Show that z = x^n f((y)/(x)) is a solution of x^2 (∂^2 z)/(∂…
Show that $z = x^n f\!\left(\dfrac{y}{x}\right)$ is a solution of \[ x^2\,\frac{\partial^2 z}{\partial x^2} + 2xy\,\frac{\partial^2 z}{\partial x\,\partial y} + y^2\,\frac{\partial^2 z}{\partial y^2} + x\,\frac{\partial z}{\partial x} + y\,\frac{\partial z}{\partial y} = n^2 z. \]
Explanation of Method
Recognise the left-hand side as $E^2z$ where $E=x\partial_x+y\partial_y$ is the Euler (homogeneity) operator, then apply Euler's theorem for homogeneous functions twice.
Step 1 · Operator identity
Direct expansion shows \[ (x\partial_x+y\partial_y)^2 z = x^2z_{xx}+2xyz_{xy}+y^2z_{yy}+xz_x+yz_y, \] so the PDE is equivalent to $E^2z=n^2z$ with $E=x\partial_x+y\partial_y$.
Step 2 · Apply Euler's theorem once
Since $z=x^nf(y/x)$ is homogeneous of degree $n$, Euler's theorem gives $Ez=nz$; directly, with $t=y/x$: $z_x=nx^{n-1}f-x^{n-2}yf'$, $z_y=x^{n-1}f'$, so \[ xz_x+yz_y = nx^nf-x^{n-1}yf'+x^{n-1}yf' = nx^nf = nz. \]
Step 3 · Apply $E$ a second time
Since $Ez=nz$: $E^2z=E(nz)=nEz=n^2z$. Hence \[ x^2z_{xx}+2xyz_{xy}+y^2z_{yy}+xz_x+yz_y = E^2z = n^2z. \]
Direct verification
Computing all second derivatives explicitly and substituting confirms this: the $f$, $f'$, and $f''$ coefficients collect as \[ \bigl[n(n-1)+n\bigr]x^nf + 0\cdot x^{n-1}yf' + 0\cdot x^{n-2}y^2f'' = n^2x^nf = n^2z. \checkmark \]
Result
Conclusion: