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Unformatted text preview: y ( x ) 1, but that is not the solution that you want here. Since the equation is separable, you have Z 1 1-y dy = Z 1 dx, and-ln | 1-y | = x + C. Since y (0) =-1, that means that C =-ln2. So | 1-y | = e-x +ln 2 = 2 e-x , and y = 1-2 e-x . b) Find the solution to the equation in part a) satisfying y (0) = 1. Solution. This is the equilibrium solution y ( x ) 1....
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