The solution of `(x dx+y dy)/(x dy-y dx)=sqrt((1-x^2-y^2)/(x^2+y^2))` is

Options

A. `sqrt(x^(2)+y^(2))= sin{(tan^(-1)y//x)+C}`
B. `sqrt(x^(2)+y^(2))=cos{(tan^(-1)y//x)+C}`
C. `sqrt(x^(2)+y^(2))=(tan(sin^(-1)y//x)+C)`
D. `y=xtan(c+sin^(-1)sqrt(x^(2)+y))`

Answer

A, D

Explanation

The differential equation can be rewritten as `(xdx+ydy)/sqrt(1-(x^(2)+y^(2)))=(xdy-ydx)/sqrt(x^(2)+y^(2))`
Since `dtan^(-1)(y//x)=(xdy-ydx)/(x^(2)+y^(2))`
and `d(x^(2)+y^(2))=2(xdx+ydy)`
We have `((1/2)(d(x^(2)+y^(2))/(sqrt(x^(2)+y^(2))sqrt(1-(x^(2)+y^(2)))))=(xdy-ydx)/(x^(2)+y^(2))=d{(tan^(-1)(y//x))}`
Put `x^(2)+y^(2)=r^(2)` in the LHS. Then
`(tdt)/(tsqrt(1-t^(2)))=d{tan^(-1)(y//x)}`
Integrating both sides, we get
`sin^(-1)t=tan^(-1)(y//x)+c`
i.e., `sin^(-1)sqrt((x^(2)+y^(2)))=tan^(-1)(y//x)+c`
i.e., `sin^(-1)sqrt((x^(2)+y^(2)))=tan^(-1)(y//x)+c`