Unformatted text preview: 29. (a) Clearly, both functions Ï† 1 ( x ) â‰¡ 0 and Ï† 2 ( x ) = ( x âˆ’ 2) 3 satisfy the initial condition, y (2) = 0. Next, we check that they also satisfy the diÂ±erential equation dy/dx = 3 y 2 / 3 . dÏ† 1 dx = d dx (0) = 0 = 3 Ï† 1 ( x ) 2 / 3 ; dÏ† 2 dx = d dx Â± ( x âˆ’ 2) 3 Â² = 3( x âˆ’ 2) 2 = 3 Â± ( x âˆ’ 2) 3 Â² 2 / 3 = 3 Ï† 2 ( x ) 2 / 3 . Hence both functions, Ï† 1 ( x ) and Ï† 2 ( x ), are solutions to the initial value problem of Exapmle 9. (b) In this initial value problem, f ( x, y ) = 3 y 2 / 3 â‡’ âˆ‚f ( x, y ) âˆ‚y = 3 2 3 y 2 / 3 âˆ’ 1 = 2 y 1 / 3 , x = 0 and y = 10 âˆ’ 7 . The function f ( x, y ) is continuous everywhere; âˆ‚f ( x, y ) /âˆ‚y is continuous in any region which does not intersect the xaxis (where y = 0). In particular, 9...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Continuous function, Boundary value problem, Lipschitz continuity

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