# 285hw1 - HW1 1. Sec. 1.1: 10. Let y (x) = x ln x. Then 1 x...

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HW1 1. Sec. 1.1: 10. Let y ( x ) = x & ln x . Then dy ( x ) dx = 1 & 1 x and d 2 y ( x ) dx 2 = 1 x 2 : We have x 2 y 00 + xy 0 & y = x 2 & 1 x 2 ± + x & 1 & 1 x ± & ( x & ln x ) = ln x: Now let y ( x ) = 1 x & ln x: Then dy ( x ) dx = & 1 x 2 & 1 x and d 2 y ( x ) dx 2 = 2 x 3 + 1 x 2 : x 2 y 00 + xy 0 & y = x 2 & 2 x 3 + 1 x 2 ± + x & & 1 x 2 & 1 x ± & & 1 x & ln x ± = ln x: 2. Sec. 1.1: 15. y ( x ) = e rx . Then y 0 ( x ) = re rx and y 00 ( x ) = r 2 e rx . We have y 00 + y 0 & 2 y = r 2 e rx + re rx & 2 e rx = ² r 2 + r & 2 ³ e rx = 0 ; whence r 2 + r & 2 = 0 , Solution is: & 2 ; 1 r 2 + r & 2 = 0 : Hence r 1 = & 2 and r 2 = 1 . Solutions: y 1 ( x ) = e & 2 x and y 2 ( x ) = e x : 1

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3. Sec. 1.1 :21. Substitute y ( x ) = Ce & x 3 into the equation y 0 + 3 x 2 y = 0 : We &nd that dy ( x ) dx = & 3 Cx 2 e x 3 ; then y 0 + 3 x 2 y = & 3 2 e x 3 + 3 x 2 & & x 3 ± = 0 : 4. Sec. 1.1: 31. The equation of the tangent line to the graph of g at the point ( x; y ) is Y & y = g 0 ( x ) ( X & x ) . Now because ( & y; x ) is on the line, we have x & y = g 0 ( x ) ( & y & x ) , whence g 0 ( x ) ( x + y ) = x & y ) g 0 ( x ) = y & x x + y : Because y = g ( x ) ; y 0 ( x ) = y & x x + y : 5. Sec. 1.1: 33. dv dt = kv 2 ; k = const: 6. Sec. 1.1: 35. The number of people who have not heard the rumor is P & N . So, dN dt = k ( P & N ) .
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## This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285hw1 - HW1 1. Sec. 1.1: 10. Let y (x) = x ln x. Then 1 x...

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