Question

If u=sin^-1(x/y)+tan^-1(y/x) prove that xdu/dx)+ydu/dy)=0

Answer 1

Answer:

$x\frac{du}{dx} +y\frac{du}{dy} =0$ ..... Proved.

Step-by-step explanation:

We have, $u=sin^{-1}(\frac{x}{y} )+tan^{-1}(\frac{y}{x} )$ ....... (1)

We know the formulas $\frac{d}{dx}(sin^{-1}x )=\frac{1}{\sqrt{1-x^{2} } }$ and $\frac{d}{dx} (tan^{-1}x )= \frac{1}{1+x^{2} }$

Now, partially differentiating equation (1) with respect to x, we get,

$\frac{du}{dx}=\frac{1}{\sqrt{1-\frac{x^{2} }{y^{2}}}}.\frac{1}{y}+\frac{1}{1+\frac{y^{2}}{x^{2}}} .(-\frac{y}{x^{2}})$

⇒ $x\frac{du}{dx}=\frac{x}{y}.\frac{y}{\sqrt{y^{2}-x^{2}}} -\frac{y}{x}.\frac{x^{2}}{x^{2}+y^{2}}$

⇒ $x\frac{du}{dx}=\frac{x}{\sqrt{y^{2}-x^{2}}}-\frac{xy}{x^{2}+y^{2}}$ ........ (2)

Again, partially differentiating equation (1) with respect to y, we get,

$\frac{du}{dy}=\frac{1}{\sqrt{1-\frac{x^{2}}{y^{2}}}}(-\frac{x}{y^{2} } )+\frac{1}{1+\frac{y^{2}}{x^{2}}} .\frac{1}{x}$

⇒ $y\frac{du}{dy} =\frac{xy}{x^{2}+y^{2}}-\frac{x}{y}.\frac{y}{\sqrt{y^{2}-x^{2}}}$

⇒ $y\frac{du}{dy}=\frac{xy}{x^{2}+y^{2}}-\frac{x}{\sqrt{y^{2}-x^{2}}}$........ (3)

Hence, adding equations (2) and (3) we get

$x\frac{du}{dx} +y\frac{du}{dy} =0$

Therefore, proved.

Answer 2

Answer:

Step-by-step explanation:

As the given equation is of degree 0.

Hence by using Euler's theorem,

Xdu/dx + ydu/dy = nu

Putting n=0 as n represents degree

We can the RHS as 0