1550078481-Ordinary_Differential_Equations__Roberts_

CSODE User's Guide 519

Iii SOLVEIVP ~(!IJl:i

SOLVEIVP calculates ana graphs the solution of the in~ial value problem: y' = f{•. y); y{c) = d on the interval [a. bl where a<= c <= b

Enter the function f{•.Yl =,.v_!ll .. :<·1_!'1 .. x+:_'JJ .. •11,_'"-------------.------~--

Enter the in~erval of integration [a. b~ Enter the initial condition y{c) = d

a =I .,,_ __ ·2.5 __, 6 =. .._ J __ 0.5 ..... c=( ___ _. d=I ,_ __ .,.,.....

Actual interval of integration is [Xmin. Xma•J Minimum and maxim ... solution values on [Xmin. Xma•)

Xmin = I ·2.5 XmaK = 5.000000E ·01 Ymin = I ·20 YmaK =·I ===OPTIONS ===i

._ __ . ·"-------' "'--------·^20 i.=-=--.....

YmaKr----...---.----.--------.----.----.--------~,

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Xmin Xmax

Figure A.14 Graph of the Direction F ield and Numerical Approximation of

the Solution to the IVP: y' = ( ~ ) + ~; y(-1) = 2

x-l x + 2 x

The differential equation of t he given initial value problem is linear- tha t i s,

y' = a(x)y+b(x) where a(x ) = 1/ ((x-l)(x+2)) and b(x ) = l/x. The function

a(x ) is not defined , and therefore, not continuous at x = -2 a nd x = 1. The

function b(x ) is not defined and not continuous at x = 0. The functions a(x )

and b(x ) a re both continuous on the intervals (-oo, -2), (-2, 0) , (0, 1) , and

(1, oo). Since -1 E (-2, 0) , t he IVP of this example has a unique solut ion on

(-2, 0) and this is the la rgest interval on which it h as a solution. The graph

and direction field indicat e the presence of vertical asymptot es in the solution

near x = - 2 and x = 0. In this exam ple, SOLVEIVP calculat ed a numerical

approxima tion to the solut ion of the initial value problem where none exist s.

For x > 0, SOLVEIVP is generating a numerical approximation to a different

init ial value problem- one which has t he same different ial equation but a
different init ial condition.