Find the partial differential equation whose general…
Find the partial differential equation whose general solution is \[ z = (x^2 + a)(y + b), \] where $a$ and $b$ are arbitrary constants.
Explanation of Method
This general solution contains two arbitrary constants $a$ and $b$, so we differentiate twice to obtain enough equations to eliminate both.
Step 1 · Compute first-order partial derivatives
Standard notation: $p = \dfrac{\partial z}{\partial x}$, $\quad q = \dfrac{\partial z}{\partial y}$. Differentiating with respect to $x$: \[ p = \frac{\partial z}{\partial x} = 2x(y + b). \tag{1} \] Differentiating with respect to $y$: \[ q = \frac{\partial z}{\partial y} = x^2 + a. \tag{2} \]
Step 2 · Eliminate the arbitrary constants
From equation (2): $x^2 + a = q$. Substitute into the original equation: \[ z = q(y + b). \tag{3} \] From equation (1): $y + b = \dfrac{p}{2x}$. Substitute into (3): \[ z = q \cdot \frac{p}{2x}. \]
Verification
Given $z = (x^2+a)(y+b)$, compute directly: \[ p = 2x(y+b), \qquad q = x^2 + a. \] Then $pq = 2x(y+b)(x^2+a) = 2x \cdot z$. Hence $2xz = pq$. $\checkmark$
Result
Eliminating the constants $a$ and $b$ yields the first-order nonlinear PDE: