Prove that every function z=x^mf(y/x) satisfies x (∂ z)/(∂…
Prove that every function $z=x^mf(y/x)$ satisfies \[ x\,\frac{\partial z}{\partial x}+y\,\frac{\partial z}{\partial y} = mz, \tag{I} \] \[ x^2\,\frac{\partial^2 z}{\partial x^2}+2xy\,\frac{\partial^2 z}{\partial x\,\partial y} +y^2\,\frac{\partial^2 z}{\partial y^2} = m(m-1)z. \tag{II} \] Hence deduce that every function $z=r\,\alpha(\varphi)+\beta(\varphi)$, where $r$ and $\varphi$ are the modulus and argument of $x+iy$, satisfies \[ x^2\,\frac{\partial^2 z}{\partial x^2}+2xy\,\frac{\partial^2 z}{\partial x\,\partial y} +y^2\,\frac{\partial^2 z}{\partial y^2} = 0. \tag{III} \]
Explanation of Method
Prove (I) directly, obtain (II) by differentiating (I) with respect to $x$ and $y$ and recombining, then split $z=r\alpha(\varphi)+\beta(\varphi)$ into a degree-1 and a degree-0 homogeneous piece and apply (II) to each, using linearity of the operator.
Step 1 · Proof of (I)
With $t=y/x$: $z_x=mx^{m-1}f-x^{m-2}yf'$, $z_y=x^{m-1}f'$, so \[ xz_x+yz_y = mx^mf-x^{m-1}yf'+x^{m-1}yf' = mx^mf = mz. \]
Step 2 · Proof of (II)
Differentiate (I) with respect to $x$: $z_x+xz_{xx}+yz_{xy}=mz_x \Rightarrow xz_{xx}+yz_{xy}=(m-1)z_x$. Differentiate (I) with respect to $y$: $xz_{xy}+yz_{yy}=(m-1)z_y$. Multiply the first by $x$, the second by $y$, and add: \[ x^2z_{xx}+2xyz_{xy}+y^2z_{yy} = (m-1)(xz_x+yz_y) = (m-1)mz = m(m-1)z. \]
Step 3 · Deduce (III)
Since $x+iy=re^{i\varphi}$, write $r=x\sqrt{1+(y/x)^2}$, so $z_1=r\alpha(\varphi)=x^1F_1(y/x)$ is homogeneous of degree $m=1$, and $z_2=\beta(\varphi)=x^0F_2(y/x)$ is homogeneous of degree $m=0$. By (II), each satisfies the equation with right-hand side $m(m-1)z=0$ (since $1\cdot0=0$ and $0\cdot(-1)=0$). By linearity of $L=x^2\partial_{xx}+2xy\partial_{xy}+y^2\partial_{yy}$: \[ Lz = Lz_1+Lz_2 = 0+0 = 0. \]
Result
Summary: