The Initial Value Problem y' = f(x, y); y(c) = d 115
b. In this case, SOLVEIVP generated a numerical approximation on the
interval [-1, 1 J which is a very good approximation of the solution
y(x) = x^3 /9 to the IVP (29b). However , the integration procedure
gave us no indication that there are an infinite number of solutions of
the IVP (29b) on the interval [-1, 1] or that the solution is not unique
once y(x) = 0. MAPLE generated a numerical approximation of the
solution to the IVP (29b) on the interval [-1, -.054] which is a very
good approximation of the solution y(x) = x^3 /9. The software gave us
no indication that the solution could be extended further to the right,
that there are an infinite number of solutions of the IVP (29b) on the
interval [-1, 1], or that the solution is not unique once y(x) = 0.
The previous examples were presented to show that computer programs
which generate numerical approximations of solutions to initial value prob-
lems sometimes generate a solution where no solution exists, sometimes gener-
ate a single solution where there are multiple solutions, and sometimes do not
produce a solution where there is one or more solutions. This is the typical be-
havior of such computer programs. It is not a flaw of the computer program.
Hence, the previous examples should vividly illustrate to you the necessity
of performing a thorough mathematical analysis for each individual initial
value problem based on the fundamental theorems prior to generating any
numerical approximation. Use your computer software numerical integration
algorithm to solve the previous examples and compare your results to ours.
EXERCISES 2.4.4
Use the fundamental existence, uniqueness, and continuation the-
orems to mathematically analyze each of the following initial value
problems. Then use SOLVEIVP or a Computer Algebra System
( CAS) such as MAPLE, Mathematica, or MATLAB to calculate
and graph a numerical solution on the interval specified. Based
on the analysis or graph specify, when possible, the subinterval on
which the solution exists and the subinterval on which the solution
is unique.
1. y' = 1 /(x - 1); y(O) = 1 on [-2, 2]
- y'=y+x; y(O)=O on [-1,1]
- y' = y/x on [-2, 2] a. y(-1) = 1 b. y(-1) = -1
- y' = y/(l - x^2 ) + fi on [-2, 2]
a. y(.5) = 1 b. y(l) = 1 c. y(2) = 1
