227_pdfsam_math 54 differential equation solutions odd

227_pdfsam_math 54 differential equation solutions odd -...

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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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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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