hw2solns - y = x , so solutions exist and are unique...

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Section 1.5 3. (2 x + 1) y 0 = 4 x + 2 y leads to y 0 - 2 y 2 x + 1 = 4 x 2 x + 1 Using P = - 2 2 x +1 and Q = 4 x 2 x +1 gives Z - 2 2 x + 1 dx = - ln | 2 x + 1 | e R p ( x ) = e - ln | 2 x +1 | = 1 | 2 x + 1 | e - R p ( x ) = e ln | 2 x +1 | = | 2 x + 1 | Z 4 x 2 x + 1 | 2 x + 1 | dx = 2 x 2 · sgn(2 x + 1) where sgn(2 x + 1) = ± - 1 if 2 x + 1 < 0 1 if 2 x + 1 0 The solution is y = | 2 x + 1 | (2 x 2 · sgn(2 x + 1) + C ) or y = (2 x + 1)(2 x 2 + C 1 ) 15. y 0 + 2 xy = 1 with y (2) = 1. Then we have A ( x ) = Z 2 x dx = x 2 e A ( x ) = e x 2 , e - A ( x ) = e - 4 x 2 Z b ( x ) e A ( x ) = Z e x 2 dx Z x 2 y ( t ) e A ( t ) = Z x 2 b ( t ) e At Section 2.1 1. Looking at the point (2 , 1), y 0 = 9 2 , which matches graph b. 2. Looking at the point (2 , 1), y 0 = 18, which matches graph c. 3. Looking at the point (2 , 1), y 0 = 2 9 , which matches graph a. 4. Looking at the point (2 , 1), y 0 = 1 18 , which matches graph d. Section 2.2 2. 1. f ( x, y ) = 3 x ( y +2) 2 / 3 , ∂f ∂y = 2 x ( y + 2) 1 / 3 , which is discontinuous when y = - 2, so a solution exists and is unique everywhere except possibly along y = - 2. 2. f ( x, y ) = ( x - y ) 1 / 5 , ∂f ∂y = - 1 5( x - y ) 4 / 5 , which is discontinuous when
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Unformatted text preview: y = x , so solutions exist and are unique everywhere except possibly along y = x . Alternatively, f ( x, y ) = ( x-y ) 5 , f y =-1 5 , so solutions exist and are unique everywhere. 3. f ( x, y ) = x 2 y-1 , f y =-x 2 y-2 , which are both discontinuous when y = 0, so solutions exist and are unique everywhere except possibly along y = 0. 4. f ( x, y ) = ( x + y )-2 , f y =-2 ( x + y ) 3 , which are both discontinuous when y =-x , so solutions exist and are unique everywhere except possibly along y =-x ....
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hw2solns - y = x , so solutions exist and are unique...

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