39_pdfsam_math 54 differential equation solutions odd

# 39_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.2 Solving for y yields y = ± 1 1 x 2 . (2.6) Since, at the initial point, x =0 , y (0) = 1 > 1, we choose the positive sign in the above expression for y .Thu s ,theso lu t ioni s y = 1 1 x 2 . Similarly we Fnd solutions for the other two initial conditions: y (0) = 1 2 C =4 y = 1 4 x 2 ; y (0) = 2 C = 1 4 y = 1 p (1 / 4) x 2 . (c) ±or the solution to the Frst initial problem in (b), y (0) = 1, the domain is the set of all values of x satisfying two conditions ( 1 x 2 0 (for existence of the square root) 1 x 2 6 = 0 (for existence of the quotient) 1 x 2 > 0 . Solving for x ,weget x 2 < 1 ⇒| x | < 1o r
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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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