\text{d}\text{t}}+{\text{a}}_0$

We assume that J  =   (−∞, ∞), we call

(82) $\text{p}\left(\lambda \right)={\lambda }^{\text{n}}+{\text{a}}_{\text{n}-1}{\lambda }^{\text{n}-1}+\cdots +{\text{a}}_1\lambda +{\text{a}}_0$

the characteristic polynomial of the differential equation (72), and we call

(83) $\text{p}\left(\lambda \right)=0$

the characteristic equation of Eq. (72). The roots of p(λ) are called the characteristic roots of Eq. (72).

We see that the study of Eq. (72) reduces to the study of the system of first-order ordinary differential equations with constant coefficients given by x′   = Ax, where

(84) $\text{A}=\left[\begin{array}{cccccc}\hfill 0& \hfill 1\hfill & 0\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill 0& \hfill 0\hfill & 1\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill & \hfill ⋮\hfill \\ \hfill 0& \hfill 0\hfill & 0\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 1\hfill \\ \hfill -{\text{a}}_0& \hfill -{\text{a}}_1\hfill & -{\text{a}}_2\hfill & \hfill -{\text{a}}_3\hfill & \hfill \cdots \hfill & \hfill -{\text{a}}_{\text{n}-1}\hfill \end{array}\right]$

The following result, which is proved in a straightforward manner, connects Eq. (72) and x′   = Ax with A given by Eq. (84):

The characteristic polynomial of A in Eq. (83) is precisely the characteristic polynomial p(λ) given by Eq. 82), that is,

$\text{p}\left(\lambda \right)=\text{d}\text{e}\text{t}\left(\lambda {\text{E}}_{\text{n}}-\text{A}\right)$

The next result enumerates a fundamental set for Eq. (72):

Let λ₁,   …,λ _s be the distinct roots of the characteristic equation (83) and suppose that λ _i has multiplicity m _i , i  =   1,   …,s, with Σ ^s _i=1 m _i   = n. Then the following set of functions is a fundamental set for Eq.(72):

(85) $\begin{array}{cc}{\text{t}}^{\text{k}}{\text{e}}^{{\lambda }_{\mathrm{i}}\text{t}},\hfill & \text{k}=0,1,\cdots ,{\text{m}}_{\text{i}}-1,\hfill \\ \hfill & \text{i}=1,\cdots ,\text{s}\hfill \end{array}$

As a specific example, consider:

(86) $\text{p}\left(\lambda \right)=\left(\lambda -2\right){\left(\lambda +3\right)}^2\left(\lambda +\text{i}\right)\left(\lambda -\text{i}\right){\left(\lambda -4\right)}^4$

Then, n  =   9, and {e ^2t , e ^−3t , te ^−3t , e ^−it , e ^+it , e ^4t , te ^4t , t ² e ^4t , t ³ e ^4t } is a fundamental set for the differential equation corresponding to the characteristic equation (86).

We conclude this section by considering adjoint equations. Corresponding to the operator L _n given in Eq. (73), we define a second linear operator L _n ⁺ of order n, which we call the adjoint of L _n , as follows. The domain of L _n ⁺ is the set of all continuous functions defined on J such that $[{\overline{\text{a}}}_{\text{j}}(\text{t})\text{y}(\text{t})]$ has j continuous derivatives on J. (Here, ${\overline{\text{a}}}_{\text{j}}(\text{t})$ denotes the complex conjugate of a _j (t).) For each function y, define:

$\begin{array}{cc}{\text{L}}_{\text{n}}^{+}\text{y}=\hfill & {(-1)}^{\text{n}}{\text{y}}^{(\text{n})}+{(-1)}^{\text{n}-1}{({\bar{\text{a}}}_{\text{n}-1}\text{y})}^{\text{n}-1}\hfill \\ \hfill & +\cdots +(-1)({\bar{\text{a}}}_1\text{y})\prime +{\bar{\text{a}}}_0\text{y}\hfill \end{array}$

The equation,

$\begin{array}{cc}\hfill {\text{L}}_{\text{n}}^{+}\text{y}=0,& \hfill \text{t}\in \text{J}\hfill \end{array}$

is called the adjoint equation to L _n y  =   0.

When Eq. (75) is written in companion form (LH) with A(t) given by Eq. (76), then the adjoint system is z′   =   −A ^*(t)z, where

${\text{A}}^{*}\left(\text{t}\right)=\left[\begin{array}{ccccc}\hfill 0& \hfill 0\hfill & \cdots \hfill & \hfill 0\hfill & \hfill -{\bar{\text{a}}}_0(\text{t})\hfill \\ \hfill 1& \hfill 0\hfill & \cdots \hfill & \hfill 0\hfill & \hfill -{\bar{\text{a}}}_1(\text{t})\hfill \\ \hfill 0& \hfill 1\hfill & \cdots \hfill & \hfill 0\hfill & \hfill -{\bar{\text{a}}}_2(\text{t})\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill & \hfill ⋮\hfill & \hfill ⋮\hfill \\ \hfill 0& \hfill 0\hfill & \cdots \hfill & \hfill 1\hfill & \hfill -{\bar{\text{a}}}_{\text{n}-1}(\text{t})\hfill \end{array}\right]$

This adjoint system can be written in component form as:

(87) $\begin{array}{cc}{{\text{z}}^{\prime }}_1={\overline{\text{a}}}_0(\text{t}){\text{z}}_{\text{n}}& \\ {{\text{z}}^{\prime }}_1=-{\text{z}}_{\text{j}-1}+{\overline{\text{a}}}_{\text{j}-1}(\text{t}){\text{z}}_{\text{n}},& 2\le \text{j}\le \text{n}\end{array}$

If ψ   =   [ψ₁,ψ₂,   …,ψ _n ]^T is a solution of Eq. (87) and if a _j ψ _n has j derivatives, then

${\psi }_{\text{n}}\prime -({\bar{\text{a}}}_{\text{n}-1}{\psi }_{\text{n}})=-{\psi }_{\text{n}-1}$

and

${\psi }_{\text{n}}″-({\bar{\text{a}}}_{\text{n}-1}{\psi }_{\text{n}})\prime =-{\psi }_{\text{n}-1}^{\prime }={\psi }_{\text{n}-1}-({\bar{\text{a}}}_{\text{n}-2}{\psi }_{\text{n}})$

or

${\psi }_{\text{n}}″-({\bar{\text{a}}}_{\text{n}-1}{\psi }_{\text{n}})\prime +({\bar{\text{a}}}_{\text{n}-2}{\psi }_{\text{n}})={\psi }_{\text{n}-2}$

Continuing in this manner, we see that ψ _n solves L ⁺ _n ψ   =   0.

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