Nonlinear Systems and Phenomena In previous chapters we have often used explicit solutions of differential equations to answer specific numerical questions. But even when a given ode is difficult or impossible to solve explicitly,it often is possible to extract qualitative information about general properties of its solutions. 口1①y元200
Nonlinear Systems and Phenomena In previous chapters we have often used explicit solutions of differential equations to answer specific numerical questions. But even when a given ode is difficult or impossible to solve explicitly, it often is possible to extract qualitative information about general properties of its solutions
Example Let x(t)denote the temperature of a body with initial tempera- turex(0)=x0.Att=0 this body is immersed in a medium with constant temperature A.Assuming Newton's law of cooling, 合=--A)k>0 conctant) 4日10y至，无2000
Example Let x(t) denote the temperature of a body with initial temperature x(0) = x0. At t = 0 this body is immersed in a medium with constant temperature A. Assuming Newton’s law of cooling, dx dt = −k(x−A) (k > 0 constant) x(t) = A+ (x0 −A)e −kt It follows immediately that lim t→∞ x(t) = A. So the temperature of the body approaches that of the surrounding medium
Example Let x(t)denote the temperature of a body with initial tempera- turex(0)=x0.At t=0 this body is immersed in a medium with constant temperature A.Assuming Newton's law of cooling, 告=-k-A刻>0 x(1)=A+(xO-A)e-k It follows immediately that limx(t)=A.So the temperature of the body approaches that of the surrounding medium. 4日10y至，无2000
Example Let x(t) denote the temperature of a body with initial temperature x(0) = x0. At t = 0 this body is immersed in a medium with constant temperature A. Assuming Newton’s law of cooling, dx dt = −k(x−A) (k > 0 constant) x(t) = A+ (x0 −A)e −kt It follows immediately that lim t→∞ x(t) = A. So the temperature of the body approaches that of the surrounding medium
r'>0 r'0 =A Phase iogram (a) (6) 口0y至，1无12000
t x x=A 0 Typicalsolution curves (a) x<A x=A x '>0 x>A x '<0 Phase diagram (b)
Example d西 d =Ay-By2,AB>0,0)=0 Solution A y- B+(分-B)eA 生0 vftl-0 0 -0 牛0 (e).t-0,B-0 (64<0,B<0 0a0
Example dy dt = Ay−By2 ,AB > 0, y(0) = y0 Solution y = A B+ ( A y0 −B)e−At t y 0 A B y2 (t)= A B y1(t)=0 dy dt <0 dy dt <0 dy dt >0 (a ) A>0, B>0 (c) t y y2 (t)= A B y1(t)=0 dy dt >0 dy dt <0 dy dt >0 (b) A<0, B<0 0 A B (d)