Find the partial differential equation whose general…
Find the partial differential equation whose general solution is \[ x = f(z) + g(y), \] where $f$ and $g$ are arbitrary differentiable functions.
Explanation of Method
The solution contains two arbitrary functions $f(z)$ and $g(y)$, so we expect a second-order PDE. We differentiate to generate enough equations to eliminate both functions and their derivatives. Standard notation: $p = \dfrac{\partial z}{\partial x}$, $q = \dfrac{\partial z}{\partial y}$, $r = \dfrac{\partial^2 z}{\partial x^2}$, $s = \dfrac{\partial^2 z}{\partial x\,\partial y}$, $t = \dfrac{\partial^2 z}{\partial y^2}$.
Step 1 · First-order partial derivatives
Differentiate $x = f(z) + g(y)$ with respect to $x$: \[ 1 = f'(z)\,p. \tag{1} \] Differentiate with respect to $y$: \[ 0 = f'(z)\,q + g'(y). \tag{2} \]
Step 2 · Second-order partial derivatives
Differentiate (1) with respect to $x$: \[ 0 = f''(z)\,p^2 + f'(z)\,r. \tag{3} \] Differentiate (2) with respect to $x$: \[ 0 = f''(z)\,p\,q + f'(z)\,s. \tag{4} \]
Step 3 · Eliminate $f'(z)$, $f''(z)$, and $g'(y)$
From (1): $f'(z) = \dfrac{1}{p}$. \tag{1'} Substitute (1') into (3): $0 = f''(z)\,p^2 + \dfrac{r}{p} \Rightarrow f''(z) = -\dfrac{r}{p^3}$. \tag{3'} Substitute (1') into (4): $0 = f''(z)\,p\,q + \dfrac{s}{p} \Rightarrow f''(z) = -\dfrac{s}{p^2 q}$. \tag{4'} Equate (3') and (4'): \[ -\frac{r}{p^3} = -\frac{s}{p^2 q}. \] Multiply both sides by $-p^3 q$: \[ rq = sp. \]
Verification
From $x = f(z)+g(y)$ we have $f'(z) = 1/p$. Differentiating (1) with respect to $y$: \[ 0 = f''(z)\,q + f'(z)\cdot\frac{\partial p}{\partial y} = f''(z)\,q + \frac{s}{p}, \] hence $f''(z) = -s/(pq)$. Differentiating (1) with respect to $x$: \[ 0 = f''(z)\,p^2 + f'(z)\,r, \] so $f''(z) = -r/p^3$. Setting these equal: \[ -\frac{r}{p^3} = -\frac{s}{pq} \;\Longrightarrow\; rq = sp. \checkmark \]
Result
Eliminating $f$ and $g$ (together with their derivatives) yields the second-order PDE, linear in the highest-order terms: