Math 20d mathlab 2 PDF

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Matlab2...

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Ex 2.1 (a, b) (c)>> dsolve('Dy=y/5','x') ans = C1*exp(x/5)

Ex 2.2 (a). >> f=@(x,y) (exp(-x)-y)*(exp(-x)+2+y) f= function_handle with value: @(x,y)(exp(-x)-y)*(exp(-x)+2+y) >> slopefield(f,[-10,10],[-10,10],40) >> hold on >> drawode(f,[-10,10],2,3) >> drawode(f,[-10,10],0,2) >> hold of 10

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(b)The calculation with e is too complicated to get a value to help us imagine the solution. Direction field can fundamentally help us to visualize the solution (general).

Ex 2.3 >> f=@(x,y)x+2*y f= function_handle with value:

@(x,y)x+2*y >> slopefield(f,[-5,5],[-5,5],20) >> hold on >> drawode(f,[-5,5],0,-1/4) >> drawode(f,[-5,5],-1/4,-1/4) >> drawode(f,[-5,5],1/4,-1/4) >> drawode(f,[-5,5],0,0) >> hold of 5 4

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As we can see from the plot, if the initial value is (0, -1/4), the solution would be a straight line, otherwise it is not. So initial value not being exactly (0, -1/4) would greatly afect the solution of diferential equation.

Ex 2.4 >> f=@(x,y) y-1 f= function_handle with value: @(x,y)y-1 >> slopefield(f,[-6,6],[-6,6],20) >> hold on >> drawode(f,[-6,6],1,1) >> drawode(f,[-6,6],0.9,1)

>> drawode(f,[-6,6],1,0.9) >> drawode(f,[-6,6],0.9,0.9) >> hold of 6

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If the initial value is not (1,1), as seen from plot above, this would greatly afect the solution.

Ex 2.5 >> f=@(t,y) 2*(1-y) f= function_handle with value: @(t,y)2*(1-y) >> slopefield(f,[0,10],[-5,5],20)

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Now when A=2 >> f=@(t,y) 2*(2-y) f= function_handle with value: @(t,y)2*(2-y) >> slopefield(f,[0,10],[-5,5],20) >> hold on >> slopefield(f,[0,10],[-5,5],20) >> hold on >> drawode(f,[0,10],3,3) >> drawode(f,[0,10],3,4) >> drawode(f,[0,10],5,7) 5

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Now when A=4 >> f=@(t,y) 2*(4-y) f= function_handle with value: @(t,y)2*(4-y) >> slopefield(f,[0,10],[-5,5],20) >> hold on >> drawode(f,[0,10],3,3) >> drawode(f,[0,10],3,4) >> drawode(f,[0,10],5,7) >> hold of 5

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As we can observe from the above plots, A should represents the temperature of surrounding environment at which the temperature of the object will eventually get to.

Ex 2.6 (1).dy/dt=0.4*(41-y) y(0)=-6 (2) A=41

(3) >> f=@(t,y) 0.4*(39-y) f= function_handle with value: @(t,y)0.4*(39-y) >> slopefield(f,[-10,40],[-20,40],50) >> hold on >> drawode(f,[-10,40],0,-6) >> hold of

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Under the given condition, it takes around 13 hours to defrost the chicken breast.

(4) >> f=@(t,y) 0.4*(69-y) f= function_handle with value: @(t,y)0.4*(69-y) >> slopefield(f,[-10,40],[-10,70],50) >> hold on

>> drawode(f,[-10,40],0,-6)

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As we can see from the graph, the curve intersects with y=39 at around t=2.5. So about 132.5=10.5 hours could be saved if the chicken breast is thawed on chicken counter instead

Ex 2.7 >> g=@(t,x,y) [2;-3] g= function_handle with value: @(t,x,y)[2;-3] >> phaseplane(g,[-5,5],[-5,5],20) >> hold on >> drawphase(g, 8, 2, 1) >> drawphase(g, 3, 5, 2) >> drawphase(g, 6, -1, -2) >> hold of

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>> g=@(t,x,y) [4;-1] g= function_handle with value: @(t,x,y)[4;-1] >> phaseplane(g,[-5,5],[-5,5],20) >> hold on >> drawphase(g, 8, 2, 1) drawphase(g, 3, 5, 2) drawphase(g, 6, -1, -2) >> hold of 5

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>> g=@(t,x,y) [4;3]

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g= function_handle with value: @(t,x,y)[4;3] >> phaseplane(g,[-5,5],[-5,5],20) hold on drawphase(g, 8, 2, 1) drawphase(g, 3, 5, 2) drawphase(g, 6, -1, -2) hold of 5 4 3

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So changing x’ and y’ to other constants values will change the overall slope rate (slope rate is associated with the multiplication of those two constant)

Ex 2.8 >> g=@(t,Y) [Y(2);-Y(1)] g= function_handle with value: @(t,Y)[Y(2);-Y(1)] >> phaseplane(g,[-10,10],[-10,10],40) >> hold onw >> drawphase(g, 20, 1, -3) >> drawphase(g, 14, 2, 5) >> drawphase(g, 10, 5, 5) >> hold of

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Ex 2.9 >> f=@(x,y) x^2+y^3 f= function_handle with value: @(x,y)x^2+y^3 >> slopefield(f,[-20,20],[-20,20],40) >> hold on >> drawode(f,[-20,20],0,1) >> drawode(f,[-20,20],0,-5) >> drawode(f,[-20,20],5,5) >> drawode(f,[-20,20],0,0) >> hold of 20

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>> g=@(t,Y) [1; Y(1)^2 + Y(2)^3]; phaseplane(g, [-10,10], [-10,10], 40) >> hold on >> drawphase(g,30,0,1) >> drawphase(g,30,0,5) >> drawphase(g,30,0,0) >> hold of 1 0 8 6 4

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Phase portrait figure is a dynamic visualization of my directional field figure. The directional field depicts all the paths while phase portrait depicts only some t interval. Overall, the resulting curves match up.

Ex 2.10 (a)>> k=@(t,Y) [Y(1)*(1-Y(2));Y(2)*(Y(1)-1)] k= function_handle with value: @(t,Y)[Y(1)*(1-Y(2));Y(2)*(Y(1)-1)] >> phaseplane(k, [-1,5], [-1,5], 20)

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Physically possible solutions only exist (in first quadrant) for both x and y greater than or equal to zero. (b) >> k=@(t,Y) [Y(1)*(1-Y(2));Y(2)*(Y(1)-1)] k= function_handle with value: @(t,Y)[Y(1)*(1-Y(2));Y(2)*(Y(1)-1)] >> phaseplane(k, [-1,5], [-1,5], 20) >> hold on >> drawphase(k,75,2,3) >> drawphase(k,75,1,2) >> drawphase(k,75,4,1) >> hold of (C)states as marked below. 5

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