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

Hint: 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,
\[x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = nu\]
\[ \Rightarrow x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = 0(u)\]
\[ \therefore x\dfrac{{du}}{{dy}} + y\dfrac{{du}}{{dy}} = 0\]