Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have given the following differential equation:

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

How can I solve them?

Thanks for helping! Greetings

share|cite|improve this question
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
up vote 2 down vote accepted

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

share|cite|improve this answer

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 $$H(x,y)=\tfrac{1}{2}(x^2+y^2).$$

share|cite|improve this answer

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

share|cite|improve this answer
your solution is not correct. In particular, the matrix exponent cannot be calculated as shown. – Artem Apr 27 '12 at 12:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.