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39_pdfsam_math 54 differential equation solutions odd

# 39_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.2 Solving for y yields y = ± 1 1 x 2 . (2.6) Since, at the initial point, x = 0, y (0) = 1 > 1, we choose the positive sign in the above expression for y . Thus, the solution is y = 1 1 x 2 . Similarly we find solutions for the other two initial conditions: y (0) = 1 2 C = 4 y = 1 4 x 2 ; y (0) = 2 C = 1 4 y = 1 (1 / 4) x 2 . (c) For the solution to the first initial problem in (b), y (0) = 1, the domain is the set of all values of x satisfying two conditions 1 x 2 0 (for existence of the square root) 1 x 2 = 0 (for existence of the quotient) 1 x 2 > 0 . Solving for x , we get x 2 < 1 | x | < 1 or 1 < x < 1 . In the same manner, we find domains for solutions to the other two initial value problems:
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