Question

A function f(x, y) is called homogeneous of degree n if it satisfies the equation f(tx,ty) = t n f(x, y) for all t, where n is a positive integer. Show that if f(x, y) is homogeneous of degree n, then x ∂ f ∂ x +y ∂ f ∂ y = n f(x, y). (Hint: Try differentiating: find ∂ ∂t t=1 of a homogeneous function f in two ways.)

Answer 1

Answer:

It is proved that $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)$.

Step-by-step explanation:

Given function is,

$f(tx, ty)=t^nf(x, y)$ for all t>0 and n is positive integers.

To show,

$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=n f(x, y)$

Let, u=tx and v=ty, then,

$f(tx, ty)=t^nf(x, y)$

$\implies f(u, v)=t^n f(x, y)\hfill (1)$

differentiate partially with respect to x we get,

[tex}\frac{\partial}{\partial x}f(u,v)=\frac{\partial}{\partial x}(t^nf(x, y))[/tex]

$\implies\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}=t^n\frac{\partial f}{\partial x}$

$\implies\frac{\partial f}{\partial u}t=t^n\frac{\partial f}{\partial x}$

Again, differentiate (1) with respect to t we get,

$\frac{\partial}{\partial t}f(u,v)=nt^{n-1}f(x,y)$

$\implies \frac{\partial f}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial t}=nt^{n-1}f(x,y)$

$\implies x\frac{\partial f}{\partial u}+y\frac{\partial f}{\partial v}=nt^{n-1}f(x,y)$  

$\implies xt^{n-1}\frac{\partial f}{\partial x}+yt^{n-1}\frac{\partial f}{\partial y}=nt^{n-1}f(x, y)$       (By putting values of $\frac{\partial f}{\partial u},\frac{\partial f}{\partial v}$)

$\therefore x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf(x,y)$

Hence proved.