Answer with Step-by-step explanation:
We are given that Laplace's equation
[tex]u_{xx}+u_{yy}=0[/tex]
We have to determine given function is solution of given laplace's equation.
If a function is solution of given Laplace's equation then it satisfy the solution.
1.[tex]u=e^{-x}cosy-e^{-y}cosx[/tex]
Differentiate w.r.t x
Then, we get
[tex]u_x=-e^{-x}cosy+e^{-y}sinx[/tex]
Again differentiate w.r.t x
[tex]u_{xx}=e^{-x}cosy+e^{-y}cosx[/tex]
Now differentiate u w.r.t y
[tex]u_y=-e^{-x}siny+e^{-y}cosx[/tex]
Again differentiate w.r.t y
[tex]u_{yy}=-e^{-x}cosy-e^{-y}cosx[/tex]
Substitute the values in given Laplace's equation
[tex]e^{-x}cosy+e^{-y}cosx-e^{-x}cosy-e^{-y}cosx=0[/tex]
Hence, given function is a solution of given Laplace's equation.
2.[tex]u=sinx coshy+cosx sinhy[/tex]
Differentiate w.r.t x
[tex]u_x=cosx coshy-sinx sinhy[/tex]
Again differentiate w.r.t x
[tex]u_{xx}=-sin x coshy-cosxsinhy[/tex]
Now, differentiate u w.r.t y
[tex]u_y=sinx sinhy+cosx coshy[/tex]
Again differentiate w.r.t y
[tex]u_{yy}=sinx coshy+cosx sinhy[/tex]
Substitute the values then we get
[tex]-sinx coshy-cosxsinhy+sinxcoshy+cosx sinhy=0[/tex]
Hence, given function is a solution of given Laplace's equation.
