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I have given the following differential equation:

$x'= - y$ and $y' = x$

How can I solve them?

Thanks for helping! Greetings

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    Isn't there a rather obvious pair of trigonometric functions that might satisfy these equations (assuming that the prime indicates differentiation with respect to some third variable)?2012-04-21

3 Answers 3

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If you differentiate $y'$, you have: $y'' = -y$ Which has the solutions: $y=C_1 \cos(t) + C_2 \sin(t),$

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Introduce the complex dependent variable $z=x+iy,$ then your ode is $z'=iz,$ where $'$ is again the differentiation w.r.t. the independent variable $t$.
The characteristic polinomial is $P(\lambda)=\lambda-i,$ so the general solution is $z(t)=\alpha.e^{it},$ for an arbitrary $\alpha\in\mathbb{C}$.

P.S. : By the way your original system is the Hamilton equation for the harmonic oscillator $H(x,y)=\tfrac{1}{2}(x^2+y^2).$

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Let $\displaystyle X(t)= \binom{x(t)}{y(t)}$ so

$ X' = \left( \begin{array}{ccc} 0 & -1 \\ 1 & 0 \\ \end{array} \right)X .$

This has solution $ X(t)= \exp\biggr( \left( \begin{array}{ccc} 0 & -t \\ t & 0 \\ \end{array} \right) \biggr) X(0)= \left( \begin{array}{ccc} 0 & e^{-t} \\ e^t & 0 \\ \end{array} \right)\binom{x(0)}{y(0)}$

so $ x(t) = y(0) e^{-t} \ \ \text{ and } \ \ y(t) = x(0) e^{t} . $

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    your solution is not correct. In particular, the matrix exponent cannot be calculated as shown.2012-04-27