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Question: If \(u={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\) , then by Euler’s theorem the value of \(x\dfrac{...

If u=tan1(yx)u={{\tan }^{-1}}\left( \dfrac{y}{x} \right) , then by Euler’s theorem the value of xux+yuy=x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=
1)sinu\sin u
2)tanu\tan u
3)00
4)cosu\cos u

Explanation

Solution

We will use Euler’s Homogeneous Theorem , which states that if ff is a homogeneous function of degree nn of variables x,y,zx,y,z then xfx+yfy+zfz=nfx\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}+z\dfrac{\partial f}{\partial z}=nf . We will first find the degree of given homogeneous function . Then we will apply the Euler’s Homogeneous Theorem and we will get our required answer.

Complete step-by-step solution:
A homogeneous function is a real valued function .
First we will learn how to check the given function is a homogeneous function or not
Let us give a function ff of two variables that is xx & yy .
Now we will change xx into txtxand yy intotyty .
So, our function will change from f(x,y)f\left( x,y \right) to f(tx,ty)f\left( tx,ty \right)
Now we will try to take the maximum positive natural power of tt common from the function and we will represent the function as tnf(x,y){{t}^{n}}f\left( x,y \right) .
If we are able to write the function in this above form then we can say that our function is a homogeneous function .
nn is called a degree of homogeneity or we can say that nn represents the degree of homogeneous function and nn should be a real number only .
Euler’s Homogeneous Theorem states that if f(x.y,z)f\left( x.y,z \right) is a homogeneous function then
xfx+yfy+zfz=nfx\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}+z\dfrac{\partial f}{\partial z}=nf
Where nn is the degree of homogeneous function .
For this question we will use Euler’s Homogeneous Theorem of two variables only ,
If f(x,y)f\left( x,y \right) is our given two variable function then xfx+yfy=nfx\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=nfwhere nn is the degree of given homogeneous function.
Given Function is
u(x.y)=tan1(yx)u\left( x.y \right)={{\tan }^{-1}}\left( \dfrac{y}{x} \right)
First we will check whether the given function is homogeneous or not.
We will change the given function u(x,y)u\left( x,y \right) into u(tx,ty)u\left( tx,ty \right) .
xx Changes into txtx$\And ychangesintochanges intotySo,ourfunctionbecomes So, our function becomes \begin{aligned}
& \\
& u\left( tx,ty \right)={{\tan }^{-1}}\left( \dfrac{ty}{tx} \right) \\
\end{aligned}Wecanalsowriteitas, We can also write it as, u\left( tx,ty \right)={{t}^{0}}{{\tan }^{-1}}\left( \dfrac{y}{x} \right)Wecanseethatpowerof We can see that power oftisis0Sowecanconcludethat,thedegreeofgivenhomogeneousfunctionis So we can conclude that, the degree of given homogeneous function is0thatis that isn=0Thenwewillsubstitutethevalueof Then we will substitute the value ofninabovestatedEulersHomogeneousTheoremoftwovariablesin above stated Euler’s Homogeneous Theorem of two variables x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=0\times {{\tan }^{-1}}\left( \dfrac{y}{x} \right) \therefore x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=0 \therefore SotherequiredanswerisSo the required answer is 0Hence,TheCorrectOptionis **Hence , The Correct Option is3$ .**

Note: Homogenous word is used for real valued functions. We will see the word homogeneous in other mathematics fields also like in the system of linear equations , differential equations. But in different topics it will represent different meanings .Homogeneous equations can be solved in a particular format.