Euler's Theorem Engineering Maths, Btech first year

Euler's Theorem

To understand Euler's Theorem, first we need to understand Homogeneous functions as Euler's Theorem is applicable only on Homogeneous functions.

Homogeneous Function :

A function $z=f(x,y)$ is said to be homogeneous if each term of $z=f(x,y)$ have same degree.
$$z=f(x,y)=a_0{x}^n+a_1{x}^{n-1}y+a_2{x}^{n-2}{y}^2+...+a_n{y}^n$$ To identify Homogeneous functions, we first need to identify degree of the functions, so here are some examples:

Functions	Degree
$x^2$	2
$1/x^2$	-2
$x^2{y}^3$	5
$a^4{x}^2$	2
$x^2/y^5$	-3

So again the homogeneous function $$z=f(x,y)=a_0{x}^n+a_1{x}^{n-1}y+a_2{x}^{n-2}{y}^2+...+a_n{y}^n$$ The above equation is lengthy and seems complicated, but it's actually simple and easy to remember. Look at the first term, forget the constant, focus on the degree of the term, it's $n$. Look at the second term, power of $x$ is $n-1$ and power of $y$ is $1$, that again gives the degree $n$, and so on, till the last term which has degree $n$ which means the function above is a homogeneous function of dgree $n$.
So, now that you know how to identify degree and then Homogeneous functions, you might be wondering how to represent Homogeneous functions? The above function can be represented in the standard form as: $$z=x^{n}f\left(\frac{y}{x}\right)$$ Or $$z=y^{n}f\left(\frac{x}{y}\right)$$ But how do we do that? Here's a function which we will reduce to the standard Homogeneous form: $$z=x^4+y^4+10x^2{y}^2$$ First identify the degree of each term, here it is 4, then take $x$(or $y$) with its power equal to the degree i.e. $x^4$ $$z=x^4(1+\frac{y^4}{x^4}+\frac{10x^2{y}^2}{x^4})$$ $$z=x^4[1+{\left(\frac{y}{x}\right)}^4+10{\left(\frac{y}{x}\right)}^2]$$ $$z=x^{4}f\left(\frac{y}{x}\right)$$ Let's look at another example: $$z=x^2+y^3+3x^3{y}^2$$ In this example, degree of each term is different, hence it's not a Homogeneous function.

Euler's Theorem

If $z$ is a homogeneous function of degree $n$ in $x,y$ then $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz$$ For a function $u$ with three variables, $$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=nu$$
Here's an example: $$z=x^5+y^5+20x^3{y}^2$$ Find $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}$$ Answer : $$z=x^5+y^5+20x^3{y}^2$$ $$=x^5[1+\frac{y^5}{x^5}+\frac{20x^3{y}^2}{x^5}]$$ $$=x^5[1+{\left(\frac{y}{x}\right)}^5+20{\left(\frac{y}{x}\right)}^2]$$ $$z=x^{5}f\left(\frac{y}{x}\right)$$ $$\Rightarrow n=5$$ By Euler's Theorem $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz$$ $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=5z$$

Sample Problem 1

If $z=xyf\left(\frac{y}{x}\right)$ prove that $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2z$$

Solution

The term f\left(\frac{y}{x} might trouble you at first glance but that thing simply means whatever the function it is, it's homogeneous. $$z=xyf\left(\frac{y}{x}\right)$$ $$=x^2\left[\frac{xy}{x^2}f\left(\frac{y}{x}\right)\right]$$ $$=x^2\left[\frac{y}{x}f\left(\frac{y}{x}\right)\right]$$ $$z=x^{2}g\left(\frac{y}{x}\right)$$ Now comparing it with standard form of homogeneous function, $$z=x^{2}g\left(\frac{y}{x}\right)$$ $$\Rightarrow n=2$$ By Euler's Theorem, By Euler's Theorem $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz$$ $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2z$$

Sample Problem 2

If $z=f\left(\frac{y}{x}\right)$ then find $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}$$

Solution :

$$z=f\left(\frac{y}{x}\right)$$ $$z=x^{0}f\left(\frac{y}{x}\right)$$ By comparing it with standard form of homogeneous function, $$\Rightarrow n=0$$ $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$$

Sample Problem 3

If $$u={\tan }^{-1}\left[\frac{x^3+y^3}{x-y}\right]$$ then prove that $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\sin 2u$$

Solution :

Here, you cannot find degree of this function unless you shift $ta{n}^{-1}$ on L.H.S. So, $$\tan u=\left[\frac{x^3+y^3}{x-y}\right]$$ $$\text{Let, }z=\tan u=\left[\frac{x^3+y^3}{x-y}\right]$$ $$z=\frac{x^3(1+\frac{y^3}{x^3})}{x(1-\frac{y}{x})}$$ $$z=x^{2}f\left(\frac{y}{x}\right)$$ By comparing it with standard form of homogeneous function, $$\Rightarrow n=2$$ By Euler's Theorem, $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz$$ $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2z⟶(1)$$ Now, $$z=\tan u$$ $$\frac{\partial z}{\partial x}={\sec }^{2}u\frac{\partial u}{\partial x}$$ $$\frac{\partial z}{\partial y}={\sec }^{2}u\frac{\partial u}{\partial y}$$ Putting values in equation (1) $$x({\sec }^{2}u\frac{\partial u}{\partial x})+y({\sec }^{2}u\frac{\partial u}{\partial y})=2\tan u$$ $${\sec }^{2}u[x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}]=2\tan u$$ $$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2\tan u\times {\cos }^{2}u$$ $$=\sin 2u$$ = R.H.S., Hence Proved

Sample Problem 4 :

If $$u={\sin }^{-1}\left(\frac{x^3+y^3+z^3}{x+y+z}\right)$$ find $$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}$$

Solution :

Let $v=\sin u⟶(1)$ $$v=\frac{x^3+y^3+z^3}{x+y+z}⟶(2)$$ $$=\frac{x^3(1+{\left(\frac{y}{x}\right)}^3+{\left(\frac{z}{x}\right)}^3)}{}$$ $$v=x^{2}f\left(\frac{y}{x}\right)$$ Comparing it with the standard homogeneous function, $$\Rightarrow n=2$$ By Euler's Theorem, $$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=nv$$ $$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=2v⟶(3)$$ $$\frac{\partial v}{\partial x}=\cos u\frac{\partial u}{\partial x}$$ $$\frac{\partial v}{\partial y}=\cos u\frac{\partial u}{\partial y}$$ $$\frac{\partial v}{\partial z}=\cos u\frac{\partial u}{\partial z}$$ Putting values in equation (3) $$x(\cos u\frac{\partial u}{\partial x})+y(\cos u\frac{\partial u}{\partial y})+z(\cos u\frac{\partial u}{\partial z})=2\sin u$$ $$x\left(\frac{\partial u}{\partial x}\right)+y\left(\frac{\partial u}{\partial y}\right)+z\left(\frac{\partial u}{\partial z}\right)=2\tan u$$

Second Order Euler's Theorem (2 variables)

If $z$ is a homogeneous function in variables $x,y$ then $$x^2\frac{{\partial }^{2}z}{\partial x^2}+2xy\frac{{\partial }^{2}z}{\partial x\partial y}+y^2\frac{{\partial }^{2}z}{\partial y^2}=n(n-1)z$$

Sample Problem 5 :

If $$u={\tan }^{-1}\left(\frac{x^3+y^3}{x-y}\right)$$ find $$x^2\frac{{\partial }^{2}z}{\partial x^2}+2xy\frac{{\partial }^{2}z}{\partial x\partial y}+y^2\frac{{\partial }^{2}z}{\partial y^2}$$

Solution :

$$\text{Let }z=\tan u$$ $$z=\frac{x^3+y^3}{x-y}⟶(1)$$ $$z=\frac{x^3(1+\frac{y^3}{x^3}))}{x(1-\frac{y}{x})}$$ $$z=x^{2}f\left(\frac{y}{x}\right)$$ Comparing it with the standard homogeneous function, $$\Rightarrow n=2$$ By Euler's Theorem, $$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nz$$ $$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2z⟶(2)$$ $$\text{As, }z=\tan u$$ $$\frac{\partial z}{\partial x}={\sec }^{2}u\frac{\partial u}{\partial x}$$ $$\frac{\partial z}{\partial y}={\sec }^{2}u\frac{\partial u}{\partial y}$$ Putting values in equation (2) $$x({\sec }^{2}u\frac{\partial u}{\partial x})+y({\sec }^{2}u\frac{\partial u}{\partial y})=2\tan u$$ $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2\tan u{\sec }^{2}u=\sin 2u⟶(3)$$ Partially differentiating equation (3) w.r.t. x, $$x\frac{{\partial }^{2}u}{\partial x^2}+\frac{\partial u}{\partial x}+y\frac{{\partial }^{2}u}{\partial y\partial x}=\cos 2u2\frac{\partial u}{\partial x}$$ $$x\frac{{\partial }^{2}u}{\partial x^2}+y\frac{{\partial }^{2}u}{\partial y\partial x}=(2\cos 2u-1)\frac{\partial u}{\partial x}$$ Multiplying with $x$ on both sides $$x^2\frac{{\partial }^{2}u}{\partial x^2}+xy\frac{{\partial }^{2}u}{\partial y\partial x}=x(2\cos 2u-1)\frac{\partial u}{\partial x}⟶(4)$$ Partially differentiating equation (3) w.r.t. y, $$x\frac{{\partial }^{2}u}{\partial x\partial y}+y\frac{{\partial }^{2}u}{\partial y^2}+\frac{\partial u}{\partial x}=\cos 2u2\frac{\partial u}{\partial y}$$ Multiplying $y$ on both sides $$xy\frac{{\partial }^{2}u}{\partial x\partial y}+y^2\frac{{\partial }^{2}u}{\partial y^2}=y(2\cos 2u-1)\frac{\partial u}{\partial y}⟶(5)$$ Adding equations (4) and (5) $$x^2\frac{{\partial }^{2}u}{\partial x^2}+2xy\frac{{\partial }^{2}u}{\partial x\partial y}+y^2\frac{{\partial }^{2}u}{\partial y^2}=(2\cos 2u-1)(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y})$$ $$=(2\cos 2u-1)\sin 2u$$