To find the solution to a differential equation of the form
$x^{'} = A x$
where A is a matrix.

Question

To find the solution to a differential equation of the formx′=Axwhere A is a matrix.

Kaphefsceasiaf8w9 · Accepted Answer

Step 1You consider the function |x(t)|2. If you want to show that it is increasing, then it&#039;s enough to show that |x(t)|22 is increasing, since |x(t)|2 is non-negative and the square-root with this co-domain is increasing.However, |x(t)|22=x(t)Tx(t) This is a product of differentiable functions of t, hence differentiable.We must use the (dot)product rule to differentiate. Recall that transposes commute with the derivative (they come out in differential quotients, and are continuous so can be exchanged with limits). When we do that , we getddtx(t)Tx(t)=[ddtx(t)T]x(t)+x(t)T[ddtx(t)]=[ddtx(t)]Tx(t)+x(t)T[ddtx(t)]=x(t)TATx(t)+x(t)TAx(t)=x(t)T[At+A]x(t)Now, AT+A is a symmetric matrix. As long as AT+A is also positive semi-definite, it is true that yT[AT+A]y≥0 for any y∈Rn (and in particular, for each x(t)) , and therefore the function |x(t)|2 is increasing (it is strictly increasing if AT+A is positive definite, since x(t)≠0 for all t so the derivative is in fact positive).

To find the solution to a differential equation

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