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It is known that the solution to a Cauchy problem of linear differential equations: $$x'(t)=A(t)x(t), \quad \textup{with}\quad x(t_0)=x_0,$$ can be presented by the matrix exponential as $\exp({\int_{t_0}^tA(s)\,ds})x_0,$ if the commutativity condition for the coefficient matrix $A(t)$ holds: $$\Big[\int_{t_0}^tA(s)\,ds,A(t)\Big]=0.$$ A natural question is whether this is true without the commutativity condition. To give a definite answer to this question, we present two classes of illustrative examples of coefficient matrices, which satisfy the chain rule $$\D \frac d {dt}\, \exp({\int_{t_0}^t A(s)\, ds})=A(t)\,\exp({\int_{t_0}^t A(s)\, ds}),$$ but do not possess the commutativity condition. The presented matrices consist of finite-times continuously differentiable entries or smooth entries.

}, issn = {2075-1354}, doi = {https://doi.org/10.4208/aamm.09-m0946}, url = {http://global-sci.org/intro/article_detail/aamm/8386.html} }It is known that the solution to a Cauchy problem of linear differential equations: $$x'(t)=A(t)x(t), \quad \textup{with}\quad x(t_0)=x_0,$$ can be presented by the matrix exponential as $\exp({\int_{t_0}^tA(s)\,ds})x_0,$ if the commutativity condition for the coefficient matrix $A(t)$ holds: $$\Big[\int_{t_0}^tA(s)\,ds,A(t)\Big]=0.$$ A natural question is whether this is true without the commutativity condition. To give a definite answer to this question, we present two classes of illustrative examples of coefficient matrices, which satisfy the chain rule $$\D \frac d {dt}\, \exp({\int_{t_0}^t A(s)\, ds})=A(t)\,\exp({\int_{t_0}^t A(s)\, ds}),$$ but do not possess the commutativity condition. The presented matrices consist of finite-times continuously differentiable entries or smooth entries.

*Advances in Applied Mathematics and Mechanics*.

*1*(4). 573-580. doi:10.4208/aamm.09-m0946