300_pdfsam_math 54 differential equation solutions odd

300_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 ∂N ∂x = ∂x (2 y x ) = 1 . Therefore, the equation is exact. We have F ( x, y ) = N ( x, y ) dy = (2 y x ) dy = y 2 xy + g ( x ); M ( x, y ) = ∂x F ( x, y ) = ∂x ( y 2 xy + g ( x ) ) = y + g ( x ) = ( e x + y ) g ( x ) = e x g ( x ) = ( e x ) dx = e x . Hence, a general solution to the phase plane equation is given implicitly by F ( x, y ) = y 2 xy e x = C or e x + xy y 2 = C = c, where c is an arbitrary constant. 11. In this problem, f ( x, y ) = 2 y and g ( x, y ) = 2 x . Therefore, the phase plane equation for given system is dy dx = 2 x 2 y = x y . Separation variables and integration yield y dy = x dx y dy = x dx 1 2 y 2 = 1 2 x 2 + C y 2 x 2 = c. Thus, the trajectories are hyperbolas if c = 0 and, for c = 0, the lines y = ± x . In the upper half-plane,
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Unformatted text preview: y > 0, we have x = 2 y > 0 and, therefore, x ( t ) increases. In the lower half-plane, x < 0 and so x ( t ) decreases. This implies that solutions flow from the left to the right in the upper half-plane and from the right to the left in the lower half-plane. See Figure B.30 in the text. 13. First, we will ±nd the critical points of this system. Therefore, we solve the system ( y − x )( y − 1) = 0 , ( x − y )( x − 1) = 0 . 296...
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