Find the partial differential equation whose general…
Find the partial differential equation whose general solution is \[ z = x^a f\!\left(\frac{y}{x}\right) + x^b \varphi\!\left(\frac{y}{x}\right), \qquad a \neq b, \] where $f$ and $\varphi$ are arbitrary differentiable functions.
Explanation of Method
Let $u = y/x$, so $f = f(u)$, $\varphi = \varphi(u)$, with $u_x = -y/x^2$, $u_y = 1/x$. We use the Euler (homogeneity) operator $E = x\partial_x + y\partial_y$, which satisfies $E[x^k h(u)] = k\,x^k h(u)$, applied twice to build a linear system for $x^a f$ and $x^b\varphi$. Standard notation: $p = z_x$, $q = z_y$, $r = z_{xx}$, $s = z_{xy}$, $t = z_{yy}$.
Step 1 · First-order partial derivatives
Differentiate $z = x^a f + x^b \varphi$ with respect to $x$: \[ p = a x^{a-1} f - x^{a-1} u f'(u) + b x^{b-1}\varphi - x^{b-1} u \varphi'(u). \tag{1} \] Differentiate with respect to $y$: \[ q = x^{a-1} f'(u) + x^{b-1}\varphi'(u). \tag{2} \]
Step 2 · Apply the Euler operator
Multiply (1) by $x$ and (2) by $y=xu$, then add: \[ xp + yq = a x^{a} f + b x^{b}\varphi. \tag{3} \] Together with the original equation $z = x^{a} f + x^{b}\varphi$ (0), this forms a $2\times 2$ linear system in $x^a f$ and $x^b\varphi$. Solving by elimination: \[ (b-a)\,x^{b}\varphi = xp + yq - az, \qquad (a-b)\,x^{a} f = xp + yq - bz. \]
Step 3 · Second-order partial derivatives
Differentiating (3) with respect to $x$ and $y$ and combining (multiplying by $x$ and $y$ respectively before adding) so that the $u f'(u)$ and $u\varphi'(u)$ terms cancel: \[ x^2 r + 2xys + y^2 t + xp + yq = a^2 x^a f + b^2 x^b\varphi. \tag{8} \]
Step 4 · Eliminate $x^a f$ and $x^b\varphi$
Substituting the expressions for $x^a f$ and $x^b \varphi$ from Step 2 into (8) and multiplying by $(b-a)$: \[ a^2(bz - xp - yq) + b^2(xp + yq - az) = (b-a)(x^2 r + 2xys + y^2 t + xp + yq). \] The left side simplifies to $-ab(b-a)z + (b-a)(a+b)(xp+yq)$. Dividing by $(b-a)\neq 0$: \[ -abz + (a+b)(xp+yq) = x^2 r + 2xys + y^2 t + xp + yq. \] \[ x^2 r + 2xys + y^2 t - (a+b-1)(xp+yq) + abz = 0. \]
Verification for special cases
For $a=1$, $b=0$: the solution is $z = xf(y/x)+\varphi(y/x)$ and the PDE becomes $x^2r+2xys+y^2t = 0$, i.e. $(xD_x+yD_y)^2 z = 0$, the classical Euler homogeneous equation. $\checkmark$
Result
Eliminating $f$ and $\varphi$ yields the second-order linear (parabolic) PDE: