$u=f(\frac{x}{y})$ as a solution to $\Delta u = 0$

Question

Find the solutions of $\Delta u=0$ of the form $u=f(\frac{x}{y})$

Since $\Delta u = u_{xx} + u_{yy} = 0$, I differentiated and got the following. $$f'' + \frac{2xy}{y^2 + x^2}f' = 0$$

I'm a little rusty with my ODEs and I'm not sure how to solve this further of if it can be explicitly solved for $f$

Answer 1

$$u_{xx}+u_{yy}=0$$ Solutions on the form $u=f(\frac{x}{y})$ :

$\frac{x}{y}=t\quad;\quad x=ty$

$u_x=\frac{1}{y}f'(t)$

$u_{xx}=\frac{1}{y^2}f''(t)$

$u_y=-\frac{x}{y^2}f'(t)$

$u_{yy}=\frac{x^2}{y^4}f''+\frac{2x}{y^3}f'=\frac{t^2y^2}{y^4}f''+\frac{2ty}{y^3}f'=\frac{t^2}{y^2}f''+\frac{2t}{y^2}f'$

$$\frac{1}{y^2}f''+\frac{t^2}{y^2}f''+\frac{2t}{y^2}f'=0$$ $$f''+t^2f''+2tf'=0$$ $$\frac{f''}{f'}=-\frac{2t}{1+t^2}$$ $$f'=\frac{c_1}{1+t^2}$$ $$f(t)=c_1\tan^{-1}(t)+c_2$$ $$u=c_1\tan^{-1}(\frac{x}{y})+c_2$$

Answer 2

Hint:

$$ \frac{2xy}{x^2 + y^2} = \frac{2\frac{x}{y}}{\left(\frac{x}{y}\right)^2+1} $$

Let $u = x/y$ then you have

$$ f''(u) + \frac{2u}{u^2+1}f'(u) = 0 $$

which is separable