For u_(xx)=u_y (*) : show u dx+u_x dy is exact and its…
For $u_{xx}=u_y\ (*)$: show $u\,dx+u_x\,dy$ is exact and its potential $v$ also solves $(*)$; verify on an example; and find a solution of the form $u=Y(y)e^{2ax^2/y}$.
Explanation of Method
Check the exactness condition $\partial_yP=\partial_xQ$ directly using $(*)$; the potential's own second $x$-derivative then reproduces the same structure, making $v$ a solution too. For the Gaussian-type ansatz, substitute and match powers of $x$.
Step 1 · Exactness
With $P=u$, $Q=u_x$: $P_y=u_y$, $Q_x=u_{xx}$, equal by $(*)$, so $dv=u\,dx+u_x\,dy$ for some $v$ with $v_x=u$, $v_y=u_x$.
Step 2 · $v$ also solves $(*)$
\[ v_{xx}=u_x=v_y, \] so $v$ satisfies the same equation.
Step 3 · Example
$u=e^{ax+a^2y}$ satisfies $u_{xx}=a^2e^{ax+a^2y}=u_y$. \checkmark Then $v_x=u$, $v_y=u_x=ae^{ax+a^2y}$ integrate to $v=\tfrac1ae^{ax+a^2y}=u/a$, which trivially also solves $(*)$ by linearity.
Step 4 · Gaussian-type solution $u=Y(y)e^{2ax^2/y}$
Substituting and matching the coefficient of $x^2$ forces $16a^2=-2a\Rightarrow a=-\tfrac18$; the remaining ($x^0$) balance gives $Y'/Y=4a/y=-1/(2y)$, so $Y=C/\sqrt y$. Hence \[ u = \frac{C}{\sqrt y}\exp\!\left(-\frac{x^2}{4y}\right). \]
Verification
Direct computation gives $u_{xx}=\frac{C}{\sqrt y}e^{-x^2/(4y)}\!\left[\frac{x^2}{4y^2}-\frac{1}{2y}\right]=u_y$. \checkmark This is (up to normalisation) the 1D heat kernel — the temperature from a point source at $x=0$, $y=0$.
Result
The general fact and the Gaussian solution are: