Homogeneous Differential Equations

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Homogeneous Differential Equations.

For a given function [math]F(x)[/math] and a given polynomial

[[math]] p(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_1t + a_0, [[/math]]

let us consider the differential equation

[[math]] \begin{equation} p(D)y = F(x). \label{eq11.4.1} \end{equation} [[/math]]

The simplification of the theory gained by enlarging the set of possible solutions to include complex-valued functions of a real variable was demonstrated in Section 3, and we shall continue to use this technique. Nevertheless, our primary concern is still that of finding real-valued solutions to real differential equations. For this reason, we shall assume throughout that the coefficients [math]a_0, \cdots, a_{n-1}[/math] of the polynomial [math]p(t)[/math] are real numbers and that [math]F(x)[/math] is a real-valued function. Associated with the differential equation (1) is the homogeneous differential equation

[[math]] \begin{equation} p(D)y = 0, \label{eq11.4.2} \end{equation} [[/math]]

called the associated homogeneous equation of [math]p(D)y = F(x)[/math]. A theorem of basic importance is the following:

Theorem

If [math]y_0[/math] is any particular solution of (1) and if [math]y[/math] is the general solution of (2), then [math]y + y_0[/math] is the general solution of (1).

Show Proof

{{{4}}}

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As a result of this theorem, our approach to the problem of solving the differential equation [math]p(D)y = F(x)[/math] will be divided into two parts. We shall first concentrate on finding the general solution of the associated homogeneous equation [math]p(D)y = 0[/math], and then consider methods of finding a particular solution to the original nonhomogeneous equation. The remainder of this section will be devoted to the first part. We begin with the second-order linear homogeneous differential equation with constant coefficients:

[[math]] \begin{equation} (D^2 + aD + b)y = 0. \label{eq11.4.3} \end{equation} [[/math]]

The general solution of this equation has been presented earlier (see page 617), but without proof. We shall supply the proof now by factoring the linear operator [math]D^2 + aD + b[/math] and solving the equation by the iterative technique of Section 3. The characteristic polynomial can be written as the product

[[math]] t^2 + at + b = (t - r_1)(t - r_2), [[/math]]

where the roots [math]r_1[/math] and [math]r_2[/math] are either both real or distinct conjugate complex numbers. Equation (3) can therefore be written

[[math]] \begin{equation} (D - r_1)(D - r_2)y = 0. \label{eq11.4.4} \end{equation} [[/math]]

Theorem

The general solution of the differential equation (3) [or equivalently, of (4)] is:

\item[(i)] [math]y = c_1e^{r_1x} + c_2e^{r_2x}, \;\mbox{if}\; r_1 \neq r_2, \;\mbox{or}[/math] \item[(ii)] [math]y = (c_1x + c_2)e^{rx}, \;\mbox{if}\; r_1 = r_2 = r,[/math]

where [math]c_1[/math] and [math]c_2[/math] are arbitrary complex numbers. Note that these solutions include all the real-valued ones, since the set of all real numbers is a subset of the set of all complex numbers.

Show Proof

Let [math]y[/math] be an arbitrary solution of (4). We define the function [math]u[/math] by setting [math]u = (D - r_2)y[/math]. Then (4) is equivalent to the two first-order linear equations:

[[math]] \left \{ \begin{array}{l} (D - r_1)u = 0, \\ (D - r_2)y = u. \end{array} \right . [[/math]]

An integrating factor for the first of these is [math]e^{-r_1 x}[/math], because, in the notation of first-order linear equations, we have [math]P(x) = -r_1[/math]. It follows that

[[math]] \frac{d}{dx} (e^{-r_1 x} u) = 0 . [[/math]]

Integration yields [math]e^{-r_1 x}u = c_1[/math], whence

[[math]] u = c_1 e^{-r_1 x}, \;\;\;\mbox{for some complex number}\; c_1. [[/math]]

Substituting the expression for [math]u[/math] into the second differential equation above, we obtain

[[math]] (D - r_2)y = c_1e^{r_1 x}. [[/math]]

This time an integrating factor is [math]e^{-r_2 x}[/math], and so

[[math]] \begin{equation} \frac{d}{dx} (e^{-r_2x} y) = c_1 e^{(r_1- r_2)x} . \label{eq11.4.5} \end{equation} [[/math]]

We now distinguish two cases. Case 1. [math]r_1 \neq r_2[/math]. Integration of (5) yields

[[math]] e^{-r_2 x} y = \frac{1}{r_1 - r_2} c_1e^{(r_1 - r_2)x} + c_2, [[/math]]

for some complex number [math]c_2[/math]. Multiplying both sides by [math]e^{r_2x}[/math] and replacing [math]\frac{c_1}{r_1 - r_2}[/math] by simply [math]c_1[/math], we get

[[math]] y = c_1e^{r_1x} + c_2e^{r_2x}. [[/math]]

Case 2. [math]r_1 = r_2 = r[/math]. Then [math]e^{(r_1 - r_2)x} = e^0 = 1[/math], and (5) reduces to

[[math]] \frac{d}{dx} (e^{-rx} y) = c_1. [[/math]]

Integrating, we obtain [math]e^{-rx}y = c_1 x + c_2[/math], for some complex number [math]c_2[/math], and it follows that

[[math]] y = (c_1x + c_2)e^{rx}. [[/math]]

We have now proved that, if [math]y[/math] is an arbitrary solution of the original differential equation (4), then there exist complex numbers [math]c_1[/math] and [math]c_2[/math] (either or both of which may perfectly well be real every real number is a special case of a complex number) such that [math]y[/math] is of form (i) if [math]r_1 \neq r_2[/math] and of form (ii) if [math]r_1 = r_2 = r[/math]. Conversely, it is a simple matter to check by substitution that, for any complex numbers [math]c_1[/math] and [math]c_2[/math], the function [math]c_1e^{r_1x} + c_2e^{r_2x}[/math] is a solution if [math]r_1 \neq r_2[/math] and the function [math](c_1x + c_2)e^{rx}[/math] is a solution if [math]r_1 = r_2 = r[/math]. This completes the proof of the theorem.

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\medskip How can we use Theorem (4.2) to obtain the general real-valued solution of the differential equation [math](D^2 + a D + b)y = (D - r_1)(D - r_2)y = 0?[/math] Suppose, to begin with, that [math]r_1[/math] and [math]r_2[/math] are both real and that [math]r_1 \neq r_2[/math]. It follows from part (i) of Theorem (4.2) that the function defined by

[[math]] \begin{equation} y = c_1e^{r_1x} + c_2e^{r_2x}, \;\;\;\mbox{for any two real numbers $c_1$ and \ltmath\gtc_2[[/math]]

,} \label{eq11.4.6} \end{equation} </math> is a solution, and it is certainly real-valued. There is only one obstacle in the way of the conclusion that (6) is the general real-valued solution. This is the a priori possibility that there might exist complex numbers [math]c_1[/math] and [math]c_2[/math], which are not both real, but are such that [math]c_1e^{r_1x} + c_2e^{r_1x}[/math] is a real-valued function. This, in fact, cannot happen, as the following argument shows: Let [math]c_1 = \gamma_1, + i\delta_1[/math], and [math]c_2 = \gamma_2 + i\delta_2[/math]. Since

[[math]] (\gamma_1 + i\delta_1)e^{r_1x} + (\gamma_2 + i\delta_2)e^{r_2x} [[/math]]

is by assumption real-valued, then so is

[[math]] (\gamma_1 + i\delta_1)e^{r_1x} + (\gamma_2 + i\delta_2)e^{r_2x} - \gamma_1 e^{r_1x} - \gamma_2 e^{r_2x} = i(\delta_1 e^{r_1x} + \delta_2 e^{r_2x}). [[/math]]

Hence

[[math]] \delta_1e^{r_1x} + \delta_2e^{r_2x} = 0 , [[/math]]

and so

[[math]] \delta_1 e^{(r_1- r_2)x} = -\delta_2. [[/math]]

This equation holds for all real values of [math]x[/math]. But, since [math]r_1 - r_2 \neq 0[/math], the left side has constant value only if [math]\delta_1 = 0[/math], which in turn immediately implies that [math]\delta_2 = 0[/math]. Hence [math]\delta_1 = \delta_2 = 0[/math], and the argument is complete. With this problem disposed of, it now follows from (4.2)(i) that, if [math]r_1[/math] and [math]r_2[/math] are real and unequal, then the general real-valued solution of the differential equation is given by (6). A similar situation arises if [math]r_1 = r_2 = r[/math]. In this case [math]r[/math] must be a real number, and it is a corollary of part (ii) of Theorem (4.2) that the function defined by

[[math]] \begin{equation} y = (c_1x + c_2)e^{rx}, \;\;\;\mbox{for any two real numbers $c_1$ and \ltmath\gtc_2[[/math]]

,} \label{eq11.4.7} \end{equation} </math> is a solution, and, of course, it is real-valued. Again, we must show that it is not possible to have complex numbers [math]c_1[/math] and [math]c_2[/math], not both real, such that [math](c_1x + c_2)e^{rx}[/math] is a real-valued function. The proof of this fact is similar to that of the analogous result in the preceding paragraph, and we leave it as an exercise. It then follows from (4.2)(ii) that the general real-valued solution is given by (7). The third and final possibility is that the roots [math]r_1[/math] and [math]r_2[/math] of the characteristic polynomial are distinct conjugate complex numbers. In this case, we need the lemma:

Theorem

If [math]r_1 = \alpha + i\beta, r_2 = \alpha - i\beta[/math], and [math]\beta \neq 0[/math], then the function defined by

[[math]] y = c_1e^{r_1x} + c_2e^{r_2x} \;\;\;\mbox{for arbitrary complex numbers $c_1$ and \ltmath\gtc_2[[/math]]

,} </math> is real-valued if and only if [math]c_1[/math] and [math]c_2[/math] are complex conjugates. Moreover, if [math]c_1 = \gamma + i \delta[/math] and [math]c_2 = - i \delta[/math], then

[[math]] y= e^{\alpha x}(2 \gamma \cos \beta x - 2 \delta \sin \beta x). [[/math]]

A proof in the “if” direction is given in detail in (8.3) on page 347. In addition, the above equation giving [math]y[/math] in terms of [math]\alpha, \beta, \gamma[/math], and [math]\delta[/math], is also derived there. The “only if” direction can be proved in the same direct manner as the analogous results for the other two cases: Let [math]c_1 = \gamma_1 + i\beta_1[/math] and [math]c_2 = \gamma_2 + i\delta_2[/math], substitute these values into [math]c_1e^{r_1 x} + c_2e^{r_2 x}[/math], and impose the condition that [math]y[/math] is real-valued. It will then follow that [math]\gamma_1 = \gamma_2[/math] and that [math]\delta_1 = -\delta_2[/math]. Again, we leave this task as an exercise. Let us replace the real constants [math]2\gamma[/math] and [math]-2\delta[/math] which appear in the equation in the last line of (4.3) by [math]c_1[/math] and [math]c_2[/math], respectively. It is then a corollary of (4.3) and (4.2)(i) that the general real-valued solution of the differential equation [math](D^2 + aD + b)y = (D - r_1)(D - r_2)y = 0[/math] is

[[math]] \begin{equation} y = e^{\alpha x}(c_1 \cos \beta x + c_2 \sin \beta x), \;\;\;\mbox{for any two real numbers $c_1$ and \ltmath\gtc_2[[/math]]

,} \label{eq11.4.8} \end{equation} </math> provided [math]r_1 = \alpha + i\beta, r_2 = \alpha - i\beta[/math], and [math]\beta \neq 0[/math]. This completes the proof that second-order, homogeneous, linear differential equations with real constant coefficients have the general solutions first described in Section 8 of Chapter 6 and again in Section 1 of this chapter. The higher-order homogeneous equations can be solved in the same way. If

[[math]] p(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_1t + a_0, [[/math]]

then the general solution of the differential equation [math]p(D)y = 0[/math] can be obtained by first factoring [math]p(D)[/math] to obtain an equivalent set of [math]n[/math] first-order linear differential equations which are then solved successively to find [math]y[/math]. As an illustration, we shall solve a third-order equation by this method. Following this example, we shall give (without proof) the form of the general real-valued solution for arbitrary order [math]n[/math]. Example Find the general solution of the differential equation

[[math]] \frac{d^3y}{dx^3} - 3 \frac{d^2y}{dx^2} + 4y = 0. [[/math]]

The characteristic polynomial is [math]p(t) = t^3 - 3t^2 + 4[/math]. Substituting -1 for [math]t[/math], we obtain [math]p(-1) = 0[/math], from which it follows that [math](t + 1)[/math] is a factor of [math]p(t)[/math]. Dividing, we find that

[[math]] t^3 - 3t^2 + 4 = (t + 1)(t^2 - 4t + 4) = (t + 1)(t - 2)^2. [[/math]]

Hence the differential equation can be written

[[math]] (D + 1)(D - 2)^2 y = 0. [[/math]]

We set [math]u_1 = (D - 2)^2y[/math] and [math]u_2 = (D - 2)y[/math] and, by so doing, obtain the equivalent set of three first-order equations

[[math]] \left \{ \begin{array}{rl} (D + 1)u_1 &= 0, \\ (D - 2)u_2 &= u_1,\\ (D - 2)y &= u_2 . \end{array} \right . [[/math]]

The general solution of the first of these is [math]u_1 = c_1e^{-x}[/math], and the second equation is therefore

[[math]] (D - 2)u_2 = c_1e^{-x}. [[/math]]

An integrating factor is [math]e^{-2x}[/math], and so

[[math]] \frac{d}{dx} (e^{-2x} u_2) = e^{-2x} c_1 e^{-x} = c_1e^{-3x}. [[/math]]

Hence

[[math]] e^{-2x} u_2 = -\frac{c_1}{3} e^{-3x} + c_2, [[/math]]

from which it follows by multiplying both sides by [math]e^{2x}[/math] and replacing [math]-\frac{c_1}{3}[/math] by simply [math]c_1[/math] that

[[math]] u_2 = c_1e^{-x} + c_2e^{2x}. [[/math]]

The third equation is now seen to be

[[math]] (D - 2)y = c_1e^{-x} + c_2e^{2x} . [[/math]]

Again, [math]e^{-2x}[/math] is an integrating factor, and we have

[[math]] \frac{d}{dx} (e^{-2x}y) = e^{-2x}(c_1e^{-x} + c_2e^{2x}) = c_1e^{-3x} + c_2. [[/math]]

Integration yields

[[math]] e^{-2x} y = -\frac{c_1}{3} e^{-3x} + c_2x + c_3. [[/math]]

Multiplying both sides by [math]e^{2x}[/math] and replacing [math]-\frac{c_1}{3}[/math] by simply [math]c_1[/math] again, we have

[[math]] y = c_1e^{-x} + (c_2x + c_3)e^{2x}, [[/math]]

where [math]c_1[/math], [math]c_2[/math], and [math]c_3[/math] are arbitrary real constants. This is the general realvalued solution and completes the example.

We now give the general solution for arbitrary order [math]n[/math]. Let

[[math]] p(t) = t^n + a_{n - 1}t^{n-1} + \cdots + a_1t + a_0, [[/math]]

and suppose that factorization into real-valued irreducible factors yields the product

[[math]] p(t) = (t - r_1)^{m_1} \cdots (t - r_k)^{m_k}(t^2 + c_1t + d_1)^{n_1} \cdots (t^2 + c_lt + d_l)^{nl}, [[/math]]

where [math]m_1, . . ., m_k[/math] and [math]n_1, . . ., n_l[/math] are positive integers, the factors [math]t - r_i[/math] are all distinct, and the factors [math]t^2 + c_jt + d_j[/math] are all distinct. For each factor [math](t-r_i)^{m_i}[/math], define the function

[[math]] \begin{eqnarray*} f_i(x) &=& (c_{i1}x^{m_i -1} + c_{i2}x^{m_i - 2} + \cdots + C_{im_i} )e^{r_ix}, \\ &&\mbox{for arbitrary real numbers}\; C_{i1}, . . ., C_{im_i}. \;\;\;\;\;\;\;\;\;\;\;\; (9) \end{eqnarray*} [[/math]]

For each factor [math](t^2 + c_jt + d_j)^{n_j}[/math], let [math]\alpha_j + i\beta_j[/math] and [math]\alpha_j - i\beta_j[/math] be the roots of [math]t^2 + c_jt + d_j[/math], and define the function

[[math]] \begin{eqnarray*} g_j(x) &=& (A_{j1} x^{n_j-1} + A_{j2} x^{n_j - 2} + \cdots + A_{jnj})e^{\alpha_j x} \cos \beta_j x \\ &+&(\beta_{j1} x^{n_j-1} + B_{j2} x^{n_j-2} + \cdots + B_{jn_j})e^{\alpha_j x} \sin \beta_j x, \\ &&\;\;\;\mbox{for arbitrary real numbers $A_{j1}, . . ., A_{jn_j}$ and $B_{j1}, . . ., B_{jn_j}$} . \;\;\;\;\;\; (10) \end{eqnarray*} [[/math]]

Then it can be proved that

Theorem

The general real-valued solution of the homogeneous differential equation [math]p(D)y = 0[/math] is the sum

[[math]] y = f_1(x) + \cdots + f_k(x) + g_1(x) + \cdots + g_l(x) . [[/math]]

Note that, since [math]m_1 + \cdots + m_k + 2n_1 + \cdots + 2n_l = n[/math], the number of arbitrary constants in the general solution is equal to [math]n[/math], the order of the differential equation. Example Find the general solution of the differential equation

[[math]] (D + 2)(D - 5)^3(D^2 + D + 1)^2 y = 0. [[/math]]

This is an equation of order 8. The polynomial [math]t^2 + t + 1[/math] is irreducible with roots equal to [math]-\frac{1}{2} + i\frac{\sqrt 3}{2}[/math] and [math]-\frac{1}{2} - i \frac{\sqrt 3}{2}[/math]. It follows directly from (4.4) that the general real-valued solution is

[[math]] \begin{eqnarray*} y &=& C_1e^{-2x} + (C_2x^2 + C_3x + C_4)e^{5x} \\ & & + (C_5x + C_6)e^{-(1/2)x} \cos \frac{\sqrt 3}{2}x + (C_7x + C_8)e^{-(1/2)x} \sin \frac{\sqrt 3}{2} x, \end{eqnarray*} [[/math]]

for any set of real numbers [math]C_1, C_2, \ldots, C_8[/math].

\end{exercise}

General references

Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.