Systems of First-Order Differential Equations 323Notice that this theo rem guarantees the existence of a unique solution on
the entire interval I on which the functions aij ( x ) and bi ( x ) are all defined
and continuous. This is the m ajor difference b etween the results which one
can obtain for linear system initial value problems versus nonlinear system
initial value problems. For linear system initial value problems we can explic-
itly determine the interval of existence and uniqueness of the solution from
the problem itself; whereas, for the nonlinear system initial value problem,
we can only conclude that there exists a unique solution on some "small" un-
det ermined interval centered at x = c. The following example illustrates the
type of result we can obtain for linear system initial value problems.EXAMPLE 2 Analyzing a Linear Initial Value Problem
for Existence and UniquenessAnalyze the linear initial value problem(15a)(15b)SOLUTION
I r,:: 2
Y1 = XY1 + yXY2 + --1 x-yI ( )^2 3
2 = tanx Y1 - x Y2 + -x 2 --+ 1Y1 (1.5) = -1, Y2(1.5) = 2.
The system (15a) is linear with au(x) = x, a 1 2(x) =ft, b1(x) = 2/(x-1),
a 21 (x) = tanx, a22(x) = - x^2 , and b2(x) = 3/(x^2 +1). The functions au(x),
a 22 (x), and b 2 (x ) are defined and continuous for all real x-that is, on the
interval Ji= (- oo, oo). The function a12(x) =ft is defined and continuous
for x ~ 0- that is, on h = [O, oo). So the interval of existence and uniqueness
I must be a subinterval of J 1 n h = [O, oo). The function b 1 (x) is defined and
continuous for x =f. l. So b 1 (x) is defined and continuous on h = (-oo, 1)
and J 4 = (1, oo). Since J 1 n h n h = [O, 1) and J1 n J2 n J4 = (1, oo) , the
interval I will be a subinterval of [O, 1) or (1, oo). Since the initial condition
is specified at c = 1.5 E (1, oo), I must be a subinterval of (1, oo). The
function a 21 (x) = t an x is defined and continuous for x =f. (2n + l)7r/2 where
n is any integer. So a 21 (x) is defined and continuous on the intervals Kn =
((2n - l)7r/ 2, (2n + l)7r/2). Since the initial condition is specified at c = 1.5
and since 1.5 E ( -7r /2, 7r /2), the linear IVP (15) has a unique solution on the