I want to solve this PDE with boundary conditions
$$
{u\_{xy}} + y{u\_y} = 0\,\,\,\,\,x > 0,y > 0\,,\,u\left( {x,0} \right) = {e^x},u\left( {0,y} \right) = \cos y
$$
I did the following
\begin{array}{l}
{u\_{xy}} + y{u\_y} = 0\,\,\,\,\,x > 0,y > 0\,,\,u\left( {x,0} \right) = {e^x},u\left( {0,y} \right) = \cos y\\
{u\_y} = v\,\,\, \Rightarrow \,\,{u\_{xy}} = {v\_x}\\
{v\_x} + yv = 0\\
{v\_x} = - yv\\
\frac{{dv}}{v} = - ydx\\
\ln v = - xy + \varphi \left( y \right)\\
v = {e^{ - xy + \varphi \left( y \right)}}\\
{u\_y} = {e^{ - xy + \varphi \left( y \right)}}\\
u = \int {{e^{ - xy + \varphi \left( y \right)}}dy} + \phi \left( x \right) = \frac{{{e^{ - xy + \varphi \left( y \right)}}}}{{ - x + \varphi '\left( y \right)}} + \phi \left( x \right)\\
u\left( {x,0} \right) = \frac{{{e^{\varphi \left( 0 \right)}}}}{{ - x + \varphi '\left( 0 \right)}} + \phi \left( x \right) = {e^x}\\
u\left( {0,y} \right) = \frac{{{e^{\varphi \left( y \right)}}}}{{\varphi '\left( y \right)}} + \phi \left( 0 \right) = \cos y
\end{array}
I have not seen terms like $\varphi '\left( y \right)$ in the general solution of previous PDEs. I can't continue. Please Help me



A PDE with boundary condition
If you write the equation as $(e^{xy}u\_y)\_x = 0$, then the boundary conditions tell you that $e^{xy} u\_y = e^0u\_y(0,y) = -\sin y$, so
$$
u\_y(x,y) = -\sin y\,e^{-xy},
$$
so the solution is
$$
u(x,y) = e^x - \int\_0^y\sin t\,e^{-xt}\,dt = e^x + {\frac {{{\rm e}^{-xy}} \left( x\,\sin y +\cos y \right) - 1 }{{x}^{2}+1}}.
$$


*Remark:* If you don't already know about the theory of [Laplace invariants of linear second order equations in the plane](https://en.wikipedia.org/wiki/Laplace_invariant) , you might want to check that out, in case you run into similar problems in the future.


