Find the general solution of xy (∂ u)/(∂ x) - √(1-y^2)(y (∂…
Find the general solution of \[ xy\,\frac{\partial u}{\partial x} - \sqrt{1-y^2}\!\left(y\,\frac{\partial u}{\partial y} - z\,\frac{\partial u}{\partial z}\right) = axy\,\frac{\partial u}{\partial z}, \] where $u = u(x,y,z)$ and $a$ is a real constant.
Explanation of Method
Rewritten as $xy\,u_x - y\sqrt{1-y^2}\,u_y + \bigl(z\sqrt{1-y^2}-axy\bigr)u_z=0$, a Lagrange PDE with $P=xy$, $Q=-y\sqrt{1-y^2}$, $R=z\sqrt{1-y^2}-axy$. The first integral is separable directly; the second follows from the trigonometric substitution $y=\sin\theta$ along a characteristic.
Step 1 · Characteristic equations
\[ \frac{dx}{xy} = \frac{dy}{-y\sqrt{1-y^2}} = \frac{dz}{z\sqrt{1-y^2}-axy}. \tag{1} \]
Step 2 · First integral
From the first two ratios, dividing by $y$: $dx/x=-dy/\sqrt{1-y^2}$. Integrating: $\ln x=-\arcsin y+\text{const}$, so \[ x\,e^{\arcsin y} = c_1. \tag{I} \]
Step 3 · Second integral
From the last two ratios: $(z\sqrt{1-y^2}-axy)\,dy=-y\sqrt{1-y^2}\,dz$, i.e. \[ \sqrt{1-y^2}\,d(yz) = axy\,dy. \] Let $\theta=\arcsin y$ (so $y=\sin\theta$, $\sqrt{1-y^2}=\cos\theta$); along a characteristic $x=c_1e^{-\theta}$ from (I). The equation becomes \[ d(z\sin\theta) = ac_1e^{-\theta}\sin\theta\,d\theta. \] Using $\int e^{-\theta}\sin\theta\,d\theta = -\tfrac12e^{-\theta}(\sin\theta+\cos\theta)$ and substituting back $c_1e^{-\theta}=x$, $\sin\theta=y$, $\cos\theta=\sqrt{1-y^2}$: \[ zy = -\frac{ax(y+\sqrt{1-y^2})}{2} + c_2 \;\Longrightarrow\; c_2 = zy + \frac{ax}{2}\!\left(y+\sqrt{1-y^2}\right). \tag{II} \]
Step 4 · General solution
The general solution $c_2=\Phi(c_1)$ is: \[ zy + \frac{ax}{2}\!\left(y+\sqrt{1-y^2}\right) = \Phi\!\left(x\,e^{\arcsin y}\right), \] where $\Phi$ is an arbitrary differentiable function.
Verification
Writing $\dot x=P$, $\dot y=Q$, $\dot z=R$ along a characteristic:
$c_1=xe^{\arcsin y}$:
\[ \dot c_1 = e^{\arcsin y}\cdot xy + \frac{xe^{\arcsin y}}{\sqrt{1-y^2}}\cdot\bigl(-y\sqrt{1-y^2}\bigr)
= xye^{\arcsin y}-xye^{\arcsin y}=0. \checkmark \]
$c_2=zy+\tfrac{ax}{2}(y+\sqrt{1-y^2})$: expanding
$\dot c_2=(c_2)_xP+(c_2)_yQ+(c_2)_zR$ term by term, the $y^2$-, $y\sqrt{1-y^2}$-, and
$yz\sqrt{1-y^2}$-terms all cancel in pairs, giving $\dot c_2=0$. $\checkmark$
Result
The general solution is: