Question

If $u = {\sin ^{ - 1}}\left( {\dfrac{x}{y}} \right) + {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$, show that $x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = 0$.

Answer 1

We are given that $u = {\sin ^{ - 1}}\left( {\dfrac{x}{y}} \right) + {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ and we need to find the value of $x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}}$ . We also know the formulas that, $\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}$ and $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$ . We will use this formula and differentiate it with respect to x and y.And then, we will add both the values we will get from this to get the final output.

Complete step by step answer:
Given that,
$u = {\sin ^{ - 1}}\left( {\dfrac{x}{y}} \right) + {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ --- (1)
We know that the formulas,
$\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}$ and $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$

Now we will use this formula and differentiating with respect to x and y.
First,
We will use partial differentiate with respect to x, we will get,
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {1 - \dfrac{{{x^2}}}{{{y^2}}}} }} \times \dfrac{1}{y} + \dfrac{1}{{1 + \dfrac{{{y^2}}}{{{x^2}}}}} \times \left( { - \dfrac{y}{{{x^2}}}} \right)$
On evaluating this, we will get,
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {\dfrac{{{y^2} - {x^2}}}{{{y^2}}}} }} \times \dfrac{1}{y} - \dfrac{1}{{\dfrac{{{x^2} + {y^2}}}{{{x^2}}}}} \times \left( {\dfrac{y}{{{x^2}}}} \right)$
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{y}{{\sqrt {{y^2} - {x^2}} }} \times \dfrac{1}{y} - \dfrac{{{x^2}}}{{{x^2} + {y^2}}} \times \left( {\dfrac{y}{{{x^2}}}} \right)$
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {{y^2} - {x^2}} }} - \dfrac{1}{{{x^2} + {y^2}}} \times y$
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {{y^2} - {x^2}} }} - \dfrac{y}{{{x^2} + {y^2}}}$
Multiply by ‘x’ on both the sides, we will get,
$\Rightarrow x\dfrac{{du}}{{dx}} = \dfrac{x}{{\sqrt {{y^2} - {x^2}} }} - \dfrac{{xy}}{{{x^2} + {y^2}}}$

Second,
We will use partial differentiate with respect to y, we will get,
$\Rightarrow \dfrac{{du}}{{dy}} = \dfrac{1}{{\sqrt {1 - \dfrac{{{x^2}}}{{{y^2}}}} }} \times \left( { - \dfrac{x}{{{y^2}}}} \right) + \dfrac{1}{{1 + \dfrac{{{y^2}}}{{{x^2}}}}} \times \left( {\dfrac{1}{x}} \right)$
$\Rightarrow \dfrac{{du}}{{dy}} = \dfrac{1}{{\sqrt {\dfrac{{{y^2} - {x^2}}}{{{y^2}}}} }} \times \left( { - \dfrac{x}{{{y^2}}}} \right) + \dfrac{1}{{\dfrac{{{x^2} + {y^2}}}{{{x^2}}}}} \times \left( {\dfrac{1}{x}} \right)$
$\Rightarrow \dfrac{{du}}{{dy}} = \dfrac{y}{{\sqrt {{y^2} - {x^2}} }} \times \left( { - \dfrac{x}{{{y^2}}}} \right) + \dfrac{{{x^2}}}{{{x^2} + {y^2}}} \times \left( {\dfrac{1}{x}} \right)$
$\Rightarrow \dfrac{{du}}{{dy}} = \dfrac{1}{{\sqrt {{y^2} - {x^2}} }} \times \left( { - \dfrac{x}{y}} \right) + \dfrac{x}{{{x^2} + {y^2}}}$
$\Rightarrow \dfrac{{du}}{{dy}} = - \dfrac{x}{{\sqrt {{y^2} - {x^2}} }} + \dfrac{x}{{{x^2} + {y^2}}}$
Rearrange this, we will get,
$\Rightarrow \dfrac{{du}}{{dy}} = \dfrac{x}{{{x^2} + {y^2}}} - \dfrac{1}{{\sqrt {{y^2} - {x^2}} }} \times \left( {\dfrac{x}{y}} \right)$
Multiply by ‘y’ on both the sides, we will get,
$\Rightarrow y\dfrac{{du}}{{dy}} = \dfrac{{xy}}{{{x^2} + {y^2}}} - \dfrac{y}{{\sqrt {{y^2} - {x^2}} }} \times \left( {\dfrac{x}{y}} \right)$
$\Rightarrow y\dfrac{{du}}{{dy}} = \dfrac{{xy}}{{{x^2} + {y^2}}} - \dfrac{x}{{\sqrt {{y^2} - {x^2}} }}$
Now,
$x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}}= \dfrac{x}{{\sqrt {{y^2} - {x^2}} }} - \dfrac{{xy}}{{{x^2} + {y^2}}} + \dfrac{{xy}}{{{x^2} + {y^2}}} - \dfrac{x}{{\sqrt {{y^2} - {x^2}} }}$
$\therefore x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}}= 0$

Hence, for given $u = {\sin ^{ - 1}}\left( {\dfrac{x}{y}} \right) + {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ the value of $x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = 0$.

Note: Another Method:Euler’s theorem for the homogeneous equation of degree ‘n’ with ‘x’ and ‘y’ as variables is as: $x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = nu$
For given,
$u = {\sin ^{ - 1}}\left( {\dfrac{x}{y}} \right) + {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
$\Rightarrow u= {\sin ^{ - 1}}\left( {\dfrac{1}{{\dfrac{y}{x}}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
$\Rightarrow u = {x^0}f\left( {\dfrac{y}{x}} \right)$
Here, u is a homogeneous function with degree 0.
Also, n = 0
Substituting the value of n in the formula, we will get,
$x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = nu$
$\Rightarrow x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = 0(u)$
$\therefore x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = 0$