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We have $$\frac{\mathbb du}{\mathbb dt}=Au(t),\,t>0 \,\,and \,u(0)=(1,1)$$

Here $A$ is a symmetric square matrix of order $2$ and trace$(A)<0 \,\,and\,\,det(A)>0$. $u(t)=(u_1(t),u_2(t))$ is the unique solution then how can we evaluate $\displaystyle\lim_{t\to\infty}u_1(t)?$

MY TRY:Actually i know how to solve system of differential equation by matrix eigen value method but here $A$ is not given so i am clueless.Thank you.

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I do not have enough rep to comment, but think about what you actually know about the eigenvalues. Does that tell you something regarding the limit? That is the clue.

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    sum of all eigen value is trace and product is determinant.But i can't see any clue regarding limit. – 2017-01-02
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    Yes and you also know how many eigenvalues you have. Can you determine their respective signs? – 2017-01-02
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    both should be -ve. – 2017-01-02
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    Now i understood .The answer will be zero.Right? – 2017-01-02
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    Yes, both eigenvalues must be negative and therefore the solution will approach 0 when time goes to infinity. – 2017-01-02