hw_1_30_09

hw_1_30_09 - Ma 221 Homework Solutions Due date: January...

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Ma 221 Homework Solutions Due date : January 30 , 2009 2 . 6p . 79 # 2 ̄ 1 ̄ , 23 , 28 ; 4 . 2p . 167 # 1 ̄ , 3 , 7 ̄ , 9 , 17 , 26 , 2 ̄ 7 ̄ 29 ( For 2 ̄ 7 ̄ & 29 you may use the Wronskian .) (Underlined Problems are not handed in) 2 . . 79 # 2 ̄ 1 ̄ , 23 , 28 For 21, 23 and 28 use the method discussed under "Bernoulli Equations" to solve the problems. 21.) dy dx y x x 2 y 2 This is a Bernoulli Equation with n 2 Let u y 1 n y 1 2 y 1 then y u 1 y u 2 u and the last equation becomes 1 u 2 du dx 1 ux x 2 u 2 du dx 1 x u x 2 This is a linear equation with P x 1 x Find x e Pdx e 1 x dx e ln| x | x 1 Multiply through by x to get 1 x du dx 1 x 2 u x d dx 1 x u x Integrate to get d dx 1 x u xdx 1 x u x 2 2 C 1 u 1 2 x 3 C 1 x Solve explicitly for y y 1 x 3 2 C 1 x 2 Cx x 3 y 0 is also a solution to the original equation. It was lost in the first step when we multiplied by u 2 (also same as dividing by y 2 ). 23.) dy dx 2 y x x 2 y 2 or after dividing by y 2 and moving the first term on the right to the left we have y 2 dy dx 2 y 1 x x 2 Let v y 1 v y 2 y and the last equation becomes v 2 v x x 2 This is a first order linear equation in v . The integrating faction for this equation is e 2 x dx x 2 1
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Multiplying the DE by this gives x 2 v 2 xv d dx x 2 v x 4 Thus x 2 v 1 5 x 5 c 1 and 1 y v x 3 5 c 1 x 2 Therefore y x 3 5 c 1 5 x 2 1 Letting C 5 c 1 we have finally y 5 x 2 x 3 C y 0 is also a solution to the original equation. It was lost in the first step when we divided by y 2 .
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hw_1_30_09 - Ma 221 Homework Solutions Due date: January...

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