A PDE with boundary condition [closed]

Question

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

Answer 1

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 , you might want to check that out, in case you run into similar problems in the future.