Find the partial differential equation whose general…
Find the partial differential equation whose general solution is \[ z = x^a f\!\left(\frac{y}{x}\right) + x^b \varphi\!\left(\frac{y}{x}\right) + x^c \psi\!\left(\frac{y}{x}\right), \] where $f$, $\varphi$, $\psi$ are arbitrary differentiable functions and $a$, $b$, $c$ are three distinct constants.
Explanation of Method
Let $u = y/x$ and $E = x\partial_x + y\partial_y$ the Euler operator, so that $E^k[x^m h(u)] = m^k x^m h(u)$. Applying $E$ repeatedly to $z$ produces a Vandermonde linear system for $x^a f$, $x^b\varphi$, $x^c\psi$, which is then eliminated using Newton's identities for symmetric polynomials in $a,b,c$.
Step 1 · Build the system via the Euler operator
Applied to $z$: \[ Ez = xp + yq = a\,x^a f + b\,x^b\varphi + c\,x^c\psi. \tag{1} \] \[ E^2 z = x^2 r + 2xys + y^2 t + xp + yq = a^2 x^a f + b^2 x^b\varphi + c^2 x^c\psi. \tag{2} \] \[ E^3 z = x^3 p_3 + 3x^2 y r_2 + 3xy^2 s_2 + y^3 t_3 + 3x^2 r + 6xys + 3y^2 t + xp + yq = a^3 x^a f + b^3 x^b\varphi + c^3 x^c\psi. \tag{3} \] Together with $z = x^a f + x^b\varphi + x^c\psi$ (0), equations (0)–(3) form the Vandermonde system \[ \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix} \begin{pmatrix} A \\ B \\ C \end{pmatrix} = \begin{pmatrix} z \\ xp+yq \\ x^2r+2xys+y^2t+xp+yq \end{pmatrix}, \] with $A=x^af$, $B=x^b\varphi$, $C=x^c\psi$, whose determinant $V=(b-a)(c-a)(c-b)\neq 0$.
Step 2 · Eliminate $A$, $B$, $C$ using the fourth equation
By Newton's relation, $a^3=(a+b+c)a^2-(ab+bc+ca)a+abc$ (and likewise for $b^3,c^3$), so \[ a^3 A + b^3 B + c^3 C = (a+b+c)(a^2A+b^2B+c^2C) - (ab+bc+ca)(aA+bB+cC) + abc(A+B+C). \] Substituting (0), (1), (2): \[ E^3 z = \sigma_1 E^2 z - \sigma_2 Ez + \sigma_3 z, \] with $\sigma_1=a+b+c$, $\sigma_2=ab+bc+ca$, $\sigma_3=abc$.
Step 3 · Expand and simplify
Expanding each operator and substituting: \begin{align*} &x^3 p_3 + 3x^2 y r_2 + 3xy^2 s_2 + y^3 t_3 + 3x^2 r + 6xys + 3y^2 t + xp + yq \\ &= \sigma_1\!\left(x^2 r + 2xys + y^2 t + xp + yq\right) - \sigma_2(xp + yq) + \sigma_3 z. \end{align*} Rearranging (moving all terms to the left): \begin{align*} &x^3 z_{xxx} + 3x^2 y\, z_{xxy} + 3xy^2 z_{xyy} + y^3 z_{yyy}\\ &\quad +(3-\sigma_1)\!\left(x^2 z_{xx} + 2xy\,z_{xy} + y^2 z_{yy}\right)\\ &\quad +(1 - \sigma_1 + \sigma_2)(xz_x + yz_y) - \sigma_3\, z = 0. \end{align*}
Consistency with Problem 8
If $\psi\equiv 0$, the third-order terms vanish and the equation reduces to the second-order result of Problem 8: \[ x^2 r + 2xys + y^2 t - (a+b-1)(xp+yq) + ab\,z = 0. \checkmark \]
Result
Eliminating $f$, $\varphi$, $\psi$ yields the third-order linear PDE (a generalized Euler–Cauchy equation):