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
This is the end of the preview. Sign up
access the rest of the document.
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.