# Am I right?: solving u" = Au

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1. Nov 12, 2015

### kostoglotov

I know how to solve $$\frac{d\vec{u}}{dt} = A\vec{u}$$, I was just watching a lecture, and the lecturer related that solving that equation is pretty much a direct analogy to $$\vec{u} = e^{At}\vec{u}(0)$$, 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...

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

edit:

OR do I need to go along the lines

$$\frac{d^2\vec{u}}{dt^2} = \lambda^2\vec{u} = A\vec{u}$$

and solve

$$(A - \lambda^2 I)\vec{u}=\vec{0}$$

?

Last edited: Nov 12, 2015
2. Nov 13, 2015

### Svein

One way to look at it: Introduce a new function v: $\frac{d\vec{u}}{dt}= \vec{v}, \frac{d\vec{v}}{dt}=A\vec{u}$.

3. Nov 13, 2015