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Differential Equation Answer 1—3 dy 4. t 1 2 the slope field of equation dt the slope field of Dy=t 2+1 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 7.(a) solution of Dy=3*y*(1-y) with y(0)=-1,0.5,1,2,3 on the slope field 2 1.5 1 0.5 the solution with initial conditiony(0)=0.5 0 y(t) -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 t (b) from the figure of (a) we can see that when the t is increasing, then the solution with initial condition y(0)=0.5 is approaches the equilibrium value 1. 10.(a) solution of Dy=(t+1)*y with y(0)=-1,0.5,1,2,3 on the slope field 2 1.5 1 0.5 the initial condition y(0)=0.5 y(t) 0 -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 t (b) from the figure of (a) we can see that when the t is increasing, then the solution with initial condition y(0)=0.5 is also increasing . 11.(a) 2 1.5 1 0.5 y(t) 0 -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 t dy from the above figure we can know that when t=1 ,the is 0, and dt dy dy when t >1, then <0 .when t<1,then >0,so we can choose the dt dt dy differential equation (4) 1 t . dt (b) 2 1.5 1 0.5 y(t) 0 -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 t from the above figure we can know that when y=1 is a equilibrium dy dy value ,and when y>1, then <0.when y<1,then >0 . so we can dt dt dy choose the differential equation (5) 1 y . dt (c) 2 1.5 1 0.5 y(t) 0 -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 t from the above figure we can know that when y=1,y=-1 is a equilibrium dy dy value ,and when y>1,then >0.when y<-1,then >0,when -1<y<1, dt dt dy dy 2 then <0 ,so we can choose the differential equation (8) y 1. dt dt (d) 2 1.5 1 0.5 y(t) 0 -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 t this slope field changes as y and as t change, so it can be equations (3) or (8) . when t=0 the slope are positive so it must be equation (3) 16. (a) by the definition the slope field we can sketch the slope field from the solution figure with initial condition y(0)=1. (b) the solution figure of y(0)=2 is the solution figure with initial condition y(0)=1 shift up to y(0)=2.because the differential equation depends only on t. 17. (a) because the equation is autonomous ,the slope field is constant on horizontal lines ,so this solution is enough information to sketch the slope field on the entire upper half plane ,also if we assume that f is continuous then the slope field on the line y=0 must be horizontal. (b) the solution with initial condition y(0)=2 is a translate to the left of the solution given. 1—8 2. dy 3 y t 5 dt t dy 3 y t 5 dt t 3 the integrating factor (t ) e t dt 1 3 t 1 product both side of equation with 3 t 1 dy 3 4 y t 2 3 t dt t 1 d( y ) t dt 2 3 t 1 d( y ) t dt 2 3 t 3 yt 1 3 C t 3 6 yt Ct 3 3 (C is a constant) 4. 2ty 4 e t dy 2 dt 2ty 4 e t dy 2 dt the integrating factor (t ) e 2 et 2 tdt 2 product both side of equation with product both side of equation with t e 2 dy 2 et 2t et y 4 dt 2 d ( t y ) 4dt e 2 d (et y ) 4dt 2 et y 4t C 2 2 y 4t et C et (C is constant) 12. dy 2t 2 y , y (0) 2 dt 1 t 2 1 t 2 dy 2t 2 y dt 1 t 2 1 t 2 2t the integrating factor (t ) e 1t dt 2 1 1 t 2 product both side of equation with product both side of equation with 1 1 t 2 1 dy 2t 2 y 1 t (1t 2) (1t 2) 2 2 2 dt 1 2 d( y) dy 1 t (1t ) 2 2 2 1 2 d( y) dt 1 t (1t ) 2 2 2 now we want to find the left side integrating let tta n dt s e cd 2 2 s e cd 2 4 sec 2 c o s d 2 1 (1 c o 2 )d s i n s 2 2 1 2t and we know that t tan tan t and sin 2 1 t 2 2 1 t so dt tan t (1t ) 1 t 2 2 2 hence the solution of equation is 1 1 t y tan t C 1 t 1 t 2 2 1 y (1 t ) tan t t C (1 t ) 2 2 by the initial condition y(0)=-2 ==> C =-2 so the solution with initial condition y(0)=-2 is 1 y (1 t ) tan t t 2(1 t ) 2 2 18. dy y 4 cost 2 dt dy y 4 cost 2 dt the integrating factor (t ) e 1dt t e product both side of equation with product both side of equation with t e t dy t t e y 4 e cost 2 e dt t t d (e y ) 4 e cost dt 2 t t d (e y ) 4 e cost dt 2 t t e y 4 e cost dt 2 t y 4e e t 2 cost dt 22. dS 0.06S 10000 dt S (0) 30000 dS 0.06S 1 0 0 0 0 dt the integrating factor (t ) e 0.06tdt 0.06t e product both side of equation with product both side of equation with 0.06 t e 0.06 t dS 0.06 t 0.06 t e 0.06S e 10000 e st 0.06 t 0.06 t d (e S ) 10000 e dt 0.06 t 0.06 t d (e S ) 10000 e dt 10000 0.06t 0.06 t e C S 0.06 e 10000 S Ce 0.06 t 0.06 by the initial condition S(0)=30000 10000 10000 30000 C C 30000 1.3667 10 5 = 0.06 0.06 10000 S (1.3667 10 ) e 5 0.06t 0.06 now we want to find out the t when S(t)=0 S (t ) 0 (1.6667 10 ) (1.3667 10 ) e 0 6 5 0.06 t 1.6667 10 1.6667 10 6 6 1 e t 0.06 t ln 1.3667 10 0.06 1.3667 105 5 t 3.3075( years) 24. dG 1 2 G dt 15 t G(0) 6 dG 1 G2 dt 15 t 1 the integrating factor (t ) e 15t 15 t dt product both side of equation with product both side of equation with 15+t dG (15 t ) G 2(15 t ) dt d ((15 t )G ) (30 2t )dt d ((15 t )G ) (30 2t )dt (15 t )G 30t t C 2 2 t 30t C G 15 t 15 t 15 t by the initial condition G(0)=6 C 6 C 90 15 2 t 30t 90 G 15 t 15 t 15 t 30t 90 2 G(t ) t and when t=15 it will full 15 t 15 450 90 2 G(15) 25.5 15 15 so 25.5 pound in the tank when it is full. 1—6 4. the plot of f(w)=w*cos(w) 4 3 2 1 f(w) 0 -1 -2 the equilibrium value -3 -4 -5 -4 -3 -2 -1 0 1 2 3 4 5 w from the plot of f(w)=w*cos(w) , we know that when n w=0,w= ,n ,f(w)=0 2 from the figure w=0, is a source w= , , is a sink 2 2 3 3 w= , , is a node 2 2 22.(a) we can discuss six situation: (1) when y=2 is node, y=-1 is sink: the solution of y(t) with initial condition y(0)=1 is decreasing from equilibrium value 2 to equilibrium value –1 as t decreasing. (2) when y=2 is node, y=-1 is source: the solution of y(t) with initial condition y(0)=1 is increasing from equilibrium value -1 to equilibrium value 2 as t increasing (3) when y=-1 is node, y=2 is sink: the solution of y(t) with initial condition y(0)=1 is like situation (2). (4) when y=-1 is node, y=2 is source: the solution of y(t) with initial condition y(0)=1 is like situation (1). (5) when y=-1 is sink, y=2 is source: the solution of y(t) with initial condition y(0)=1 is like situation (1). (6) when y=-1 is source, y=2 is sink: the solution of y(t) with initial condition y(0)=1 is like situation (2). (b) like situation (a)(2). 24. from the figure there are three equilibrium value denoted A,B,C A>B>C A is node ; B is source ; C is sink. 26. from the figure there are four equilibrium value denoted A,B,C,D A>B>C>D A is sink ; B is node ; C is source ; D is sink. 36.(a) the plot of 1/(y 2-y-2) 20 10 0 f(y) -10 -20 -30 -40 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 y dy dy y 1 , y2 , dt dt dy dy y2 , y 1 , dt dt dy 1 (b) solve the equation dt ( y 2)( y 1) ( y 2)( y 1)dy dt 2 ( y y 2)dy dt 3 2 y y 2y t C 3 2 1 y ( 0) 2 13 C 12 3 2 y y 2y t 13 3 2 12 plot the figure of the solution 3 2.5 2 1.5 1 y(t) 0.5 0 -0.5 -1 -1.5 -2 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 t

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