Matlab20D #4.docx - Exercise 4.1 a > B=[1.2 2.5;4 0.7 b...

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Exercise 4.1 a) >> B=[1.2 2.5;4 0.7]; b) >> [eigvec,eigval]=eig(B) eigvec = 0.6501 -0.5899 0.7599 0.8075 eigval = 4.1221 0 0 -2.2221
Exercise 4.2 a) A = [3 4;-1 -2]; b) >> A=[3 4;-1 -2]; >> [eigvec,eigval]=eig(A) eigvec = 0.9701 -0.7071 -0.2425 0.7071 eigval = 2 0 0 -1 c) v ( t ) = c 1 e 2 t ( 0.9701 0.2425 ) + c 2 e 1 t ( 0.7071 0.7071 ) =( 0.9701 c 1 e 2 t 0.7071 c 2 e t 0.2425 c 1 e 2 t + 0.7071 c 2 e t ) As t gets large, the solution will approach + ¿ ) d) >> [email protected](t,Y)[3*Y(1)+4*Y(2);-1*Y(1)- 2*Y(2)]; >> P=[-3,3;-3,0;-3,2;-2,-3;-1,3;0,- 3;1,4;2,-3;4,-2;4,0;4,2]; >> phaseplane(g,[-4.5,4.5],[-4.5,4.5],16) >> hold on >> for i=1:size(P,1) drawphase(g,50,P(i,1),P(i,2)) end >> hold off The plot supports my answer to c) -5 -4 -3 -2 -1 0 1 2 3 4 5 -4 -3 -2 -1 0 1 2 3 4
Exercise 4.3 a) >> A=[2.7 -1;4.1 3.7]; >> [eigvec,eigval]=eig(A) eigvec = -0.1093 + 0.4291i -0.1093 - 0.4291i 0.8966 + 0.0000i 0.8966 + 0.0000i eigval = 3.2000 + 1.9621i 0.0000 + 0.0000i 0.0000 + 0.0000i 3.2000 - 1.9621i b) v ( t ) = c 1 e ( 3.2 + 1.9621 i ) t ( 0.1093 + 0.4291 i 0.8966 + 0 i ) + c 2 e ( 3.2 1.9621 i ) t ( 0.1093 0.4291 i 0.8966 + 0 i ) =( ( 0.1093 + 0.4291 i ) c 1 e ( 3.2 + 1.9 ( 0.8966 + 0 i ) c 1 e ( 3.2 + 1.9 c) >> [email protected](t,Y)[2.7*Y(1)-Y(2);4.1*Y(1)+3.7*Y(2)]; >> P=[-3,3;-3,0;-3,2;-2,-3;-1,3;0,-3;1,4;2,-3;4,-2;4,0;4,2]; >> phaseplane(g,[-4.5,4.5],[-4.5,4.5],16) >> hold on >> for i=1:size(P,1) drawphase(g,50,P(i,1),P(i,2)) end >> hold off The imaginary part does not affect the solutions in their direction, but the real value does. When real value is positive, the solution directs

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