Q1rs - y x â‰ 1 but that is not the solution that you want...

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Quiz I 1. Find the general solution to dy dx + 3 x y = x 2 + 1. Solution. This is in standard form. So a ( x ) = 3 /x , A ( x ) = R a ( x ) dx = 3ln | x | and the integrating factor is e A = | x | 3 = ± x 3 . Either choice will work but x 3 is easier, and you get x 3 dy dx + 3 x 2 y = x 5 + x 3 , d dx ( x 3 y ) = x 5 + x 3 , and x 3 y = x 6 / 6 + x 4 / 4 + C So the general solution is y = x 3 / 6 + x/ 4 + C/x 3 . If you saw that x 3 would be an integrating factor right at the beginning, all the better!

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2 2. a) Find the solution to dy dx = (1 - y ) satisfying y (0) = - 1. Solution. This diﬀerential equation has the equilibrium solution
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Unformatted text preview: y ( x ) â‰¡ 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 ) â‰¡ 1....
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Q1rs - y x â‰ 1 but that is not the solution that you want...

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