1550078481-Ordinary_Differential_Equations__Roberts_

Applications of the Initial Value Problem y' = f(x, y); y(c) = d 119

Hence, the integral

y(x) = 1x J(t) dt

is equivalent to y(x) being the solution of the initial value problem

y' = f(x); y(a) = 0

and

lb J(t) dt

is equivalent to y(b), where y(b) is the solution to the initial value problem

y' = f(x); y(a) = 0 evaluated at b.

EXAMPLE 1 A Continuous Integrand Whose Antiderivative Is

Not an Elementary Function

Compute f 1

2

sinx^2 dx.

SOLUTION
The function f(x) = sinx^2 is defined and continuous on the interval [1, 2].
So J(x) is integrable on [1, 2], but its antiderivative F(x) cannot be expressed
as an elementary function. However, as we have just seen, the antiderivative
F(x) satisfies the initial value problem

(3) y' = sinx^2 ; y(l) = 0

and F(2) = f 1

2
sin x^2 dx. We used the program SOLVEIVP to solve the initial
value problem (3) numerically on the interval [1, 2]. Detailed instructions for
running the computer program SOLVEIVP is contained in Appendix A. We

found that J~ sinx^2 dx is approximately equal to 0.4945103.

The two MAPLE statements

sol:=dsolve( {diff(y(x), x) = sin(x A 2), y(l) = O} , numeric):

sol(2);

produces the value .494508245219554054 as its numerical approximation to

J 12 sinx^2 dx. This result was calculated using the default MAPLE numerical

integration procedure Runge-Kutta-Fehlberg 4(5), RKF45.

Use your computer software to solve the initial value problem (3) on the
interval [1, 2]. How does your result compare with the ones we obtained above?