Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

solve for $x,y,z$: $$\frac{dx}{x^{2}+a^{2}}=\frac{dy}{xy-az}=\frac{dz}{xz+ay}$$

please give a hint. I am not able to formulate the steps required to proceed solving this one.

share|cite|improve this question

We have $$ \begin{align} \frac{dy}{xy-az} & = \frac{dz}{xz+ay}\\ \frac{dy/y}{x-a (z/y)} & = \frac{dz/z}{x+a (y/z)} \end{align} $$ This gives a motivation to let $z = ky$ where $k$ is a constant. $$ \begin{align} \frac{dy/y}{x-a k} & = \frac{dy/y}{x+a/k} \end{align} $$ This gives us that $k = \pm i$. Let $k=i$. This gives us that $$\begin{align} \frac{dx}{x^2 + a^2} & = \frac{dy/y}{x - ia}\\ \frac{dx}{x + ia} & = \frac{dy}{y}\\ y & = c(x + ia) \end{align} $$ Hence, we get $$ \begin{align} z & = ic(x+ia)\\ y & = c(x+ia) \end{align} $$ and $$ \begin{align} z & = -ic(x-ia)\\ y & = c(x-ia) \end{align} $$ I don't know to justify my motivation why I chose $z = ky$ instead of $z=k(y)y$.

share|cite|improve this answer

Since the request is for a hint, I promote my comment to an answer. Write your system as $$\frac{dy}{dx}=\frac{xy-az}{x^{2}+a^{2}}$$

$$\frac{dz}{dx}=\frac{xz+ay}{x^{2}+a^{2}}$$ Maple does show non-constant solutions for this.

share|cite|improve this answer

Your Answer

 
discard

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.