14_pdfsam_math 54 differential equation solutions odd

# 14_pdfsam_math 54 differential equation solutions odd - or...

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Chapter 1 both functions, f ( x, y )and ∂f ( x, y ) /∂y , are continuous in the rectangle R = ± ( x, y ): 1 <x< 1 , (1 / 2)10 7 <y< (2)10 7 ² containing the initial point (0 , 10 7 ). Thus, it follows from Theorem 1 that the given initial value problem has a unique solution in an interval about x 0 . 31. (a) To try to apply Theorem 1 we must Frst write the equation in the form y 0 = f ( x, y ). Here f ( x, y )=4 xy 1 and ∂f ( x, y ) /∂y = 4 xy 2 . Neither f nor ∂f/∂y are continuous or deFned when y = 0. Therefore there is no rectangle containing ( x 0 , 0) in which both f and ∂f/∂y are continuous, so Theorem 1 cannot be applied. (b) Suppose for the moment that there is such a solution y ( x )w ith y ( x 0 )=0and x 0 6 =0 . Substituting into the di±erential equation we get y ( x 0 ) y 0 ( x 0 ) 4 x 0
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Unformatted text preview: or y ( x ) 4 x = 0 4 x = 0 . Thus x = 0, which is a contradiction. (c) Taking C = 0 in the implicit solution 4 x 2 y 2 = C given in Example 5 on page 9 gives 4 x 2 y 2 = 0 or y = 2 x . Both solutions y = 2 x and y = 2 x satisfy y (0) = 0. EXERCISES 1.3: Direction Fields, page 22 1. (a) or y = 2 x , dy dx = d dx ( 2 x ) = 2 a n d 4 x y = 4 x 2 x = 2 , x 6 = 0 . Thus y = 2 x and y = 2 x are solutions to the dierential equation dy/dx = 4 x/y on any interval not containing the point x = 0. (b) , (c) See igures B.1 and B.2 in the answers of the text. 10...
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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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