Example 3.4 To illustrate the process of using the matrizant, consider the initial value problem (3.15) c This equation is solved in Maple by finding the matrizant below. > restart: > with(linalg):with(plots): Warning, the protected names norm and trace have been redefined and unprotected Warning, the name changecoords has been redefined > N:=1; Enter the number of terms used in calculating the matrizant. (Usually 6 terms are sufficient). > nvars:=6; > Eq:=diff(c(t),t)=-t*c(t); > A:=matrix(1,1,[-t]); > Y0:=matrix(1,1,[1]); > id:=Matrix(N,N,shape=identity); Define the two dummy variables X1 and X2 > X1:=matrix(N,N);X2:=matrix(N,N); A dummy variable t1 is used in the integration. For matrix integration, Maple's 'map' command should be used. > X1:=map(int,subs(t=t1,evalm(A)),t1=0..t); > mat := evalm(id + X1) ; We now have the first two terms of the matrizant. The next step is to find the next five terms. A 'do loop' can be written to find the matrizant: > for i from 2 to nvars do S:=evalm( subs(t=t1,evalm(A))&*subs(t=t1,evalm(X1)) ):X2:= map(int,S,t1=0..t):mat := evalm(mat +X2) : X1:=evalm(X2):od : evalm(mat) ; > sol:=evalm(mat&*Y0); > C:=sol[1,1]; Thus, the process yields a series solution in t for C. This solution can be compared to the series solution given by Maple's 'dsolve' command: > dsolve({Eq,c(0)=1},c(t),type=series); By default, Maple gives a series solution accurate to the order of t 6 . The order of the series solution can be increased by using Maple's order specification as: > Order:=14; > dsolve({Eq,c(0)=1},c(t),type=series); We observe that the series solution obtained using the 'matrizant' method matches exactly with the series solution given by Maple's 'dsolve' command.