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I have eqn: $\frac{dx}{dt} = -y(t)$ and $\frac{dy}{dt} = x(t)$

I know that $(x(0),y(0))= (1,0)$.

I want to solve eqn and show that it admits an invariant $I = x(t)^2 + y(t)^2$.

I know $x' = -y$,

$y' = x$,

$x^{\prime\prime} = -y' = -x$

I know general solution of $x" = -x$ is $x = a\sin x = b\cos x$.

I know $x(0) = a\sin 0 + b\cos 0 = 1$ So $b = 1$

How can I show $a = 1$? (I think it should!)

I tried $x' = a\cos x - b\sin x$ since $y = -x$ but it just gives $ a = 0$.

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    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-11-18
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    Please, try to make the title of your question more informative. E.g., *Why does $a imply $a+c?* is much more useful for other users than *A question about inequality.* From [How can I ask a good question?](http://meta.math.stackexchange.com/a/589/): *Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader.*2012-11-18

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