Page 116 - Schaum's Outline of Differential Equations

CHAP. 11] THE METHOD OF UNDETERMINED COEFFICIENTS 99

2
11.8. Determine the form of a particular solution for /' = 9x + 2x-\.
2
Here (f>(x) = 9x + 2x- 1, and the solution of the associated homogeneous differential equation y"=Q is
2
y h = c\x + c 0. Since <j>(x) is a second-degree polynomial, we first try y p = A 2x + Ape + A 0. Note, however, that this
assumed solution has terms, disregarding multiplicative constants, in common with y h: in particular, the first-power
m
term and the constant term. Hence, we must determine the smallest positive integer m such that x (A 2x 2 + AIX + A Q)
has no terms in common with y h.
For m = 1, we obtain

which still has a first-power term in common with y h. For m = 2,we obtain

which has no terms in common with y h; therefore, we assume an expression of this form for y p.

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11.9. Solve/' = 9x + 2x-l.
Using the results of Problem 11.8, we have y h = c^x + c 0 and we assume

Substituting (1) into the differential equation, we obtain

from which A 2 = 3/4, A l = 1/3, and A Q = -1/2. Then (1) becomes

and the general solution is

The solution also can be obtained simply by twice integrating both sides of the differential equation with
respect to x.

Sx
11.10. Solve / - 5y = 2e .
5
From Problem 10.3, y h = c^. Since <j>(x) = 2e ^, it would follow from Eq. (11.2) that the guess for y p should
5x
be y p = A 0e . Note, however, that this y p has exactly the same form as y h; therefore, we must modify y p. Multiplying
by x (m = 1), we obtain
y p

As this expression has no terms in common with y h; it is a candidate for the particular solution. Substituting (1) and
5x
5x
5x
y' p = A 0e + 5A 0x^ x into the differential equation and simplifying, we obtain A 0e = 2e , from which A 0 = 2.
5x
Equation (1) becomes y p = Ixe"^, and the general solution is y = (cj + 2x)e .
11.11. Determine the form of a particular solution of

Here <j>(x) = (x — 1) sin x + (x + 1) cos x, and from Problem 10.3, we know that the solution to the associated
x
homogeneous problem y' — 5y = 0 is y h = c^ . An assumed solution for (x — 1) sin x is given by Eq. (11.5) (with
a = 0) as