1550078481-Ordinary_Differential_Equations__Roberts_

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64 Ordinary Differential Equations

2.3.3 Solution of the Linear Equation y' = a(x)y + b(x)

Throughout this subsection, we will assume that the functions a(x) and
b(x) are both defined and continuous on some interval (a, {3). By the existence
and uniqueness theorem for linear first-order initial value problems stated in
section 2 .2, we are assured that for a ny c E (a , {3) the initial value problem

(21) y' = a(x)y + b(x); y(c) = d


has a unique solution on (a, {3). The purpose of this section is to present
a technique for finding the solution of the differential equation appearing in
(21), namely,

(22) y' = a(x)y + b(x)

and to express the solution of (22) as an equation which involves two qu adra-
tures (integrals).

If b(x ) is identically equ al to zero on the interval (a, {3), then equation (22)

reduces to

(23) y' = a(x)y.

Equation (23) is called a homogeneous linea r first-order differential equa-
tion. (To indicate mathematically that "b(x) is identically equal to zero on

the interval (a,{3),'' we write "b(x ) = 0 on the interval (a,{3)." ) Notice that


equation (23) is separable. We rewrite y' as dy/ dx, multiply by dx, and divide
by y to obtain t he separation of varia bles

Integrating, we find

dy

- = a(x) dx provided y i= 0.

y

j; = j a(x) dx or ln IYI = j a(x) dx + C provided y i= 0


and where C is a n arbitrary constant. Exponentiation yields
(24) IYI = ef a(x) dx+C = ec ef a(x) dx.

The left-hand side and the right-hand side of equation (24) are both positive.

Since IYI = ±y and since e^0 is an arbitrary positive constant, we m ay remove

the absolute value appearing in (24), if we also replace e^0 by a new constant

K which may be positive or negative. Observing that y(x) = 0 is a solution

of equation (23) y' = a(x)y, we also may allow K to assume the value zero.

Thus, we find the explicit solution of (23) on the interval (a, {3) is


(25) y(x) = Kef a(x)dx

where K is an arbitrary constant. Thus, we h ave proved the following theo-
rem.

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