Math, asked by sandeepponnam28, 4 days ago

Let f(x, y) = (x 2 + y 2 ) n + x tan−1 ( y x ) + φ( y x ), where n is a positive integer and φ is a twice differentiable function. By using the Euler’s theorem show that x 2 fxx + 2xyfxy + y 2 fyy = 2n(2n − 1)(x 2 + y 2 ) n .​

Answers

Answered by keshavk1607
1

Answer:

See the solution in the images

Step-by-step explanation:

Attachments:
Answered by yogeshgangwar044
0

Answer:

It is prove that x^{2} f x x+2 x y f x y+y^{2} f y y=2 n(2 n-1)\left(x^{2}+y^{2}\right)^{2 n}  using the Euler's theorem.

Step-by-step explanation:

Given: $f(x, y)=\left(x^{2}+y^{2}\right)^{n}+x \tan ^{-1}\left(\frac{y}{x}\right)+\phi\left(\frac{y}{x}\right)$, where n is a positive integer and \phi is a twice differentiable function.

To Prove: $f(x, y)=\left(x^{2}+y^{2}\right)^{n}+x \tan ^{-1}\left(\frac{y}{x}\right)+\phi\left(\frac{y}{x}\right)$ using Euler's theorem.

Solution:

According to Euler's theorem, if $f$ is a homogeneous function of $x, y, z$ of degree $n$ then,

$x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}+z \frac{\partial f}{\partial z}=n f$

By calculating further the equation will be,

$x^{2} \frac{\partial^{2} f}{\partial x^{2}}+2 x y \frac{\partial^{2} f}{\partial x \partial y}+y^{2} \frac{\partial^{2} f}{\partial y^{2}}=n(n-1) f$

or, $x^{2} f_{x x}+2 x y f_{x y}+y^{2} f_{y y}=n(n-1) f\;\;\;\;\;\;...(1)$

Now, Apply the above equation in the given equation.

The given equation is $f(x, y)=\left(x^{2}+y^{2}\right)^{n}+x \tan ^{-1}\left(\frac{y}{x}\right)+\phi\left(\frac{y}{x}\right)$

Let assume $P(x, y)=f(x, y)-x \tan ^{-1}\left(\frac{y}{x}\right)-\Phi\left(\frac{y}{x}\right)$

or, $p(x, y)=\left(x^{2}+y^{2}\right)^{n}$     (By homogeneous function of degree 2n)

Now, from the solved equation (1) above the equation will be,

x^{2} P_{2 x}+2 x y P_{x y}+y^{2} P_{y y}=2 n(2 n-1) \rho

or, x^{2} P x x+2 x y P x y+y^{2} P y y=2 n(2 n-1)\left(x^{2}+y^{2}\right)^{2 n}

Hence, it is prove that x^{2} f x x+2 x y f x y+y^{2} f y y=2 n(2 n-1)\left(x^{2}+y^{2}\right)^{2 n} by using the Euler’s theorem.

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