Find the partial differential equation whose general…
Find the partial differential equation whose general solution is \[ z = (x+y)\,\Phi\!\left(x^2 - y^2\right), \] where $\Phi$ is an arbitrary differentiable function of one argument.
Explanation of Method
Here $\Phi$ is an arbitrary function of a single expression $t = x^2 - y^2$. We differentiate with respect to $x$ and $y$ to produce two equations involving $\Phi(t)$ and $\Phi'(t)$, then eliminate both. Standard notation: $p = \dfrac{\partial z}{\partial x}$, $\quad q = \dfrac{\partial z}{\partial y}$.
Step 1 · Compute first-order partial derivatives
Differentiating with respect to $x$: \[ p = \Phi(t) + (x+y)\,\Phi'(t)\cdot 2x. \tag{1} \] Differentiating with respect to $y$: \[ q = \Phi(t) + (x+y)\,\Phi'(t)\cdot(-2y). \tag{2} \]
Step 2 · Eliminate $\Phi'(t)$
Subtract (2) from (1): \[ p - q = (x+y)\,\Phi'(t)\cdot 2x - (x+y)\,\Phi'(t)\cdot(-2y). \] \[ p - q = (x+y)\,\Phi'(t)\,(2x + 2y) = 2(x+y)^2\,\Phi'(t). \tag{3} \] Hence: $\Phi'(t) = \dfrac{p-q}{2(x+y)^2}$. \tag{3'}
Step 3 · Eliminate $\Phi(t)$
From the original equation: $\Phi(t) = \dfrac{z}{x+y}$. \tag{4} Substitute (3') and (4) into equation (1): \[ p = \frac{z}{x+y} + (x+y)\cdot\frac{p-q}{2(x+y)^2}\cdot 2x. \] \[ p = \frac{z}{x+y} + \frac{x(p-q)}{x+y}. \] Multiply through by $(x+y)$: \[ p(x+y) = z + x(p-q). \] \[ px + py = z + xp - xq. \] \[ py + xq = z. \]
Verification via Lagrange's Method
The characteristic equations for $yp + xq = z$ are:
\[ \frac{dx}{y} = \frac{dy}{x} = \frac{dz}{z}. \]
First integral from $\dfrac{dx}{y} = \dfrac{dy}{x}$: $x\,dx = y\,dy \Rightarrow x^2 - y^2 = C_1$.
Second integral from $\dfrac{dx}{y} = \dfrac{dz}{z}$, using the property of equal ratios with multipliers:
\[ \frac{dx + dy}{y + x} = \frac{dz}{z} \;\Longrightarrow\; \frac{d(x+y)}{x+y} = \frac{dz}{z} \;\Longrightarrow\; \frac{z}{x+y} = C_2. \]
The general solution is $C_2 = \Phi(C_1)$, i.e.
\[ \frac{z}{x+y} = \Phi\!\left(x^2 - y^2\right) \;\Longrightarrow\; z = (x+y)\,\Phi\!\left(x^2-y^2\right), \]
which coincides with the given relation. $\checkmark$
Result
Eliminating the arbitrary function $\Phi$ yields the first-order linear (Lagrange-type) PDE: