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13_pdfsam_math 54 differential equation solutions odd

13_pdfsam_math 54 differential equation solutions odd -...

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Exercises 1.2 25. Writing dx dt = 4 t x = 4 tx 1 , we see that f ( t, x ) = 4 tx 1 and ∂f ( t, x ) /∂x = ( 4 tx 1 ) /∂x = 4 tx 2 . The functions f ( t, x ) and ∂f ( t, x ) /∂x are not continuous only when x = 0. Therefore, they are continuous in any rectangle R that contains the point (2 , π ), but does not intersect the t -axis; for instance, R = { ( t, x ) : 1 < t < 3 , 2 π < x < 0 } . Thus, Theorem 1 applies, and the given initial problem has a unique solution. 26. Here f ( x, y ) = 3 x 3 y 1 and ∂f ( x, y ) /∂y = 1 3 ( y 1) 2 / 3 . Unfortunately, ∂f/∂y is not continuous or defined when y = 1. So there is no rectangle containing (2 , 1) in which both f and ∂f/∂y are continuous. Therefore, we are not guaranteed a unique solution to this initial value problem. 27. Rewriting the differential equation in the form dy/dx = x/y , we conclude that f ( x, y ) = x/y . Since f is not continuous when y = 0, there is no rectangle containing the point (1 , 0) in which f is continuous. Therefore, Theorem 1 cannot be applied.
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Unformatted text preview: 29. (a) Clearly, both functions φ 1 ( x ) ≡ 0 and φ 2 ( x ) = ( x − 2) 3 satisfy the initial condition, y (2) = 0. Next, we check that they also satisfy the di±erential equation dy/dx = 3 y 2 / 3 . dφ 1 dx = d dx (0) = 0 = 3 φ 1 ( x ) 2 / 3 ; dφ 2 dx = d dx ± ( x − 2) 3 ² = 3( x − 2) 2 = 3 ± ( x − 2) 3 ² 2 / 3 = 3 φ 2 ( x ) 2 / 3 . Hence both functions, φ 1 ( x ) and φ 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 ⇒ ∂f ( x, y ) ∂y = 3 2 3 y 2 / 3 − 1 = 2 y 1 / 3 , x = 0 and y = 10 − 7 . The function f ( x, y ) is continuous everywhere; ∂f ( x, y ) /∂y is continuous in any region which does not intersect the x-axis (where y = 0). In particular, 9...
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