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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.

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2 Answers 2

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 $z=k(y)y$.

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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.

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