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5.5 Nonhomogeneous Equations and Undetermined Coefficients

We learned in Section 5.3 how to solve homogeneous linear equations with constant coefficients, but we saw in Section 5.4 that an external force in a simple mechanical system contributes a nonhomogeneous term to its differential equation. The general nonhomogeneous nth-order linear equation with constant coefficients has the form

$a_{n} y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{1} y ′ + a_{0} y = f (x) .$ (1)

By Theorem 5 of Section 5.2, a general solution of Eq. (1) has the form

$y = y_{c} + y_{p}$ (2)

where the complementary function $y_{c} (x)$ is a general solution of the associated homogeneous equation

$a_{n} y^{(n)} + a_{n - 1} y^{(n - 1)} + \dots + a_{1} y^{′} + a_{0} y = 0,$ (3)

and $y_{p} (x)$ is a particular solution of Eq. (1). Thus our remaining task is to find $y_{p}$ .

The method of undetermined coefficients is a straightforward way of doing this when the given function f(x) in Eq. (1) is sufficiently simple that we can make an intelligent guess as to the general form of $y_{p} .$ For example, suppose that f(x) is a polynomial of degree m. Then, because the derivatives of a polynomial are themselves polynomials of lower degree, it is reasonable to suspect a particular solution

$y_{p} (x) = A_{m} x^{m} + A_{m - 1} x^{m - 1} + \dots + A_{1} x + A_{0}$

that is also a polynomial of degree m, but with as yet undetermined coefficients. We may, therefore, substitute this expression for $y_{p}$ into Eq. (1), and then—by equating coefficients of like powers of x on the two sides of the resulting equation—attempt to determine the coefficients $A_{0}, A_{1}, \dots,$ $A_{m}$ so that $y_{p}$ will, indeed, be a particular solution of Eq. (1).

Similarly, suppose that

$f (x) = a \cos k x + b \sin k x .$

Then it is reasonable to expect a particular solution of the same form:

$y_{p} (x) = A \cos k x + B \sin k x,$

a linear combination with undetermined coefficients A and B. The reason is that any derivative of such a linear combination of $\cos k x$ and $\sin k x$ has the same form. We may therefore substitute this form of $y_{p}$ in Eq. (1), and then—by equating coefficients of $\cos k x$ and $\sin k x$ on both sides of the resulting equation—attempt to determine the coefficients A and B so that $y_{p}$ will, indeed, be a particular solution.

It turns out that this approach does succeed whenever all the derivatives of f(x) have the same form as f(x) itself. Before describing the method in full generality, we illustrate it with several preliminary examples.

Example 1

Find a particular solution of $y ″ + 3 y ′ + 4 y = 3 x + 2$ .

Solution

Here $f (x) = 3 x + 2$ is a polynomial of degree 1, so our guess is that

$y_{p} (x) = A x + B .$

Then $y_{p} ′ = A$ and $y_{p} ″ = 0,$ so $y_{p}$ will satisfy the differential equation provided that

$(0) + 3 (A) + 4 (A x + B) = 3 x + 2,$

that is,

$(4 A x) + (3 A + 4 B) = 3 x + 2$

for all x. This will be true if the x-terms and constant terms on the two sides of this equation agree. It therefore suffices for A and B to satisfy the two linear equations $4 A = 3$ and $3 A + 4 B = 2$ that we readily solve for $A = \frac{3}{4}$ and $B = - \frac{1}{16} .$ Thus we have found the particular solution

$y_{p} (x) = \frac{3}{4} x - \frac{1}{16} .$

Example 2

Find a particular solution of $y ″ - 4 y = 2 e^{3 x}$ .

Solution

Any derivative of $e^{3 x}$ is a constant multiple of $e^{3 x},$ so it is reasonable to try

$y_{p} (x) = A e^{3 x} .$

Then $y_{p} ″ = 9 A e^{3 x},$ so the given differential equation will be satisfied provided that

$9 A e^{3 x} - 4 (A e^{3 x}) = 2 e^{3 x};$

that is, $5 A = 2,$ so that $A = \frac{2}{5} .$ Thus our particular solution is $y_{p} (x) = \frac{2}{5} e^{3 x}$ .

Example 3

Find a particular solution of $3 y ″ + y ′ - 2 y = 2 \cos x$ .

Solution

A first guess might be $y_{p} (x) = A \cos x,$ but the presence of $y ′$ on the left-hand side signals that we probably need a term involving $\sin x$ as well. So we try

$\begin{array}{r} y_{p} (x) & = & A \cos x & + & B \sin x, \\ y_{p}^{′} (x) & = & - A \sin x & + & B \cos x, \\ y_{p}^{′ ′} (x) & = & - A \cos x & - & B \sin x . \end{array}$

Then substitution of $y_{p}$ and its derivatives into the given differential equation yields

$3 (- A \cos x - B \sin x) + (- A \sin x + B \cos x) - 2 (A \cos x + B \sin x) = 2 \cos x,$

that is (collecting coefficients on the left),

$(- 5 A + B) \cos x + (- A - 5 B) \sin x = 2 \cos x .$

This will be true for all x provided that the cosine and sine terms on the two sides of this equation agree. It therefore suffices for A and B to satisfy the two linear equations

$\begin{aligned} - 5 A & + & B & = & 2, \\ - A & - & 5 B & = & 0 \end{aligned}$

with readily found solution $A = - \frac{5}{13}, B = \frac{1}{13} .$ Hence a particular solution is

$y_{p} (x) = - \frac{5}{13} \cos x + \frac{1}{13} \sin x .$

The following example, which superficially resembles Example 2, indicates that the method of undetermined coefficients is not always quite so simple as we have made it appear.

Example 4

Find a particular solution of $y ″ - 4 y = 2 e^{2 x}$ .

Solution

If we try $y_{p} (x) = A e^{2 x},$ we find that

$y_{p}^{″} - 4 y_{p} = 4 A e^{2 x} - 4 A e^{2 x} = 0 \neq 2 e^{2 x} .$

Thus, no matter how A is chosen, $A e^{2 x}$ cannot satisfy the given nonhomogeneous equation. In fact, the preceding computation shows that $A e^{2 x}$ satisfies instead the associated homogeneous equation. Therefore, we should begin with a trial function $y_{p} (x)$ whose derivative involves both $e^{2 x}$ and something else that can cancel upon substitution into the differential equation to leave the $e^{2 x}$ term that we need. A reasonable guess is

$y_{p} (x) = A x e^{2 x},$

for which

$y_{p}^{′} (x) = A e^{2 x} + 2 A x e^{2 x} and y_{p}^{″} (x) = 4 A e^{2 x} + 4 A x e^{2 x} .$

Substitution into the original differential equation yields

$(4 A e^{2 x} + 4 A x e^{2 x}) - 4 (A x e^{2 x}) = 2 e^{2 x} .$

The terms involving $x e^{2 x}$ obligingly cancel, leaving only $4 A e^{2 x} = 2 e^{2 x},$ so that $A = \frac{1}{2} .$ Consequently, a particular solution is

$y_{p} (x) = \frac{1}{2} x e^{2 x} .$

The General Approach

Our initial difficulty in Example 4 resulted from the fact that $f (x) = 2 e^{2 x}$ satisfies the associated homogeneous equation. Rule 1, given shortly, tells what to do when we do not have this difficulty, and Rule 2 tells what to do when we do have it.

The method of undetermined coefficients applies whenever the function f(x) in Eq. (1) is a linear combination of (finite) products of functions of the following three types:

A polynomial in x;
An exponential function $e^{r x};$ (4)
cos kx or sin kx.

Any such function, for example,

$f (x) = (3 - 4 x^{2}) e^{5 x} - 4 x^{3} \cos 10 x,$

has the crucial property that only finitely many linearly independent functions appear as terms (summands) in f(x) and its derivatives of all orders. In Rules 1 and 2 we assume that $L y = f (x)$ is a nonhomogeneous linear equation with constant coefficients and that f(x) is a function of this kind.

Note that this rule is not a theorem requiring proof; it is merely a procedure to be followed in searching for a particular solution $y_{p} .$ If we succeed in finding $y_{p},$ then nothing more need be said. (It can be proved, however, that this procedure will always succeed under the conditions specified here.)

In practice we check the supposition made in Rule 1 by first using the characteristic equation to find the complementary function $y_{c},$ and then write a list of all the terms appearing in f(x) and its successive derivatives. If none of the terms in this list duplicates a term in $y_{c},$ then we proceed with Rule 1.

Example 5

Find a particular solution of

$y ″ + 4 y = 3 x^{3} .$ (5)

Solution

The (familiar) complementary solution of Eq. (5) is

$y_{c} (x) = c_{1} \cos 2 x + c_{2} \sin 2 x .$

The function $f (x) = 3 x^{3}$ and its derivatives are constant multiples of the linearly independent functions $x^{3}, x^{2},$ x, and 1. Because none of these appears in $y_{c},$ we try

$\begin{aligned} y_{p} & = & A x^{3} + B x^{2} + C x + D, \\ y_{p}^{′} & = & 3 A x^{2} + 2 B x + C, \\ y_{p}^{″} & = & 6 A x + 2 B . \end{aligned}$

Substitution in Eq. (5) gives

$\begin{aligned} y_{p}^{″} + 4 y_{p} & = & (6 A c + 2 B) + 4 (A x^{3} + B x^{2} + C x + D) \\ = & 4 A x^{3} + 4 B x^{2} + (6 A + 4 C) x + (2 B + 4 D) = 3 x^{3} . \end{aligned}$

We equate coefficients of like powers of x in the last equation to get

$\begin{aligned} 4 A & = & 3, & 4 B & = & 0, \\ 6 A + 4 C & = & 0, & 2 B + 4 D & = & 0 \end{aligned}$

with solution $A = \frac{3}{4}, B = 0, C = - \frac{9}{8},$ and $D = 0 .$ Hence a particular solution of Eq. (5) is

$y_{p} (x) = \frac{3}{4} x^{3} - \frac{9}{8} x .$

Example 6

Solve the initial value problem

$\begin{array}{l} y ″ - 3 y ′ + 2 y = 3 e^{- x} - 10 \cos 3 x; \\ y (0) = 1, y^{′} (0) = 2. \end{array}$ (6)

Solution

The characteristic equation $r^{2} - 3 r + 2 = 0$ has roots $r = 1$ and $r = 2,$ so the complementary function is

$y_{c} (x) = c_{1} e^{x} + c_{2} e^{2 x} .$

The terms involved in $f (x) = 3 e^{- x} - 10 \cos 3 x$ and its derivatives are $e^{- x},$ cos 3x, and sin 3x. Because none of these appears in $y_{c},$ we try

$\begin{array}{r} y_{p} & = & A e^{- x} & + & B \cos 3 x & + & C \sin 3 x, \\ y_{p}^{′} & = & A e^{- x} & - & 3 B \sin 3 x & + & 3 C \cos 3 x, \\ y_{p}^{″} & = & A e^{- x} & - & 9 B \cos 3 x & - & 9 C \sin 3 x . \end{array}$

After we substitute these expressions into the differential equation in (6) and collect coefficients, we get

$\begin{aligned} y_{p}^{″} - 3 y_{p}^{′} + 2 y_{p} & = & 6 A e^{- x} + (- 7 B - 9 C) \cos 3 x + (9 B - 7 C) \sin 3 x \\ = & 3 e^{- x} - 10 \cos 3 x . \end{aligned}$

We equate the coefficients of the terms involving $e^{- x},$ those involving cos 3x, and those involving sin 3x. The result is the system

$\begin{array}{r} 6 A & = & 3, \\ - 7 B - 9 C & = & - 10, \\ 9 B - 7 C & = & 0 \end{array}$

with solution $A = \frac{1}{2}, B = \frac{7}{13},$ and $C = \frac{9}{13} .$ This gives the particular solution

$y_{p} (x) = \frac{1}{2} e^{- x} + \frac{7}{13} \cos 3 x + \frac{9}{13} \sin 3 x,$

which, however, does not have the required initial values in (6).

To satisfy those initial conditions, we begin with the general solution

$\begin{aligned} y (x) & = & y_{c} (x) + y_{p} (x) \\ = & c_{1} e^{x} + c_{2} e^{2 x} + \frac{1}{2} e^{- x} + \frac{7}{13} \cos 3 x + \frac{9}{13} \sin 3 x, \end{aligned}$

with derivative

$y ′ (x) = c_{1} e^{x} + 2 c_{2} e^{2 x} - \frac{1}{2} e^{- x} - \frac{21}{13} \sin 3 x + \frac{27}{13} \cos 3 x .$

The initial conditions in (6) lead to the equations

$\begin{aligned} y (0) & = & c_{1} + c_{2} + \frac{1}{2} + \frac{7}{13} = 1, \\ y ′ (0) & = & c_{1} + 2 c_{2} - \frac{1}{2} + \frac{27}{13} = 2 \end{aligned}$

with solution $c_{1} = - \frac{1}{2}, c_{2} = \frac{6}{13} .$ The desired particular solution is therefore

$y (x) = - \frac{1}{2} e^{x} + \frac{6}{13} e^{2 x} + \frac{1}{2} e^{- x} + \frac{7}{13} \cos 3 x + \frac{9}{13} \sin 3 x .$

Example 7

Find the general form of a particular solution of

$y^{(3)} + 9 y ′ = x \sin x + x^{2} e^{2 x} .$ (7)

Solution

The characteristic equation $r^{3} + 9 r = 0$ has roots $r = 0, r = - 3 i$ , and $r = 3 i .$ So the complementary function is

$y_{c} (x) = c_{1} + c_{2} \cos 3 x + c_{3} \sin 3 x .$

The derivatives of the right-hand side in Eq. (7) involve the terms

$\begin{array}{c} \cos x, \sin x, x \cos x, x \sin x, \\ e^{2 x}, x e^{2 x}, and x^{2} e^{2 x} . \end{array}$

Because there is no duplication with the terms of the complementary function, the trial solution takes the form

$y_{p} (x) = A \cos x + B \sin x + C x \cos x + D x \sin x + E e^{2 x} + F x e^{2 x} + G x^{2} e^{2 x} .$

Upon substituting $y_{p}$ in Eq. (7) and equating coefficients of like terms, we get seven equations determining the seven coefficients A, B, C, D, E, F, and G.

The Case of Duplication

Now we turn our attention to the situation in which Rule 1 does not apply: Some of the terms involved in f(x) and its derivatives satisfy the associated homogeneous equation. For instance, suppose that we want to find a particular solution of the differential equation

${(D - r)}^{3} y = (2 x - 3) e^{r x} .$ (8)

Proceeding as in Rule 1, our first guess would be

$y_{p} (x) = A e^{r x} + B x e^{r x} .$ (9)

This form of $y_{p} (x)$ will not be adequate because the complementary function of Eq. (8) is

$y_{c} (x) = c_{1} e^{r x} + c_{2} x e^{r x} + c_{3} x^{2} e^{r x},$ (10)

so substitution of (9) in the left-hand side of (8) would yield zero rather than $(2 x - 3) e^{r x}$ .

To see how to amend our first guess, we observe that

${(D - r)}^{2} [(2 x - 3) e^{r x}] = [D^{2} (2 x - 3)] e^{r x} = 0$

by Eq. (13) of Section 5.3. If y(x) is any solution of Eq. (8) and we apply the operator ${(D - r)}^{2}$ to both sides, we see that y(x) is also a solution of the equation ${(D - r)}^{5} y = 0 .$ The general solution of this homogeneous equation can be written as

$y (x) = \underset{y_{c}}{\underset{︸}{c_{1} e^{r x} + c_{2} x e^{r x} + c_{3} x^{2} e^{r x}}} + \underset{y_{p}}{\underset{︸}{A x^{3} e^{r x} + B x^{4} e^{r x}}} .$

Thus every solution of our original equation in (8) is the sum of a complementary function and a particular solution of the form

$y_{p} (x) = A x^{3} e^{r x} + B x^{4} e^{r x} .$ (11)

Note that the right-hand side in Eq. (11) can be obtained by multiplying each term of our first guess in (9) by the least positive integral power of x (in this case, $x^{3}$ ) that suffices to eliminate duplication between the terms of the resulting trial solution $y_{p} (x)$ and the complementary function $y_{c} (x)$ given in (10). This procedure succeeds in the general case.

To simplify the general statement of Rule 2, we observe that to find a particular solution of the nonhomogeneous linear differential equation

$L y = f_{1} (x) + f_{2} (x),$ (12)

it suffices to find separately particular solutions $Y_{1} (x)$ and $Y_{2} (x)$ of the two equations

$L y = f_{1} (x) and L y = f_{2} (x),$ (13)

respectively. For linearity then gives

$L [Y_{1} + Y_{2}] = L Y_{1} + L Y_{2} = f_{1} (x) + f_{2} (x),$

and therefore $y_{p} = Y_{1} + Y_{2}$ is a particular solution of Eq. (12). (This is a type of “superposition principle” for nonhomogeneous linear equations.)

Now our problem is to find a particular solution of the equation $L y = f (x),$ where f(x) is a linear combination of products of the elementary functions listed in (4). Thus f(x) can be written as a sum of terms each of the form

$P_{m} (x) e^{r x} \cos k x or P_{m} (x) e^{r x} \sin k x,$ (14)

where $P_{m} (x)$ is a polynomial in x of degree m. Note that any derivative of such a term is of the same form but with both sines and cosines appearing. The procedure by which we arrived earlier at the particular solution in (11) of Eq. (8) can be generalized to show that the following procedure is always successful.

In practice we seldom need to deal with a function f(x) exhibiting the full generality in (14). The table in Fig. 5.5.1 lists the form of $y_{p}$ in various common cases, corresponding to the possibilities $m = 0, r = 0,$ and $k = 0$ .

On the other hand, it is common to have

$f (x) = f_{1} (x) + f_{2} (x),$

where $f_{1} (x)$ and $f_{2} (x)$ are different functions of the sort listed in the table in Fig. 5.5.1. In this event we take as $y_{p}$ the sum of the trial solutions for $f_{1} (x)$ and $f_{2} (x),$ choosing s separately for each part to eliminate duplication with the complementary function. This procedure is illustrated in Examples 8 through 10.

Example 8

Find a particular solution of

$y^{(3)} + y ″ = 3 e^{x} + 4 x^{2} .$ (16)

FIGURE 5.5.1.

Substitutions in the method of undetermined coefficients.

`f`(`x`)	$y_{p}$
$\begin{array}{c} P_{m} (x) = b_{0} + b_{1} x + b_{2} x^{2} + \dots + b_{m} x^{m} \\ a \cos k x + b \sin k x \\ e^{r x} (a \cos k x + b \sin k x) \\ P_{m} (x) e^{r x} \\ P_{m} (x) (a \cos k x + b \sin k x) \end{array}$	$\begin{array}{c} x^{s} (A_{0} + A_{1} x + A_{2} x^{2} + \dots + A_{m} x^{m}) \\ x^{s} (A \cos k x + B \sin k x) \\ x^{s} e^{r x} (A \cos k x + B \sin k x) \\ x^{s} (A_{0} + A_{1} x + A_{2} x^{2} + \dots + A_{m} x^{m}) e^{r x} \\ x^{s} [(A_{0} + A_{1} x + \dots + A_{m} x^{m}) \cos k x \\ + (B_{0} + B_{1} x + \dots + B_{m} x^{m}) \sin k x] \end{array}$

Solution

The characteristic equation $r^{3} + r^{2} = 0$ has roots $r_{1} = r_{2} = 0$ and $r_{3} = - 1,$ so the complementary function is

$y_{c} (x) = c_{1} + c_{2} x + c_{3} e^{- x} .$

As a first step toward our particular solution, we form the sum

$(A e^{x}) + (B + C x + D x^{2}) .$

The part $A e^{x}$ corresponding to $3 e^{x}$ does not duplicate any part of the complementary function, but the part $B + C x + {D x}^{2}$ must be multiplied by $x^{2}$ to eliminate duplication. Hence we take

$\begin{aligned} y_{p} & = & A e^{x} + B x^{2} + C x^{3} + D x^{4}, \\ y_{p}^{′} & = & A e^{x} + 2 B x + 3 C x^{2} + 4 D x^{3}, \\ y_{p}^{″} & = & A e^{x} + 2 B + 6 C x + 12 D x^{2}, and \\ y_{p}^{(3)} & = & A e^{x} + 6 C + 24 D x . \end{aligned}$

Substitution of these derivatives in Eq. (16) yields

$2 A e^{x} + (2 B + 6 C) + (6 C + 24 D) x + 12 D x^{2} = 3 e^{x} + 4 x^{2} .$

The system of equations

$\begin{aligned} 2 A & = & 3, & 2 B + 6 C & = & 0, \\ 6 C + 24 D & = & 0, & 12 D & = & 4 \end{aligned}$

has the solution $A = \frac{3}{2}, B = 4, C = - \frac{4}{3},$ and $D = \frac{1}{3} .$ Hence the desired particular solution is

$y_{p} (x) = \frac{3}{2} e^{x} + 4 x^{2} - \frac{4}{3} x^{3} + \frac{1}{3} x^{4} .$

Example 9

Determine the appropriate form for a particular solution of

$y^{″} + 6 y^{′} + 13 y = e^{- 3 x} \cos 2 x .$

Solution

The characteristic equation $r^{2} + 6 r + 13 = 0$ has roots $- 3 \pm 2 i,$ so the complementary function is

$y_{c} (x) = e^{- 3 x} (c_{1} \cos 2 x + c_{2} \sin 2 x) .$

This is the same form as a first attempt $e^{- 3 x} (A \cos 2 x + B \sin 2 x)$ at a particular solution, so we must multiply by x to eliminate duplication. Hence we would take

$y_{p} (x) = e^{- 3 x} (A x \cos 2 x + B x \sin 2 x) .$

Example 10

Determine the appropriate form for a particular solution of the fifth-order equation

${(D - 2)}^{3} (D^{2} + 9) y = x^{2} e^{2 x} + x \sin 3 x .$

Solution

The characteristic equation ${(r - 2)}^{3} (r^{2} + 9) = 0$ has roots $r = 2, 2, 2, 3 i,$ and $- 3 i,$ so the complementary function is

$y_{c} (x) = c_{1} e^{2 x} + c_{2} x e^{2 x} + c_{3} x^{2} e^{2 x} + c_{4} \cos 3 x + c_{5} \sin 3 x .$

As a first step toward the form of a particular solution, we examine the sum

$[(A + B x + C x^{2}) e^{2 x}] + [(D + E x) \cos 3 x + (F + G x) \sin 3 x] .$

To eliminate duplication with terms of $y_{c} (x),$ the first part—corresponding to $x^{2} e^{2 x}$ —must be multiplied by $x^{3},$ and the second part—corresponding to $x \sin 3 x$ —must be multiplied by x. Hence we would take

$y_{p} (x) = (A x^{3} + B x^{4} + C x^{5}) e^{2 x} + (D x + E x^{2}) \cos 3 x + (F x + G x^{2}) \sin 3 x .$

Variation of Parameters

Finally, let us point out the kind of situation in which the method of undetermined coefficients cannot be used. Consider, for example, the equation

$y^{″} + y = \tan x,$ (17)

which at first glance may appear similar to those considered in the preceding examples. Not so; the function $f (x) = \tan x$ has infinitely many linearly independent derivatives

$\sec^{2} x, 2 \sec^{2} x \tan x, 4 \sec^{2} x \tan^{2} x + 2 \sec^{4} x, \dots .$

Therefore, we do not have available a finite linear combination to use as a trial solution.

We discuss here the method of variation of parameters, which—in principle (that is, if the integrals that appear can be evaluated)—can always be used to find a particular solution of the nonhomogeneous linear differential equation

$y^{(n)} + p_{n - 1} (x) y^{(n - 1)} + \dots + p_{1} (x) y^{′} + p_{0} (x) y = f (x),$ (18)

provided that we already know the general solution

$y_{c} = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n}$ (19)

of the associated homogeneous equation

$y^{(n)} + p_{n - 1} (x) y^{(n - 1)} + \dots + p_{1} (x) y ′ + p_{0} (x) y = 0.$ (20)

Here, in brief, is the basic idea of the method of variation of parameters. Suppose that we replace the constants, or parameters, $c_{1}, c_{2}, \dots,$ $c_{n}$ in the complementary function in Eq. (19) with variables: functions $u_{1}, u_{2}, \dots,$ $u_{n}$ of x. We ask whether it is possible to choose these functions in such a way that the combination

$y_{p} (x) = u_{1} (x) y_{1} (x) + u_{2} (x) y_{2} (x) + \dots + u_{n} (x) y_{n} (x)$ (21)

is a particular solution of the nonhomogeneous equation in (18). It turns out that this is always possible.

The method is essentially the same for all orders $n ≧ 2,$ but we will describe it in detail only for the case $n = 2 .$ So we begin with the second-order nonhomogeneous equation

$L [y] = y ″ + P (x) y ′ + Q (x) y = f (x)$ (22)

with complementary function

$y_{c} (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x)$ (23)

on some open interval I where the functions P and Q are continuous. We want to find functions $u_{1}$ and $u_{2}$ such that

$y_{p} (x) = u_{1} (x) y_{1} (x) + u_{2} (x) y_{2} (x)$ (24)

is a particular solution of Eq. (22).

One condition on the two functions $u_{1}$ and $u_{2}$ is that $L [y_{p}] = f (x) .$ Because two conditions are required to determine two functions, we are free to impose an additional condition of our choice. We will do so in a way that simplifies the computations as much as possible. But first, to impose the condition $L [y_{p}] = f (x),$ we must compute the derivatives $y_{p} ′$ and $y_{p} ″ .$ The product rule gives

$y_{p}^{′} = (u_{1} y_{1}^{′} + u_{2} y_{2}^{′}) + (u_{1}^{′} y_{1} + u_{2}^{′} y_{2}) .$

To avoid the appearance of the second derivatives $u_{1} ″$ and $u_{2} ″,$ the additional condition that we now impose is that the second sum here must vanish:

$u_{1}^{′} y_{1} + u_{2}^{′} y_{2} = 0.$ (25)

Then

$y_{p}^{′} = u_{1} y_{1}^{′} + u_{2} y_{2}^{′},$ (26)

and the product rule gives

$y_{p}^{″} = (u_{1} y_{1}^{″} + u_{2} y_{2}^{″}) + (u_{1}^{′} y_{1}^{′} + u_{2}^{′} y_{2}^{′}) .$ (27)

But both $y_{1}$ and $y_{2}$ satisfy the homogeneous equation

$y ″ + P y ′ + Q y = 0$

associated with the nonhomogeneous equation in (22), so

$y_{i}^{″} = - P y_{i}^{′} - Q y_{i}$ (28)

for $i = 1, 2 .$ It therefore follows from Eq. (27) that

$y_{p}^{″} = (u_{1}^{′} y_{1}^{′} + u_{2}^{′} y_{2}^{′}) - P \cdot (u_{1} y_{1}^{′} + u_{2} y_{2}^{′}) - Q \cdot (u_{1} y_{1} + u_{2} y_{2}) .$

In view of Eqs. (24) and (26), this means that

$y_{p}^{″} = (u_{1}^{′} y_{1}^{′} + u_{2}^{′} y_{2}^{′}) - P y_{p}^{′} - Q y_{p};$

hence

$L [y_{p}] = u_{1}^{′} y_{1}^{′} + u_{2}^{′} y_{2}^{′} .$ (29)

The requirement that $y_{p}$ satisfy the nonhomogeneous equation in (22)—that is, that $L [y_{p}] = f (x)$ —therefore implies that

$u_{1}^{′} y_{1}^{′} + u_{2}^{′} y_{2}^{′} = f (x) .$ (30)

Finally, Eqs. (25) and (30) determine the functions $u_{1}$ and $u_{2}$ that we need. Collecting these equations, we obtain a system

$\begin{aligned} u_{1}^{′} y_{1} + u_{2}^{′} y_{2} & = & 0, \\ u_{1}^{′} y_{1}^{′} + u_{2}^{′} y_{2}^{′} & = & f (x) \end{aligned}$ (31)

of two linear equations in the two derivatives $u_{1} ′$ and $u_{2} ′ .$ Note that the determinant of coefficients in (31) is simply the Wronskian $W (y_{1}, y_{2}) .$ Once we have solved the equations in (31) for the derivatives $u_{1} ′$ and $u_{2} ′,$ we integrate each to obtain the functions $u_{1}$ and $u_{2}$ such that

$y_{p} = u_{1} y_{1} + u_{2} y_{2}$ (32)

is the desired particular solution of Eq. (22). In Problem 63 we ask you to carry out this process explicitly and thereby verify the formula for $y_{p} (x)$ in the following theorem.

Example 11

Find a particular solution of the equation $y ″ + y = \tan x$ .

Solution

The complementary function is $y_{c} (x) = c_{1} \cos x + c_{2} \sin x,$ and we could simply substitute directly in Eq. (33). But it is more instructive to set up the equations in (31) and solve for $u_{1} ′$ and $u_{2} ′,$ so we begin with

$\begin{aligned} y_{1} & = & \cos x, & y_{2} & = & \sin x, \\ y_{1}^{′} & = & - \sin x, & y_{2}^{′} & = & \cos x . \end{aligned}$

Hence the equations in (31) are

$\begin{aligned} (u_{1}^{′}) (\cos x) + (u_{2}^{′}) (\sin x) & = & 0, \\ (u_{1}^{′}) (- \sin x) + (u_{2}^{′}) (\cos x) & = & \tan x . \end{aligned}$

We easily solve these equations for

$\begin{aligned} u_{1}^{′} & = & - \sin x \tan x = - \frac{\sin^{2} x}{\cos x} = \cos x - \sec x, \\ u_{2}^{′} & = & \cos x \tan x = \sin x . \end{aligned}$

Hence we take

$u_{1} = \int (\cos x - \sec x) d x = \sin x - \ln | \sec x + \tan x |$

and

$u_{2} = \int \sin x d x = - \cos x .$

(Do you see why we choose the constants of integration to be zero?) Thus our particular solution is

$\begin{aligned} y_{p} (x) & = & u_{1} (x) y_{1} (x) + u_{2} (x) y_{2} (x) \\ = & (\sin x - \ln | \sec x + \tan x |) \cos x + (- \cos x) (\sin x); \end{aligned}$

that is,

$y_{p} (x) = - (\cos x) \ln | \sec x + \tan x | .$

5.5 Problems

In Problems 1 through 20, find a particular solution $y_{p}$ of the given equation. In all these problems, primes denote derivatives with respect to x.

$y ″ + 16 y = e^{3 x}$
$y ″ - y ′ - 2 y = 3 x + 4$
$y ″ - y ′ - 6 y = 2 \sin 3 x$
$4 y ″ + 4 y ′ + y = 3 x e^{x}$
$y ″ + y ′ + y = \sin^{2} x$
$2 y ″ + 4 y ′ + 7 y = x^{2}$
$y ″ - 4 y = \sinh x$
$y ″ - 4 y = \cosh 2 x$
$y ″ + 2 y ′ - 3 y = 1 + x e^{x}$
$y ″ + 9 y = 2 \cos 3 x + 3 \sin 3 x$
$y^{(3)} + 4 y ′ = 3 x - 1$
$y^{(3)} + y ′ = 2 - \sin x$
$y ″ + 2 y ′ + 5 y = e^{x} \sin x$
$y^{(4)} - 2 y ″ + y = x e^{x}$
$y^{(5)} + 5 y^{(4)} - y = 17$
$y ″ + 9 y = 2 x^{2} e^{3 x} + 5$
$y ″ + y = \sin x + x \cos x$
$y^{(4)} - 5 y ″ + 4 y = e^{x} - x e^{2 x}$
$y^{(5)} + 2 y^{(3)} + 2 y ″ = 3 x^{2} - 1$
$y^{(3)} - y = e^{x} + 7$

In Problems 21 through 30, set up the appropriate form of a particular solution $y_{p},$ but do not determine the values of the coefficients.

$y ″ - 2 y ′ + 2 y = e^{x} \sin x$
$y^{(5)} - y^{(3)} = e^{x} + 2 x^{2} - 5$
$y ″ + 4 y = 3 x \cos 2 x$
$y^{(3)} - y ″ - 12 y ′ = x - 2 x e^{- 3 x}$
$y ″ + 3 y ′ + 2 y = x (e^{- x} - e^{- 2 x})$
$y ″ - 6 y ′ + 13 y = x e^{3 x} \sin 2 x$
$y^{(4)} + 5 y ″ + 4 y = \sin x + \cos 2 x$
$y^{(4)} + 9 y ″ = (x^{2} + 1) \sin 3 x$
${(D - 1)}^{3} (D^{2} - 4) y = x e^{x} + e^{2 x} + e^{- 2 x}$
$y^{(4)} - 2 y ″ + y = x^{2} \cos x$

Solve the initial value problems in Problems 31 through 40.

$y ″ + 4 y = 2 x; y (0) = 1,$ $y ′ (0) = 2$
$y ″ + 3 y ′ + 2 y = e^{x}; y (0) = 0,$ $y ′ (0) = 3$
$y ″ + 9 y = \sin 2 x; y (0) = 1,$ $y ′ (0) = 0$
$y ″ + y = \cos x; y (0) = 1,$ $y ′ (0) = - 1$
$y ″ - 2 y ′ + 2 y = x + 1; y (0) = 3,$ $y ′ (0) = 0$
$y^{(4)} - 4 y ″ = x^{2}; y (0) = y ′ (0) = 1,$ $y ″ (0) = y^{(3)} (0) = - 1$
$y^{(3)} - 2 y ″ + y ′ = 1 + x e^{x}; y (0) = y ′ (0) = 0,$ $y ″ (0) = 1$
$y ″ + 2 y ′ + 2 y = \sin 3 x; y (0) = 2,$ $y ′ (0) = 0$
$y^{(3)} + y ″ = x + e^{- x}; y (0) = 1, y ′ (0) = 0,$ $y ″ (0) = 1$
$y^{(4)} - y = 5; y (0) = y ′ (0) = y ″ (0) = y^{(3)} (0) = 0$
Find a particular solution of the equation

$y^{(4)} - y^{(3)} - y ″ - y ′ - 2 y = 8 x^{5} .$
Find the solution of the initial value problem consisting of the differential equation of Problem 41 and the initial conditions

$y (0) = y ′ (0) = y ″ (0) = y^{(3)} (0) = 0.$
1. Write
  
  $\cos 3 x + i \sin 3 x = e^{3 i x} = {(\cos x + i \sin x)}^{3}$
  
  by Euler’s formula, expand, and equate real and imaginary parts to derive the identities
  
  $\begin{aligned} \cos^{3} x & = & \frac{3}{4} \cos x + \frac{1}{4} \cos 3 x, \\ \sin^{3} x & = & \frac{3}{4} \sin x - \frac{1}{4} \sin 3 x . \end{aligned}$
2. Use the result of part (a) to find a general solution of
  
  $y ″ + 4 y = \cos^{3} x .$

Use trigonometric identities to find general solutions of the equations in Problems 44 through 46.

$y ″ + y ′ + y = \sin x \sin 3 x$
$y ″ + 9 y = \sin^{4} x$
$y ″ + y = x \cos^{3} x$

In Problems 47 through 56, use the method of variation of parameters to find a particular solution of the given differential equation.

$y ″ + 3 y ′ + 2 y = 4 e^{x}$
$y ″ - 2 y ′ - 8 y = 3 e^{- 2 x}$
$y ″ - 4 y ′ + 4 y = 2 e^{2 x}$
$y ″ - 4 y = \sinh 2 x$
$y ″ + 4 y = \cos 3 x$
$y ″ + 9 y = \sin 3 x$
$y ″ + 9 y = 2 \sec 3 x$
$y ″ + y = \csc^{2} x$
$y ″ + 4 y = \sin^{2} x$
$y ″ - 4 y = x e^{x}$
You can verify by substitution that $y_{c} = c_{1} x + c_{2} x^{- 1}$ is a complementary function for the nonhomogeneous second-order equation

$x^{2} y ″ + x y ′ - y = 72 x^{5} .$

But before applying the method of variation of parameters, you must first divide this equation by its leading coefficient $x^{2}$ to rewrite it in the standard form

$y ″ + \frac{1}{x} y ′ - \frac{1}{x^{2}} y = 72 x^{3} .$

Thus $f (x) = 72 x^{3}$ in Eq. (22). Now proceed to solve the equations in (31) and thereby derive the particular solution $y_{p} = 3 x^{5}$ .

In Problems 58 through 62, a nonhomogeneous second-order linear equation and a complementary function $y_{c}$ are given. Apply the method of Problem 57 to find a particular solution of the equation.

$x^{2} y ″ - 4 x y ′ + 6 y = x^{3}; y_{c} = c_{1} x^{2} + c_{2} x^{3}$
$x^{2} y ″ - 3 x y ′ + 4 y = x^{4}; y_{c} = x^{2} (c_{1} + c_{2} \ln x)$
$4 x^{2} y ″ - 4 x y ′ + 3 y = 8 x^{4/3}; y_{c} = c_{1} x + c_{2} x^{3/4}$
$x^{2} y ″ + x y ′ + y = \ln x; y_{c} = c_{1} \cos (\ln x) + c_{2} \sin (\ln x)$
$(x^{2} - 1) y ″ - 2 x y ′ + 2 y = x^{2} - 1; y_{c} = c_{1} x + c_{2} (1 + x^{2})$
Carry out the solution process indicated in the text to derive the variation of parameters formula in (33) from Eqs. (31) and (32).
Apply the variation of parameters formula in (33) to find the particular solution $y_{p} (x) = - x \cos x$ of the nonhomogeneous equation $y ″ + y = 2 \sin x$ .

5.5 Application Automated Variation of Parameters

The variation of parameters formula in (33) is especially amenable to implementation in a computer algebra system when the indicated integrals would be too tedious or inconvenient for manual evaluation. For example, suppose that we want to find a particular solution of the nonhomogeneous equation

$y ″ + y = \tan x$

of Example 11, with complementary function $y_{c} (x) = c_{1} \cos x + c_{2} \sin x .$ Then the Maple commands

y1 := cos(x):
y2 := sin(x):
f := tan(x):
W := y1*diff(y2,x) - y2*diff(y1,x):
W := simplify(W):
yp := -y1*int(y2*f/W,x) + y2*int(y1*f/W,x):
simplify(yp);

implement (33) and produce the result

$y_{p} (x) = - (\cos x) \ln (\frac{1 + \sin x}{\cos x})$

equivalent to the result $y_{p} (x) = - (\cos x) \ln (\sec x + \tan x)$ found in Example 11. The analogous Mathematica commands

y1 = Cos[x];
y2 = Sin[x];
f = Tan[x];
W = y1*D[y2,x] - y2*D[y1,x]   //  Simplify
yp = -y1*Integrate[y2*f/W,x] + y2*Integrate[y1*f/W,x];
Simplify[yp]

produce the result

$y_{p} (x) = - (\cos x) \ln (\frac{\cos (x / 2) + \sin (x / 2)}{\cos (x / 2) - \sin (x / 2)}),$

which (by the usual difference-of-squares technique) also is equivalent to the result found in Example 11.

To solve similarly a second-order linear equation $y ″ + P (x) y ′ + Q (x) y = f (x)$ whose complementary function $y_{c} (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x)$ is known, we need only insert the corresponding definitions of $y_{1} (x), y_{2} (x),$ and f(x) in the initial lines shown here. Find in this way the indicated particular solution $y_{p} (x)$ of the nonhomogeneous equations in Problems 1 through 6.

1. $y ″ + y = 2 \sin x$	$y_{p} (x) = - x \cos x$
2. $y ″ + y = 4 x \sin x$	$y_{p} (x) = x \sin x - x^{2} \cos x$
3. $y ″ + y = 12 x^{2} \sin x$	$y_{p} (x) = 3 x^{2} \sin x + (3 x - 2 x^{3}) \cos x$
4. $y ″ - 2 y ′ + 2 y = 2 e^{x} \sin x$	$y_{p} (x) = - x e^{x} \cos x$
5. $y ″ - 2 y ′ + 2 y = 4 x e^{x} \sin x$	$y_{p} (x) = e^{x} (x \sin x - x^{2} \cos x)$
6. $y ″ - 2 y ′ + 2 y = 12 x^{2} e^{x} \sin x$	$y_{p} (x) = e^{x} [3 x^{2} \sin x + (3 x - 2 x^{3}) \cos x]$

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Table of Contents for 5.5 Nonhomogeneous Equations and Undetermined Coefficients

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Table of Contents for
5.5 Nonhomogeneous Equations and Undetermined Coefficients