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Question

Solution of the equation $$\displaystyle xdx+ydy+\frac { xdy-ydx }{ { x }^{ 2 }+{ y }^{ 2 } } =0$$


A
y=xtan(c+x2+y22)
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B
x=ytan(c+x2+y22)
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C
y=xtan(cx2y22)
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D
None of these
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Solution

The correct option is D $$\displaystyle y=x\tan { \left( \frac { c-{ x }^{ 2 }-{ y }^{ 2 } }{ 2 } \right) } $$
We have, $$\displaystyle xdx+ydy+\frac { xdy-ydx }{ { x }^{ 2 }+{ y }^{ 2 } } =0$$
$$\displaystyle \frac { 1 }{ 2 } d\left( { x }^{ 2 }+{ y }^{ 2 } \right) +d\left( \tan ^{ -1 }{ \left( \frac { y }{ x }  \right)  }  \right) =0$$
Integrating both sides, we get
$$\displaystyle \frac { 1 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 } \right) +\tan ^{ -1 }{ \left( \frac { y }{ x }  \right)  } =\frac { c }{ 2 } $$
$$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }+2\tan ^{ -1 }{ \frac { y }{ x }  } =c$$
$$\displaystyle \therefore y=x\tan { \left( \frac { c+{ x }^{ 2 }+{ y }^{ 2 } }{ 2 }  \right)  } $$ is the required solution.

Mathematics

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