# hw1solns - Section 1.1 dy d dy 2 y = 2 =(2 = 0 Substituting...

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Section 1.1 2. y = 2, dy dx = d dx (2) = 0. Substituting y and dy dx into the ODE, we get 0 = x 3 (2 - 2) 2 , which is true for all x . Thus y = 2 is a solution to dy dx = x 3 ( y - 2) 2 on ( -∞ , ). 6. y ( x ) = - 1 x - 3 , dy dx = ( x - 3) - 2 = 1 ( x - 3) 2 . y and dy dx do not exist when x - 3 = 0 x = 3. Substituting y and dy dx into the ODE, we get 1 ( x - 3) 2 = - 1 x - 3 2 = ( - 1) 2 ( x - 3) 2 = 1 ( x - 3) which is true for { x | x = 3 } . Thus y = - 1 x - 3 is a solution to the ODE on ( -∞ , 3). Section 1.2 7. We have ( x 2 - 1) y + 2 xy 2 = 0 and y ( 2) = 1. Then ( x 2 - 1) dy dx = - 2 xy 2 dy y 2 = 2 x dx x 2 - 1 - dy y 2 = 2 x x 2 - 1 dx 1 y = ln | x 2 - 1 | + C y = 1 ln | x 2 - 1 | + C IC: 1 1 = ln | 2 - 1 | + C C = 1 y = 1 ln | x 2 - 1 | + 1 9. y = 10 x + y dy dx = 10 x 10 y dy 10 y = 10 x dx 10 - y dy = 10 x dx Use a u = e u ln a . Then - 10 - y = 10 x + C 1 10 y = - 1 10 x + C 1 y = log 10 |- 10 x + C |

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13. We have ( x + 2 y ) y = 1, y (0) = - 2. Let u = 2 y + x . Then du dx
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Unformatted text preview: u ± u-1 2 ² = 1 This is separable: u-1 = 2 u ⇒ u = 2 u + 1 = 2 + u u ⇒ Z u 2 + u du = Z dx Note that u 2 + u = u + 2-2 2 + u = 1-2 2 + u Then Z ± 1-2 2 + u ² du = Z dx ⇒ u-2 ln | 2 + u | = x + C Substitute back u = x + 2 y : ( x + 2 y )-2 ln | 2 + x + 2 y | = x + C Apply IC: 0 + 2(-2)-2 ln | 2 + 0 + 2(-2) | = 0 + C-4-2 ln |-2 | = C The solution is given in implicit form as 2 y-2 ln | 2 + x + 2 y | + 4 + 2 ln 2 = 0...
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• Fall '06
• wirkus
• Derivative, Trigraph, UCI race classifications, Substitute good, Logarithm, Hugh Jackman

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