# Is the function f(x,y)=x+y+xy a solution of Laplace's equation: (d^(2)u)/(dx^(2))+(d^(2)u)/(dy^(2))?

lfryerda | Certified Educator

To determine if the function `f(x,y)=x+y+xy` is a solution of Laplace's equation `{partial^2 f}/{partial x^2}+{partial^2 f}/{partial y^2}=0` , we need to find the appropriate partial derivatives of `f(x,y)` , add them together and see if they add to zero.  If so, then the function is a solution of Laplace's equation.

Recall that the partial derivative of a function is performed while pretending that all the other variables are a constant.

`{partial f}/{partial x}=1+y`

`{partial^2 f}/{partial x^2}=0`

similarly, we get with the partial derivatives with respect to y:

`{partial f}/{partial y}=1+x`

`{partial^2 f}/{partial y^2}=0`

Since `{partial^2 f}/{partial x^2}+{partial^2 f}/{partial y^2}=0` , then f(x,y) is a solution of Laplace's equation.

quantatanu | Student

Laplac's equation for a function u(x,y) is:

[(del/del x)^2 + (del/del y)^2] u = 0  ------------------------(1)

Where (del/del x) is the partial derivative with respect to x:

that is (del/del x)y=0, (del/del y)x=0,

Given,

u = f(x,y) = x + y +xy

Putting it in the L.H.S of equation (1) we get,

[(del/del x)^2 + (del/del y)^2] ( x + y + xy )

= (del/del x) (del/del x)  ( x + y + xy )

+ (del/del y)(del/del y) ( x + y + xy )

= (del/del x) ( 1 + y ) +  (del/del y) ( 1 + x )

= 0

Hence given f(x,y) satisfies Laplace's Equation.