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Example 3

(a)    Find the general solution of the equation y' + 3t-1 y = 0 from Example 2, but for the interval t < 0 instead of t > 0.

(b)    Find the particular solution of the initial value problem with y(-2) = 1.

SOLUTION

(a)    This time we integrate with a negative t,

∫ 3t-1 dt = 3 ln |t| + B = 3 ln(-t) + B.

The general solution for t < 0 is thus

y(t) = Ce-3 ln (-t) = C(-t)-3 = -Ct-3,

or

y(t) = At-3,

where A is the constant -C.

(b)    The particular solution with the initial value y(-2) = 1 is found by solving for A:

1 = A(-2)-3, A = -8, y(t) = -8t-3.

The solution to this example is shown in Figure 14.2.3.

14_differential_equations-40.gif

Figure 14.2.3 Example 3

For some purposes, it is useful to describe the solution of a differential equation using a definite integral from some point a to t, instead of using an indefinite integral. In the definite integral form, the general solution of the linear homogeneous differential equation

(1)

y' + p(t) y = 0

is

14_differential_equations-41.gif

where a is any point in the interval I, and C is any real number. This formula is helpful in a problem where one cannot evaluate the integral of p(t) exactly and must use a numerical approximation. The formula holds because the integral

14_differential_equations-42.gif

is an antiderivative of p(t) by the Fundamental Theorem of Calculus. The choice of the endpoint a does not matter because a change in the value of a will be absorbed by a change in the value of the constant C. If we are given an initial value y(t0) = y0, the particular solution can again be found by substituting and solving for C. If we take a = t0, the constant C will be equal to y0,

14_differential_equations-43.gif

Thus the particular solution of the initial value problem (1) with y(a) = y0 is given by

14_differential_equations-44.gif


Last Update: 2006-11-17