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Unformatted text preview: 2. Solve the initial value problem y = (1 + 3 x 2 ) / (3 y 2 6 y ) , y (0) = 1 and determine the interval in which the solution is valid. As before, we first manipulate the equation to the form (3 y 2 6 y ) dy = (1 + 3 x 2 ) dx. By the initial condition, one has Z y 1 (3 y 2 6 y ) dy = Z x (1 + 3 x 2 ) dx, y 3 3 y 2 fl fl y 1 = ( x + x 3 ) fl fl x , y 3 3 y 2 + 2 = x + x 3 . To find the interval of definition, we ought to look for points where the integral curve has a vertical tangent. Clearly this case occurs only when y = 0 or y = 2 . Considering the initial condition, which states that y (0) = 1 falls between 0 and 2, we have 0 < y < 2 thus 2 < y 3 3 y 2 + 2 < 2 . In other words, we obtain 2 < x + x 3 < 2 hence 1 < x < 1 . 2 3. Consider the initial value problem y = ty (4 y ) / (1 + t ) , y (0) = y > . (a) Determine how the solution behaves as t → ∞ . (b) If y = 2, find the time T at which the solution first reaches the value 3 . 99 ....
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This note was uploaded on 01/28/2011 for the course MATH 150 taught by Professor T.qian during the Spring '09 term at HKUST.
 Spring '09
 T.Qian

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