168
and
Ordinary Differential Equati ons6
y"(x) = 4 + 4.
x
Substituting these expressions for y , y' , and y" into t he DE (7), we find
6 2 1x^2 y" + xy' - 4y = x^2 (4 + 4 ) + x(4x - 3) - 4(2x^2 + 2)
x x x
2 6 2 2 2 4
= 4x + - + x2 4x - -x 2 - 8x - -x2= 0 , provided x-/= 0.Since y , y' , and y" are defined for x-/= 0, y(x ) is a solution of the DE (7) on
(-oo, 0) and (0 , oo). Evaluating y and y' a t x = 1, yields y(l) = 2 + 1 = 3
1
and y'(l) = 4 - 2 = 2. Hen ce, y( x) = 2x^2 + 2 is the unique solution of the
x
IVP (9) on the interval (0 , oo ).
Suppose that y 1 (x ) and y 2 (x ) are both solutions of the same differential
equation and suppose that c 1 and c 2 a re two arbitrary const ants, then the
sum c 1 y 1 (x ) + c 2 y 2 (x ) is called a linear combination of t he two solutions.
The following example illustrates t hat a linear combination of two solutions
of a homogeneous linear differential equation is also a solution.
EXAMPLE 3 Linear Combination Solution of a
Homogeneous Linear Equationa. Show that y 1 (x ) = e-^2 x and Y2(x ) =ex are solutions on (-00,00) of
the homogeneous linear different ial equation
(10) y" + y' - 2y = o.b. Show that the linear combination y( x ) = c 1 y 1 (x) + c2y2(x ) = c 1 e-^2 x +
c2ex, where c1 and c2 are arbitrary constants, is also a solution to the
DE (10) on (-oo, oo ).SOLUTION
a. Differentiating Y1 = e-^2 x twice, we find Yi = -2e-^2 x and y{ = 4e-^2 x.
Substitution into (10), yieldsy" + y^1 - 2y = (4e-^2 x) + (- 2e-^2 x) - 2( e-^2 x) = (4 - 2 - 2) e -^2 x = 0.
Since Y1,yi, and y{ a re defined on (-00,00) , y 1 = e-^2 x is a solution of
the DE (10) on ( -oo, oo). Likewise, differentiating y 2 = ex twice, we
find y?, = ex and y~ = ex. Substitution into equation (10) yields