In one line
For a homogeneous function \(u\) of degree \(n\), \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=n(n-1)u\); it follows from differentiating Euler's theorem once more.
Explained simply
Euler's theorem gives a first-order shortcut. This deduction gives a second-order shortcut: instead of computing three second derivatives and combining them, you can write the answer immediately from the degree. It is like a ready-made formula sheet for a whole family of functions.
Technically, if \(u(x,y)\) is homogeneous of degree \(n\) with continuous second partial derivatives, then \(x^2\dfrac{\partial^2u}{\partial x^2}+2xy\dfrac{\partial^2u}{\partial x\partial y}+y^2\dfrac{\partial^2u}{\partial y^2}=n(n-1)u\). The syllabus lists deductions "without proof", but the idea is short and worth knowing.
A handy way to remember it: the left side looks like the expansion of \((x\partial_x+y\partial_y)^2\) without the first-order terms, and the right side \(n(n-1)\) looks like the second derivative of \(t^n\) at \(t=1\). If you add Euler's first-order result to both sides you get the neat form \(n^2u\), which some questions ask for directly.
How it works
- Start from Euler: \(xu_x+yu_y=nu\).
- Differentiate with respect to \(x\): \(u_x+xu_{xx}+yu_{xy}=nu_x\). Multiply by \(x\).
- Differentiate with respect to \(y\): \(xu_{xy}+u_y+yu_{yy}=nu_y\). Multiply by \(y\).
- Add: \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}+(xu_x+yu_y)=n(xu_x+yu_y)\).
- Replace \(xu_x+yu_y=nu\): the left extra term is \(nu\), the right is \(n^2u\). So the sum equals \(n^2u-nu=n(n-1)u\).
Key formula
\(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=n(n-1)u\). Combined with Euler: \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}+xu_x+yu_y=n^2u\).
Quick example
If \(u=x^3+y^3\) (degree 3), the result predicts \(3\cdot2\,u=6u\).
Check: \(u_{xx}=6x\), \(u_{xy}=0\), \(u_{yy}=6y\). So \(x^2(6x)+0+y^2(6y)=6x^3+6y^3=6u\). Correct.
Shortcut use: for \(u=x^4y^2\sin^{-1}\dfrac yx\) (degree \(6\)), \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=30u\) and \(xu_x+yu_y=6u\), with no differentiation needed.
Common mistakes
- Writing \(n(n+1)u\) instead of \(n(n-1)u\).
- Forgetting the factor 2 on the mixed term \(xyu_{xy}\).
- Using this deduction directly for \(u=\sin^{-1}(\dots)\) when only \(\sin u\) is homogeneous; that needs the \(G(u)\) deduction.
How MU asks it
Asked as 5/6-mark parts: "If \(u=x^3y^2\sin^{-1}\dfrac yx\), find \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}\)" or "If \(u=\dfrac{x^3+y^3}{\sqrt x+\sqrt y}\), find the value of \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}+xu_x+yu_y\)" (degree 5/2, answer \(n^2u=\frac{25}{4}u\)). For full marks: show the degree with the \(t\)-test, quote the deduction, substitute.