Exercises 5.4 that solutions to the system ﬂow into (0 , 0) on one of them and ﬂow away from (0 , 0) on the other. This is the case of a saddle (unstable) point. (f) The only remaining picture is the asymptotically stable spiral. (One can also get a diagram .-% for solution’s ﬂows with just one matching picture in Figure 5.7.) 31. (a) Setting y0 = v and so y0 = v0 , we transform given equation to a ±rst order system dy dx = v, dv dx = f ( y ) . (b) By the chain rule, dv dy = dv dx · dx dy = dv dx ± dy dx = f ( y ) v ⇒ dv dy = f ( y ) v . This equation is separable. Separation variables and integration yield vdv = f ( y ) dy ⇒ Z vdv = Z f ( y ) dy ⇒ 1 2 v 2 = F ( y )+ K, where F ( y ) is an antiderivative of f ( y ). Substituting back v = y0 gives the required.
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