1550078481-Ordinary_Differential_Equations__Roberts_

The Initial Value Problem y' = f(x, y); y(c) = d 67

we obtain the function

(30) Yi(x) = ef a(x)dx_

Remember that y 1 (x) is a solution of the homogeneous differential equation

y' = a(x)y on the interval (a,/3). Observe that y 1 (x) is positive on (a,/3)

and that any constant times y 1 (x) is also a solution of y' = a(x)y on (a,/3).

Let us see if it is possible to find a nonconstant function v ( x) defined on (a, JJ)

such that the function

(31) y(x) = v(x)y1(x)

is a particular solution on (a, /3) of the nonhomogeneous differential equation

(28) y' = a(x)y + b(x). Differentiating (31), we find

(32) y' ( x) = v' ( x )Y1 ( x) + v( x )y~ ( x).

Substituting (31) and (32) into (28), we see that v(x) must satisfy

(33) v'(x)y 1 (x) + v(x)y~(x) = a(x)(v(x)y 1 (x)) + b(x).

Since the function y 1 ( x) satisfies the homogeneous linear differential equation

y~(x) = a(x)y 1 (x), it follows that v(x)y~(x) = v(x)a(x)y 1 (x). Consequently,

(33) reduces to

(34) v^1 (x)y1(x) = b(x).

Since y 1 (x) is positive on the interval (a,/3), the function v'(x) = b(x)/y 1 (x)

is defined for all x E (a,/3) and since b(x) and y 1 (x) are continuous on (a,/3),

v(x) = i x b(t) dt

Y1 (t)

exists on (a, /3). Therefore, a particular solution to the nonhomogeneous
differential equation (28) is

(35)

i

x b(t)

yp(x) = v(x)y1(x) = Y1(x) Yi(t) dt.

Substituting (30) and (35) into (29), we find the solution of the nonhomoge-
neous differential equation (28) is

z(x) = Ky1(x) + v(x)y1(x) = Y1(x)(K + v(x))

where K is any arbitrary constant, y 1 (x) = ef"' a(t) dt, and v(x) = r b((t)) dt.
Y1 t
Consequently, we have proved the following theorem.