If f x-y -1-x-y -then it is homogenous function of degree-

The correct option is D 1 f( kx,ky ) = 1 / kx+ky = k ^ 1 / x+y = k ^ 1 f( x,y ) #160; Therefore, degree is 1 ...

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Degree of a Differential Equations

If f x,y =1...

Question

If $f (x, y) = \frac{1}{x + y}$ ,then it is homogenous function of degree:

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f (x, y) = \frac{1}{x + y}

f (x, y, z) = \sqrt{y z} + \sqrt{z x} + \sqrt{x y}

x f_{x} + y f_{y} + z f_{z} = f (x, y, z)

F (u) = f (x, y, z)

1

x, y, z

x u_{x} + y u_{y} + z u_{z} = \frac{n . F (u)}{F^{'} (u)}

f (x, y) = x^{3} l n [\sqrt{x + y} / \sqrt{x - y}]

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z}

v = f (x, y)

n

I .

x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = n v

I I .

x^{2} \frac{\partial^{2} v}{\partial x^{2}} + 2 x y \frac{\partial^{2} v}{\partial x \partial y} + y^{2} \frac{\partial^{2} v}{\partial y^{2}} = n (n - 2) v

I I I . x \frac{\partial^{2} v}{\partial x^{2}} + y \frac{\partial^{2} v}{\partial x \partial y} = (n - 1) \frac{\partial v}{\partial x}

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