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Solve the initial value problem:

$\frac{du}{dt}= \pmatrix{1&2\\-1&1}u, u(0) = \pmatrix{1\\0}$

What I got is,

Eigenvalues : $\lambda = 1 \pm \sqrt{2}i$

Eigenvectors : $v_1 = \pmatrix{1\\-\frac{1}{\sqrt {2}i}},v_2 = \pmatrix{1\\\frac{1}{\sqrt {2}i}}$

Then, $[e^tcos\sqrt{2}t + ie^tsin\sqrt{2}t]\pmatrix{1\\-\frac{1}{\sqrt {2}i}}$ the problem I am having is the initial conditions. Can anyone show me how to do it?

The answer to this problem is

$u(t) = \pmatrix{e^tcos\sqrt{2}t\\-\frac{1}{\sqrt {2}}e^tsin\sqrt{2}t}$ I do not know how they got that.

1 Answers 1

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You write "eigenvalue" in the singular, but you (correctly) give two eigenvalues; then you give only a single eigenvector. The two different eigenvalues have different eigenvectors. You can only satisfy the initial conditions by using both eigenvectors.

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    Okay, I am going to reattempt the problem with your suggestion and I will fix my grammar mistake.2012-11-29