1550078481-Ordinary_Differential_Equations__Roberts_

N-th Order Linear Differential Equations 213

SOLUTION

The associated homogeneous equation y^11 - 3y' + 2y = 0 has a uxiliary

equation p(r) = r^2 - 3r + 2 = 0. Since the two roots of this equation are 1

and 2, the complementary solution of equation (3) is

Si nce the nonhomogeneity b(x) = 5, a constant, we guess that the particular

solution has the form Yp(x) = A, where A is an unspecified constant. Dif-

ferentiating Yp(x) =A twice, we find y~(x) = 0 and y~(x) = 0. Substituting

Yp(x) and its derivatives into equation (3), we find A must satisfy 2A = 5.

Hence, A= 5/2 and a particular solut ion of (3) is Yp(x) = 5/2. Therefore, the

general solution of (3) is

EXAMPLE 2 Solution of a Nonhomogeneous Linear

Differential Equation with Constant Coefficients

Solve the nonhomogeneous linear differential equation

(4) y^11 - 3y^1 +2y = 4e^3 x.

SOLUTION

As in example 1, the auxiliary equation of the associated homogeneous

equation y^11 - 3y' + 2y = 0 has roots 1 and 2. So the complementary solution

of equation ( 4) is

Yc(x) = c1ex + c2e^2 x.

Since the nonhomogeneity b(x) = 4e^3 x, we guess that the particular solution

has the form YP ( x) = Ae^3 x, where A is an unspecified constant. Differentiating

Yp(x) = Ae^3 x twice, we find y~(x) = 3Ae^3 x and y~(x) = 9Ae^3 x. Substituting

Yp(x) and its derivatives into equation (4), we find A must satisfy

9Ae^3 x - 3(3Ae^3 x) + 2(Ae^3 x) = 4e^3 x or 2Ae^3 x = 4e^3 x.

Hence, A = 2 and a particular solut ion of ( 4) is Yp ( x) = 2e^3 x. Therefore, the

general solution of ( 4) is