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