# Show that z=ln((x^2)+ (y^2)) is a solution of Laplace’s equation: δ^2z/δx^2 + δ^2/δy^2 = 0

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To show that `z(x,y)=ln(x^2+y^2)` is a solution of the pde `{partial^2 z}/{partial x^2}+{partial^2 z}/{partial y^2}=0` , we need to calculate the left side of the pde using the function `z(x,y)` and then show that it is identically zero.

`{partial z}/{partial x}={2x}/{x^2+y^2}`

`{partial^2z}/{partial x^2}={2(x^2+y^2)-2x{2x}}/{(x^2+y^2)^2}`

`={-2x^2+2y^2}/{(x^2+y^2)^2}`

`{partial z}/{partial y}={2y}/{x^2+y^2}`

`{partial^2z}/{partial y^2}={2(x^2+y^2)-2y(2y)}/{(x^2+y^2)^2}`

`={2x^2-2y^2}/{(x^2+y^2)^2}`

Now adding `{partial^2z}/{partial x^2}` and `{partial^2z}/{partial y^2}` we get

`1/{(x^2+y^2)^2}(-2x^2+2y^2+2x^2-2y^2)`

which equals zero since all the terms in the numerator cancel out.

**Therefore the function satisfies the Laplace equation.**