The Initial Value Problem y' = f(x, y); y(c) = d 57we set x = - 7 in the general solution ( 4), solve the resulting equation
for C, and substitute this value into (4). Doing so, we find the solution
to the IVP (7) to be(8) y(x) = ln -lx+21
5- +3.
Since the initial condition, y(-7) = 3, is specified at x = - 7 E (-oo, -2),
equation (8) is the unique solution of the IVP (7) on the interval
(-oo, -2).Many pages in calculus texts and much student effort is devoted to methods
for calculating and expr essing antiderivatives of continuous function in terms
of elementary functions. Expressions for antiderivatives h ave b een collected
and appear in tables of integr als. However , the reader should be aware that
there are many relatively simple continuous functions whose antiderivatives
cannot b e expressed as an elementary function. For example, the following
list of functions which are defined and continuous on certain intervals do not
h ave antiderivatives that can be expressed as elementary functions:
- x2
e ,
x
1lnx'
sinx^2 ,sinx
xsin^2 x
x1
xtanx,When it is impossible or impractical to express the antiderivative of a function
f(x ) as an elementary function, t hen one must use series or numerical integra-
tion techniques to calculate the integral of f(x). In some cases, even if an a n-
tiderivative for f(x) can b e expressed as a n elementary function, it may still b e
simpler to approximat e the integral of f(x ) than to evaluate the antiderivative.
EXERCISES 2 .3 .1
Solve the given initial value problem by first finding the solution
of the d ifferential equation and then determining the value of the
constant of integration which satisfies the initial condition. A lso
specify the interval on which each solution exists.
1
l. y' = 3x + l ; y(l) = 2 2. y' = x + - ; y(l) = 2
x3. y' = 2sinx; y(n) = 1 4. y' = xsinx; y(n/2) = 1
1y(2) = 1
1- y'=--; 6. y' -. y(O) =^1