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51_pdfsam_math 54 differential equation solutions odd

# 51_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.3 25. (a) This is a linear problem and so an integrating factor is µ ( x )=exp ±Z 2 xdx ² =exp ( x 2 ) . Multiplying the equation by this integrating factor yields e x 2 dy dx +2 xe x 2 y = e x 2 D x ³ ye x 2 ´ = e x 2 x Z 2 D t ³ ye t 2 ´ dt = x Z 2 e t 2 dt, where we have changed the dummy variable x to t and integrated with respect to t from 2 (since the initial value for x in the initial condition is 2) to x . Thus, since y (2) = 1, ye x 2 e 4 = x Z 2 e t 2 dt y = e x 2 e 4 + x Z 2 e t 2 dt = e 4 x 2 + e x 2 x Z 2 e t 2 dt . (b) We will use Simpson’s rule (page A.3 of the Appendix B) to approximate the deFnite integral found in part (a) with upper limit x = 3. Simpson’s rule requires an even number of intervals, but we don’t know how many are required to obtain the desired 3 place accuracy. Rather than make an error analysis, we will compute the approximate
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