Given the partial differential equation xp+yq=0 . 1° Find…

Given the partial differential equation $xp+yq=0$.
1° Find its general integral.
2° Among the functions satisfying the given equation, find those that also satisfy $p^2+q^2=\dfrac{a^2}{x^2+y^2}$.

Step 1 · Part 1: General integral

\[ \frac{dx}{x}=\frac{dy}{y}=\frac{dz}{0}. \] From $dz=0$: $z=c_1$. From $dx/x=dy/y$: $y/x=c_2$. General solution: \[ z = \Phi\!\left(\frac{y}{x}\right), \] where $\Phi$ is an arbitrary differentiable function.

Step 2 · Part 2: Impose $p^2+q^2=a^2/(x^2+y^2)$

With $t=y/x$: $p=-\dfrac{y}{x^2}\Phi'(t)$, $q=\dfrac{\Phi'(t)}{x}$, so \[ p^2+q^2 = \frac{[\Phi'(t)]^2}{x^2}\!\left(\frac{y^2}{x^2}+1\right) = \frac{(x^2+y^2)[\Phi'(t)]^2}{x^4}. \] Setting this equal to $a^2/(x^2+y^2)$ and using $x^2+y^2=x^2(1+t^2)$: \[ [\Phi'(t)]^2 = \frac{a^2}{(1+t^2)^2} \;\Longrightarrow\; \Phi'(t)=\pm\frac{a}{1+t^2}. \] Integrating: $\Phi(t)=\pm a\arctan t+C$.

Verification

With $z=a\arctan(y/x)+C$: $p=-\dfrac{ay}{x^2+y^2}$, $q=\dfrac{ax}{x^2+y^2}$, so $xp+yq=0$ $\checkmark$, and $p^2+q^2=\dfrac{a^2(x^2+y^2)}{(x^2+y^2)^2}=\dfrac{a^2}{x^2+y^2}$. $\checkmark$