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## Homework Statement

If ##z=x\ln(x+r)-r## where ##r^2=x^2+y^2##, prove that

$$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{x+y}$$

## Homework Equations

## The Attempt at a Solution

Since ##r^2=x^2+y^2##, ##∂r/∂x=x/r## and ##∂r/∂y=y/r##.

Differentiating z w.r.t x partially,

$$\frac{∂z}{∂x}=\ln(x+r)+x\cdot\left(\frac{1}{x+r}\right)\cdot \left(1+\frac{∂r}{∂x}\right)-\frac{∂r}{∂x}$$

Using ##∂r/∂x=x/r##

$$\frac{∂z}{∂x}=\ln(x+r)$$

$$\frac{∂^2z}{∂x^2}=\frac{1}{x+r}\cdot\left(1+\frac{x}{r}\right)=\frac{1}{r}$$

Differentiating z w.r.t y partially,

$$\frac{∂z}{∂y}=\frac{x}{x+r}\cdot \frac{∂r}{∂y}-\frac{∂r}{∂y}=\frac{∂r}{∂y}\cdot \frac{-r}{x+r}$$

Using ##∂r/∂y=y/r##

$$\frac{∂z}{∂y}=\frac{-y}{x+r}$$

$$\frac{∂^2z}{∂y^2}=-\frac{(x+r)-y(∂r/∂y)}{(x+r)^2}=-\frac{rx+r^2-y^2}{r(x+r)^2}$$

Since ##r^2-y^2=x^2##

$$\frac{∂^2z}{∂y^2}=-\frac{x}{r(x+r)}$$

Adding the second order derivatives,

$$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{r}-\frac{x}{r(x+r)}=\frac{1}{x+r}$$

Where did I go wrong?

Any help is appreciated. Thanks!