Use separation of variables to find product solutions for the given PDE $$ y u_{xy} + u = 0$$
Let $u(x,y) = X(x)T(y)$, then this PDE yields $$y X'(x)T'(y) + X(x) T(y) = 0$$
I am not sure how to proceed from here any suggestions are greatly appreciated.
For the equation $y \, u_{x y} + u = 0$ the proposer has provided $u(x,y) = f(x) g(y)$ for which \begin{align} y \, f' \, g + f \, g &= 0 \\ \frac{f'}{f} \, \frac{y g'}{g} &= -1. \end{align} Now, in order to separate the variables constants need to be chosen such that the product is equal to $-1$, say $ -\lambda \cdot (1/\lambda)$, or $$ \frac{f'}{f} \, \frac{y g'}{g} = - \lambda \cdot \frac{1}{\lambda} = -1$$ leading to \begin{align} f' + \lambda \, f &= 0 \\ y \, g' - \frac{1}{\lambda} \, g &= 0. \end{align} The solutions of these equations are \begin{align} f(x) &= c_{0} \, e^{-\lambda \, x} \\ g(y) &= c_{1} \, y^{\frac{1}{\lambda}}. \end{align} In terms of $u(x,y)$ it is seen that $$u(x,y) = c_{2} \, e^{-\lambda \, x} \, y^{\frac{1}{\lambda}}.$$
