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9780199212033

Which has the general solution 1 2 y 2 ax e x c ii

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which has the general solution 1 2 y 2 = ax e x + C . (ii) Case (a) , 0. The system has no equilibrium points. From (i), d y/ d x is never zero, negative for y> 0 and positive for y< 0. Some phase paths are shown in Figure 1.12. Case (b) , a = 0. The system has no equilibrium points. As in (a), d d x is never zero. Some phase paths are shown in Figure 1.13. Case (c) , a> 0. This equation has one equilibrium point at x = ln a . The potential V (x) (see Section 1.3) of this conservative system is V = Z ( a + e x ) d x =− + e x , which has the expected stationary value at x = ln a . Since V ±± ( ln a) = e ln a = 0, the stationary point is a minimum which implies a centre in the phase diagram. Some phase paths are shown in Figure 1.14 for a = 2.

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10 Nonlinear ordinary differential equations: problems and solutions –2 –1 1 2 x –2 –1 1 2 y Figure 1.13 Problem 1.4: a = 0. –1 1 2 x –2 –1 1 2 y Figure 1.14 Problem 1.4: a> 0. 1.5 Sketch the phase diagrams for the equation ¨ x e x = a , for a< 0, a = 0, and 0. 1.5. ¨ x e x = a . The differential equation of the phase paths is given by y d y d x = a + e x , which has the general solution 1 2 y 2 = ax + e x + C . Case (a) , 0. There is a single equilibrium point, at x = ln ( a) . The potential V (x) (see Section 1.3 in NODE) of this conservative system is V = Z ( a e x ) d x =− e x ,
1 : Second-order differential equations in the phase plane 11 which has the expected stationary value at x = ln ( a) . Since V ±± ( ln ( a)) =− e ln ( = a< 0, the stationary point is a maximum, indicating a saddle at x = ln ( . Some phase paths are shown in Figure 1.15. Case (b) , a> 0. The equation has no equilibrium points. Some typical phase paths are shown in Figure 1.16. Case (c) , a = 0. Again the equation has no equilibrium points, and the phase diagram has the main features indicated in Figure 1.16 for the case 0, that is, phase paths have positive slope for y> 0 and negative slope for y< 0. –1 1 2 x –2 –1 1 2 y 2 Figure 1.15 Problem 1.5: 0. –3 –1 1 2 x – 2 – 1 1 2 y Figure 1.16 Problem 1.5: 0.

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12 Nonlinear ordinary differential equations: problems and solutions 1.6 The potential energy V (x) of a conservative system is continuous, and is strictly increasing for x< 1, zero for | x |≤ 1, and strictly decreasing for x> 1. Locate the equilibrium points and sketch the phase diagram for the system. 1.6. From Section 1.3, a system with potential V has the governing equation ¨ x =− d V d x . Equilibrium points occur where ¨ x = 0, or where d V (x)/ d x = 0, which means that all points on the x axis such that | x 1 are equilibrium points. Also, the phase paths are given by 1 2 y 2 = V + C . Therefore the paths in the interval | x 1 are the straight lines y = C . Since V is strictly increasing for 1, the paths must resemble the left-hand half of a centre at x 1. In the same way the paths for 1 must be the right-hand half of a centre. A schematic phase diagram is shown in Figure 1.17.
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which has the general solution 1 2 y 2 ax e x C ii Case a 0...

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