# Section 9.7# Example 1 with(linalg):A := matrix([[1,-2,2],[-2,1,2],[2,2,1]]);

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g := vector([-9,0,-18]);

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a := vector(3); b := vector(3);

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xp := evalm(t*a+b);

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de := -map(diff,xp,t) + (A&*xp+t*g)=[0,0,0];# This is equivalent to the 2nd to last eqn on P.551

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de := evalm(%);

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map(diff,lhs(de),t)=[0,0,0]; # This is the "coefficient" of t, or eqn (3), P. 552

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eq1 := %;

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# The following verifies eq1 is equivalent to eqn(3) in the text:evalm(A&*a+g=[0,0,0]);

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eq2 := subs(t=0,lhs(de))=[0,0,0]; # This is the "coefficient" of 1, or eqn (4), P 552

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# The following verifies eq2 is equivalent to eqn(4):evalm(A&*b-a=[0,0,0]);

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# The following solve for a and b:linsolve(A,-g);

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a := %;

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linsolve(A,a);

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b := %;

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xp := evalm(a*t+b);

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# Example 2X := matrix([[3*exp(t),exp(-t)],[exp(t),exp(-t)]]);

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I := matrix([[1,0],[0,1]]);

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XI := augment(X,I);

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#We will use elementary row operations to reduce this to the form [I:X^(-1]:mulrow(XI,1,1/(3*exp(t)));#This multiplies row 1 by 1/(3*exp(t))

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addrow(%,1,2,-exp(t));#This adds -exp(t) times row 1 to row 2

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mulrow(%,2,3*exp(t)/2); #This multiplies row 2 by 3*exp(t)/2

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simplify(%);

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addrow(%,2,1,-1/3*exp(-2*t));

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simplify(%);

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# This right 2 by 2 matrix is the inverse of X.# Another way is to use Maple's "inverse" command:evalm(X);

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inverse(X);

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Xinv := simplify(%);

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# We next compute the integral in formula (13):subs(t=s,evalm(Xinv));

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Xinvs := %;

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fs := vector([exp(2*s),1]);

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evalm(Xinvs&*fs);

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simplify(%);

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map(int,%,s=0..t);

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temp := %; # Save this while we work on the first term in formula (13)

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subs(t=0,evalm(Xinv));

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simplify(%);

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evalm(%&*[-1,0]);

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# Next, we add this to temp, and then left-multiply the sum by X:evalm(%+temp);

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evalm(X &* %);

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simplify(%);

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