1
$\begingroup$

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.

$\endgroup$
0
1
$\begingroup$

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 $z=k(y)y$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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