solution to a differential equation

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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It's in fact a differential system. You can differentiate in both sides of the first equation, use the second to get an equation involving only $x$. – Davide Giraudo Apr 21 '12 at 9:16
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)? – user22805 Apr 21 '12 at 9:24

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. – Artem Apr 27 '12 at 12:07