I know how to solve [tex]\frac{d\vec{u}}{dt} = A\vec{u}[/tex], I was just watching a lecture, and the lecturer related that solving that equation is pretty much a direct analogy to [tex]\vec{u} = e^{At}\vec{u}(0)[/tex], in so far as all we need to do after that is understand exactly what it means to take the exponential of a matrix, which is fine...(adsbygoogle = window.adsbygoogle || []).push({});

My question is this, do I solve [tex]\frac{d^2\vec{u}}{dt^2} = A\vec{u}[/tex] with [tex]\vec{u}=A^{-1}e^{At}\frac{d\vec{u}}{dt}(0)[/tex]?

edit:

OR do I need to go along the lines

[tex]\frac{d^2\vec{u}}{dt^2} = \lambda^2\vec{u} = A\vec{u}[/tex]

and solve

[tex](A - \lambda^2 I)\vec{u}=\vec{0}[/tex]

?

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# Am I right?: solving u" = Au

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