Need Euler!

An expression of the form $a_{o}x^n+a_{1}x^{n-1}y+a^{2}x^{n-2}y^2+...+a_{n}y^{n}$ is a homogeneous function of degree $n$ which can be rewritten as

$\large{x^{n}[a_{o}+a_{1}(y/x)+a_{2}(y/x)^2+...+a_{n}(y/x)^{n}]}$

In general a function $f(x,y,z,t,...)$ is said to be a homogeneous function of degree $n$ in $x,y,z,t,..$ if it can be expressed in the form $\large{x^{n} \phi(y/x,z/x,t/x,..)}$ .

Euler theorem on homogeneous functions

If $u$ be a homogeneous function of degree $n$ in $x$ and $y$ . then

$\large{x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=nu}$

Proof

$u=x^{n}f(y/x)$

$\large{\therefore \frac{\partial u}{\partial x}=nx^{n-1}f(y/x)-yx^{n-2}f^{\prime}(y/x) \\ \frac{\partial u}{\partial y}=x^{n-1}f^{\prime}(y/x)}$

Hence $\large{x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=nx^{n}f(y/x)=nu}$

Now, let $\csc u=z$

$\large{z={ x }^{ \frac { 1 }{ 12 } }\left( \sqrt { \frac { 1+\sqrt { \frac { y }{ x } } }{ 1+\sqrt [ 3 ]{ \frac { y }{ x } } } } \right) }$

$\large{\Rightarrow z=x^{\frac{1}{12}} \phi(y/x)}$

$\large{x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=\frac{z}{12}}$

Also $\large{\frac{\partial z}{\partial x}=- \csc u \cot u \frac{\partial u}{\partial x}}$

Thus we have

$\large{x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=-\frac{\tan u}{12} \qquad (1) }$

Partial differentiation equation $(1)$ w.r.t $x$

$\large{ \frac{\partial u}{\partial x}+x \frac{\partial^{2} u}{\partial x^{2}}+y \frac{\partial^2 u}{\partial x \partial y}=-\frac{1}{12} (\sec^2 u) \frac{\partial u}{\partial x}}$

$\large{\Rightarrow x \frac{\partial^{2} u}{\partial x^{2}}+y \frac{\partial^2 u}{\partial x \partial y}=- \left(\frac{\sec^2 u}{12}+1 \right) \frac{\partial u}{\partial x} \qquad (2)}$

Similarly partial differentiation of equation $(1)$ w.r.t $y$ gives

$\large{\Rightarrow y \frac{\partial^{2} u}{\partial y^{2}}+x \frac{\partial^2 u}{\partial y \partial x}=- \left(\frac{\sec^2 u}{12}+1 \right) \frac{\partial u}{\partial y} \qquad (3)}$

Note that here $\large{\frac{\partial^2 u}{\partial y \partial x}= \frac{\partial^2 u}{\partial x \partial y}}$ You can easily check this yourself.

Now multiply $(2)$ with $x$ and $(3)$ with $y$ and add the both resulting equations, we get

$\large{\Rightarrow x^2 \frac{\partial^{2} u}{\partial x^{2}}+2xy \frac{\partial^2 u}{\partial x \partial y} +y^2 \frac{\partial^{2} u}{\partial y^{2}}=- \left(\frac{\sec^2 u}{12}+1 \right) \left[x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}\right]}$

From $(1)$ we have

$\large{\Rightarrow x^2 \frac{\partial^{2} u}{\partial x^{2}}+2xy \frac{\partial^2 u}{\partial x \partial y} +y^2 \frac{\partial^{2} u}{\partial y^{2}}=- \left(\frac{\sec^2 u}{12}+1 \right). \left(-\frac{\tan u}{12} \right)}$

$\large{\Rightarrow x^2 \frac{\partial^{2} u}{\partial x^{2}}+2xy \frac{\partial^2 u}{\partial x \partial y} +y^2 \frac{\partial^{2} u}{\partial y^{2}}=\frac{\tan u}{144} \left(13+\tan^2 u \right)}$