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**Unformatted text preview: **Exercises 4.6
Diﬀerentiating y (t) yields y (t) = −c1 e
−t et + c2 e + 2
t t 1 e−x dx + x et 2 e−t t e−t + 2 t 1 ex dx − x e−t 2 et t , where we have used the product rule and the fundamental theorem of calculus to diﬀerentiate the last two terms of y (t). We now plug in the second initial condition into the equation we just found for y (t) to obtain y (1) = −c1 e−1 + c2 e1 + Solving the system c1 e−1 + c2 e1 = 0, −c1 e−1 + c2 e1 = −2 yields c2 = −e−1 and c1 = e1 . Therefore, the solution to our problem is given by y (t) = e
1−t −e−1 2 e1 1 + e1 2 e−1 1 = −c1 e−1 + c2 e1 − 11 + = −2. 22 −e t−1 et + 2 t 1 e−t e−x dx − x 2 t 1 ex dx . x (4.8) Simpson’s rule is implemented on the software package provided free with the text (see also the discussion of the solution to Problem 25 in Exercises 2.3). Simpson’s rule requires an even number of intervals, but we don’t know how many are required to obtain the 2-place accuracy desired. We will compute the approximate value of y (t) at t = 2 using 2, 4, 6, . . . intervals for Simpson’s rule until the approximate value changes by less than ﬁve in the third place. For n = 2, we divide [1, 2] into 4 equal subintervals. Thus each interval will be of length (2 − 1)/4 = 1/4. Therefore the integrals are approximated by
2 1 2 1 e1 e1.25 e1.5 e1.75 e2 ex dx ≈ +4 +2 +4 + ≈ 3.0592 , x 12 1 1.25 1.5 1.75 2 1 e−1 e−1.25 e−1.5 e−1.75 e−2 e−x dx ≈ +4 +2 +4 + ≈ 0.1706 . x 12 1 1.25 1.5 1.75 2 223 1 ...

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