Find the partial differential equation whose general…
Find the partial differential equation whose general solution is given implicitly by \[ \Phi\!\left(\frac{z}{x^2},\; x - y\right) = 0, \] where $\Phi$ is an arbitrary differentiable function of two arguments.
Explanation of Method
Set $u = \dfrac{z}{x^2}$, $v = x - y$, and differentiate $\Phi(u,v)=0$ with respect to $x$ and $y$, then eliminate $\Phi_u$ and $\Phi_v$. Standard notation: $p = \dfrac{\partial z}{\partial x}$, $\quad q = \dfrac{\partial z}{\partial y}$.
Step 1 · Differentiate with respect to $x$
\[ \Phi_u\,\frac{\partial u}{\partial x} + \Phi_v\,\frac{\partial v}{\partial x} = 0. \tag{1} \] \[ \frac{\partial u}{\partial x} = \frac{p \cdot x^2 - z \cdot 2x}{x^4} = \frac{px - 2z}{x^3}, \qquad \frac{\partial v}{\partial x} = 1. \] Substituting: \[ \Phi_u \cdot \frac{px - 2z}{x^3} + \Phi_v = 0. \tag{1'} \]
Step 2 · Differentiate with respect to $y$
\[ \Phi_u\,\frac{\partial u}{\partial y} + \Phi_v\,\frac{\partial v}{\partial y} = 0. \tag{2} \] \[ \frac{\partial u}{\partial y} = \frac{q}{x^2}, \qquad \frac{\partial v}{\partial y} = -1. \] Substituting: \[ \Phi_u \cdot \frac{q}{x^2} - \Phi_v = 0. \tag{2'} \]
Step 3 · Eliminate $\Phi_u$ and $\Phi_v$
From (2'): $\Phi_v = \Phi_u \cdot \dfrac{q}{x^2}$. Multiply (1') by $x^3$: \[ \Phi_u(px - 2z) + \Phi_v \cdot x^3 = 0. \tag{A} \] Substitute $\Phi_v$ into (A): \[ \Phi_u(px - 2z) + \Phi_u \cdot \frac{q}{x^2} \cdot x^3 = 0. \] \[ \Phi_u\!\left[(px - 2z) + qx\right] = 0. \] Since $\Phi_u \neq 0$ in general, divide by $\Phi_u$: \[ px - 2z + qx = 0. \]
Verification via Lagrange's Method
The characteristic equations for $xp + xq = 2z$ are:
\[ \frac{dx}{x} = \frac{dy}{x} = \frac{dz}{2z}. \]
First integral from $\dfrac{dx}{x} = \dfrac{dy}{x}$: $dx = dy \Rightarrow x - y = C_1$.
Second integral from $\dfrac{dx}{x} = \dfrac{dz}{2z}$:
\[ 2\ln x = \ln z + \text{const} \;\Longrightarrow\; \frac{z}{x^2} = C_2. \]
The general solution is $\Phi(C_1, C_2) = 0$, i.e.
\[ \Phi\!\left(x - y,\;\frac{z}{x^2}\right) = 0, \]
which coincides with the given relation. $\checkmark$
Result
Eliminating the arbitrary function $\Phi$ yields the first-order linear (Lagrange-type) PDE: