Find the partial differential equation whose general…
Find the partial differential equation whose general solution is \[ z = \tfrac{1}{2}(a^2+2)\,x^2 + axy + bx + \Phi(y+ax), \] where $a$ and $b$ are arbitrary constants and $\Phi$ is an arbitrary differentiable function.
Explanation of Method
This solution contains two arbitrary constants $a$, $b$ and one arbitrary function $\Phi$. Let $\xi = y + ax$, so that $\Phi = \Phi(\xi)$, $\Phi' = \Phi'(\xi)$, etc. Standard notation: $p = z_x$, $q = z_y$, $r = z_{xx}$, $s = z_{xy}$, $t = z_{yy}$, $p_3 = z_{xxx}$, $s_2 = z_{xxy}$, $t_2 = z_{xyy}$.
Step 1 · First-order partial derivatives
Differentiate with respect to $x$ ($\xi_x = a$): \[ p = (a^2+2)\,x + ay + b + a\,\Phi'(\xi). \tag{1} \] Differentiate with respect to $y$ ($\xi_y = 1$): \[ q = ax + \Phi'(\xi). \tag{2} \]
Step 2 · Second-order partial derivatives
Differentiate (1) with respect to $x$: \[ r = (a^2+2) + a^2\Phi''(\xi). \tag{3} \] Differentiate (2) with respect to $x$ (or equivalently (1) with respect to $y$): \[ s = a + a\,\Phi''(\xi). \tag{4} \] Differentiate (2) with respect to $y$: \[ t = \Phi''(\xi). \tag{5} \]
Step 3 · Eliminate $\Phi''(\xi)$ and $a$
From (5): $\Phi''(\xi) = t$. \tag{5'} Substitute (5') into (4): $s = a + at \Rightarrow s = a(1+t) \Rightarrow a = \dfrac{s}{1+t}$. \tag{6} Substitute (5') into (3): $r = (a^2 + 2) + a^2 t = a^2(1+t) + 2$. \tag{7} Substitute (6) into (7): \[ r = \frac{s^2}{(1+t)^2}\cdot(1+t) + 2 = \frac{s^2}{1+t} + 2. \] Multiply through by $(1+t)$: \[ r(1+t) = s^2 + 2(1+t). \] So far $\Phi''$ and $a$ have been eliminated; since $b$ never appears in equations (3)–(5), it is already absent from this relation.
Verification
From the derivation, $\Phi''(\xi)=t$, $a=s/(1+t)$, and \[ r = a^2(1+t)+2 = \frac{s^2}{(1+t)^2}(1+t)+2 = \frac{s^2}{1+t}+2, \] so $r(1+t) = s^2 + 2(1+t)$. $\checkmark$ The constants $a$ and $b$ and the function $\Phi$ have all been successfully eliminated.
Result
Eliminating the constants $a$, $b$ and the function $\Phi$ yields the second-order PDE: