Answer to Question #294670 in Differential Equations for miski

Question

verify the function u(x,y)=e^xsiny is one of the solution of uxx+uyy=0

Answer 1

$\displaystyle u=u(x,y)=e^x\sin y\\$

Now,

$\displaystyle \frac{\partial u}{\partial x}=u_x=e^x\sin y,\ \frac{\partial^2u}{\partial x^2}=u_{xx}=e^x\sin y\\ \frac{\partial u}{\partial y}=u_y=e^x\cos y,\ \frac{\partial^2u}{\partial y^2}=u_{yy}=-e^x\sin y$

$\displaystyle u_{xx}+u_{yy}=e^x\sin y+(-e^x\sin y)=e^x\sin y-e^x\sin y=0$

Thus, $\displaystyle u=u(x,y)=e^x\sin y\\$ is a solution of the PDE $u_{xx}+u_{yy}=0$