Find the most general u=u(r) satisfying (∂ u)/(∂ x)(∂^2u)/(∂…

Find the most general $u=u(r)$ satisfying $\frac{\partial u}{\partial x}\frac{\partial^2u}{\partial y^2} - \frac{\partial u}{\partial y}\frac{\partial^2u}{\partial x^2} = 0$ under the substitution $r=y/x$.

Step 1: Chain Rule Derivatives

Here $\partial r/\partial x=-y/x^2$ and $\partial r/\partial y=1/x$. The first partials of $u$ are

\[ \frac{\partial u}{\partial x} = -\frac{y}{x^2}\,u', \qquad \frac{\partial u}{\partial y} = \frac{u'}{x}. \]

Differentiating the second identity again with respect to $y$ (using $\partial r/\partial y=1/x$ once more): \[\frac{\partial^2u}{\partial y^2} = \frac{1}{x}\frac{\partial u'}{\partial y} = \frac{1}{x}\left(\frac{u''}{x}\right) = \frac{u''}{x^2}.\]

Differentiating the first identity with respect to $x$ (product rule on the three factors $-y$, $u'$, $x^{-2}$, and the chain rule on $u'$): \[\frac{\partial^2u}{\partial x^2} = -y\left[u''\left(-\frac{y}{x^2}\right)x^{-2} + u'\left(-\frac{2}{x^3}\right)\right] = \frac{2y}{x^3}\,u' + \frac{y^2}{x^4}\,u''.\]

Step 2: Substitution into the PDE

Substituting the four derivatives above, \[\left(-\frac{y}{x^2}u'\right)\left(\frac{u''}{x^2}\right) - \left(\frac{u'}{x}\right)\left(\frac{2y}{x^3}u'+\frac{y^2}{x^4}u''\right) = 0,\]

which expands to

\[ -\frac{y\,u'u''}{x^4} - \frac{2y\,(u')^2}{x^4} - \frac{y^2\,u'u''}{x^5} = 0. \]

Factoring out $y/x^4$ (nonzero for $y\neq0$, $x\neq0$) and using $r=y/x$: \[-u'u'' - 2(u')^2 - \frac{y}{x}u'u'' = 0 \;\Longrightarrow\; -(1+r)\,u'u'' - 2(u')^2 = 0.\]

For a nonconstant solution $u'\neq0$; dividing through by $-u'$ gives an ODE in $r$ alone: $(1+r)\,u'' + 2u' = 0$.

Step 3: Solving the ODE

Multiplying by $(1+r)$ turns the left side into an exact derivative: \[\frac{d}{dr}\Bigl[(1+r)^2u'\Bigr] = (1+r)^2u'' + 2(1+r)u' = (1+r)\bigl[(1+r)u''+2u'\bigr] = 0.\]

Hence $(1+r)^2u'=C_1$, so

\[ u' = \frac{C_1}{(1+r)^2}. \]

Integrating once more, $u = -\frac{C_1}{1+r} + C_2$.

Step 4: Back-Substitution

With $r=y/x$, $1+r=(x+y)/x$, so

\[ u = -\frac{C_1x}{x+y} + C_2. \]

Renaming $-C_1\to C_1$: $u = \frac{C_1x}{x+y} + C_2$.

Step 5: Verification

Write $f=x/(x+y)$, so $u=C_1f+C_2$. Direct computation gives

\[ f_x = \frac{y}{(x+y)^2}, \qquad f_y = -\frac{x}{(x+y)^2}, \qquad f_{xx} = -\frac{2y}{(x+y)^3}, \qquad f_{yy} = \frac{2x}{(x+y)^3}. \]

Since $u_x=C_1f_x$, $u_y=C_1f_y$, $u_{xx}=C_1f_{xx}$, $u_{yy}=C_1f_{yy}$, the PDE's left side equals $C_1^2(f_xf_{yy}-f_yf_{xx})$: \[f_xf_{yy} - f_yf_{xx} = \frac{y}{(x+y)^2}\cdot\frac{2x}{(x+y)^3} - \left(-\frac{x}{(x+y)^2}\right)\left(-\frac{2y}{(x+y)^3}\right) = \frac{2xy}{(x+y)^5} - \frac{2xy}{(x+y)^5} = 0.\ \checkmark\]

The two terms cancel identically, confirming the PDE holds for every $C_1,C_2$ away from the line $x+y=0$.

\[u = \frac{C_1x}{x+y} + C_2\]