# Q2_130 - Math 415 Name: Quiz 2A Find the solution of y= 3...

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Math 415 Name: Quiz 2A Find the solution of y 0 = 3 x 2 + e x 2 y - 10 , y (0) = 1 , and determine (approximately) where the solution is deﬁned. The equation separates as (2 y - 10) dy = (3 x 2 + e x ) dx and so has implicit solution y 2 - 10 y = x 3 + e x + C . Using the initial condition tells us that C = - 10, so including this and completing the square gives us y 2 - 10 y + 25 = x 3 + e x + 15, or y ( x ) = 5 ± x 3 + e x + 15 , a choice of two functions We plug the initial condition into the function again to ﬁnd out that it is the minus case: y ( x ) = 5 - p x 3 + e x + 15 . Now this solution exists when x 3 + e x + 15 0, and graphing x 3 + e x + 15 tells us that this is true for x bigger than approximately - 2 . 47. Solve the initial value problem y 0 = 4 ty 2 , y (0) = y 0 , and explain how the interval in which the solution exists depends on the initial value y 0 . The equation separates as y - 2 dy = 4 tdt , which has solution - y - 1 = 2 t 2 + C , or, with the initial condition, - 1 y

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## This note was uploaded on 07/26/2011 for the course MATH 415 taught by Professor Costin during the Spring '07 term at Ohio State.

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Q2_130 - Math 415 Name: Quiz 2A Find the solution of y= 3...

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