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Unformatted text preview: MATH315 2011 (Winter) (Ordinary Differential Equations) Solutions to WrittenAssignment #1 February 5, 2011 Solution of 1(a): One can rewrite the given equation as y ( y 2 1) 4 / 3 dy = xdx, which is a separable equation, so that a simple integration yields ( y 2 1) 1 / 3 = 1 / 3( x 2 + c ) . The initial condition determines c as c = 3( b 2 1) 1 / 3 . Therefore y = 1 + ( 1 3 x 2 + ( b 2 1) 1 / 3 ) 3 . Note that we have to choose the + sign since b > 0. In order to find the interval of definition, we need to know for which values of x , the expression under the radical is nonnegative. That is, when is ( 1 3 x 2 + ( b 2 1) 1 / 3 ) 3  1? To this end, we first make the following simple observation. For A R { } , A 3  1 A > or A  1 . Now we consider two cases: (I) b > 1. Then obviously b 2 1 > 0. With A = [ 1 3 x 2 + ( b 2 1) 1 / 3 ] , note that A >  x  < 3 ( b 2 1) 1 / 3 ; and that A  1  x  3 + 3 ( b 2 1) 1 / 3 . 1 Therefore, in this case, the interval of definition is { x R :  x  < 3 ( b 2 1) 1 / 3 } { x R :  x  3 + 3 ( b 2 1) 1 / 3 } . (II) 0 < b < 1. Then obviously, 0 < 1 b 2 < 1 . We can write A = [ 1 3 x 2  b 2 1  1 / 3 ] . In this case, A > 0 cannot happen(why?)....
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 Winter '09
 Equations

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