Question

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

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

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More