Find the partial differential equation whose general…
Find the partial differential equation whose general solution is given implicitly by \[ \Phi\!\left(\frac{z}{x^3},\;\frac{y}{x}\right) = 0, \] where $\Phi$ is an arbitrary (differentiable) function of two arguments.
Explanation of Method
When a general solution is written as $\Phi(u,v)=0$ for two known expressions $u = u(x,y,z)$ and $v = v(x,y,z)$, the corresponding first-order PDE is obtained by eliminating the arbitrary function $\Phi$. We differentiate $\Phi(u,v)=0$ with respect to $x$ and $y$ (treating $z$ as a function $z(x,y)$), then eliminate $\Phi_u$ and $\Phi_v$ from the resulting system.
Step 1 · Setup
Denote the two arguments by \[ u = \frac{z}{x^3}, \qquad v = \frac{y}{x}. \] Standard notation: $p = \dfrac{\partial z}{\partial x}$, $\quad q = \dfrac{\partial z}{\partial y}$.
Step 2 · Differentiate $\Phi(u,v)=0$ with respect to $x$
By the chain rule: \[ \Phi_u\,\frac{\partial u}{\partial x} + \Phi_v\,\frac{\partial v}{\partial x} = 0. \tag{1} \] Compute the partial derivatives of $u$ and $v$ with respect to $x$: \[ \frac{\partial u}{\partial x} = \frac{p\cdot x^3 - z\cdot 3x^2}{x^6} = \frac{px - 3z}{x^4}, \] \[ \frac{\partial v}{\partial x} = \frac{\partial}{\partial x}\!\left(\frac{y}{x}\right) = -\frac{y}{x^2}. \] Substituting into (1): \[ \Phi_u \cdot \frac{px - 3z}{x^4} \;-\; \Phi_v \cdot \frac{y}{x^2} = 0. \tag{1'} \]
Step 3 · Differentiate $\Phi(u,v)=0$ with respect to $y$
\[ \Phi_u\,\frac{\partial u}{\partial y} + \Phi_v\,\frac{\partial v}{\partial y} = 0. \tag{2} \] \[ \frac{\partial u}{\partial y} = \frac{q}{x^3}, \qquad \frac{\partial v}{\partial y} = \frac{1}{x}. \] Substituting into (2): \[ \Phi_u \cdot \frac{q}{x^3} \;+\; \Phi_v \cdot \frac{1}{x} = 0. \tag{2'} \]
Step 4 · Eliminate $\Phi_u$ and $\Phi_v$
Multiply equation (1') by $x^4$ and equation (2') by $x^3$ to clear denominators: \[ \Phi_u(px - 3z) - \Phi_v \cdot x^2 y = 0, \tag{A} \] \[ \Phi_u\, q + \Phi_v \cdot x^2 = 0. \tag{B} \] From (B): $\Phi_v \cdot x^2 = -\Phi_u\, q \Rightarrow \Phi_v = -\dfrac{\Phi_u\, q}{x^2}$. Substitute into (A): \[ \Phi_u(px - 3z) - \left(-\frac{\Phi_u\, q}{x^2}\right) x^2 y = 0. \] \[ \Phi_u\!\left[(px - 3z) + qy\right] = 0. \] Since $\Phi_u \neq 0$ in general (otherwise the relation $\Phi=0$ would be trivially degenerate), we divide by $\Phi_u$: \[ (px - 3z) + qy = 0. \]
Verification via Lagrange's Method
The characteristic equations for $xp + yq = 3z$ are:
\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dz}{3z}. \]
First integral from $\dfrac{dx}{x} = \dfrac{dy}{y}$: $\dfrac{y}{x} = C_1$.
Second integral from $\dfrac{dx}{x} = \dfrac{dz}{3z}$:
\[ 3\ln x = \ln z + \text{const} \;\Longrightarrow\; \frac{z}{x^3} = C_2. \]
The general solution is $\Phi(C_1, C_2) = 0$, i.e.
\[ \Phi\!\left(\frac{y}{x},\;\frac{z}{x^3}\right) = 0, \]
which coincides with the given relation (the order of arguments is immaterial
since $\Phi$ is arbitrary). $\checkmark$
Result
Eliminating the arbitrary function $\Phi$ yields the first-order linear (Lagrange-type) PDE: