52 Ordinary Differential Equationsand continuous functions of x on (a, (3) is reached. For linear initial value
problems, it is always the left-hand and right-hand boundary of the largest
rectangle that the solution approaches. Hence, for linear initial value problems
we have the following theorem.
AN EXISTENCE AND UNIQUENESS THEOREM FOR
LINEAR FIRST-ORDER INITIAL VALUE PROBLEMS
If the functions a( x) and b( x) are defined and continuous on the interval
(a, (3) and if c E (a, (3), then the linear initial value problem(11) y' = a(x)y + b(x); y(c) = d
has a unique solution on the interval (a, (3).The existence and uniqueness theorem for the linear IVP (11) states that
it always has a solution on any interval (a, (3) containing c on which both
a(x) and b(x) are continuous for any choice of the initial valued. Further-
more, the linear IVP (11) has only one solution on the entire interval (a, (3).
Moreover, because of the uniqueness of solutions throughout the rectangle R,
no two distinct solution curves can cross one another. This existence and
uniqueness theorem and the previous two examples illustrate one of the im-
portant differences between the types of results we can expect for linear initial
value problems versus nonlinear initial value problems. For the linear initial
value problem, we are able to explicitly determine the interval of existence
and uniqueness of the solution from the differential equation and initial con-
dition prior to producing a solution; whereas, for the nonlinear initial value
problem we are, in general, unable to do so. For this reason when producing
a numerical solution to a nonlinear initial value problem, it is often desirable
to be able to monitor the solution as it is being generated.
EXAMPLE 5 Determination of the Interval of Existence
and Uniqueness
On what interval does each of the following linear initial value problems
have a unique solution?a. y'=3x^2 y+ex; y(-2)=4
b · y' --(x-l)(x+2) y +!·, x y( -1) = 2
c. y' = (cotx)y + tanx; y(2) = 3