The solution of the equation `(dy)/(dx)=(x+y)/(x-y)`, is

Options

A. ` C (x^(2)+y^(2))^(1//2) + e^(tan^(-1)(y//x))=0`
B. ` C (x^(2)+y^(2))^(1//2)=e^(tan^(-1)(y//x))`
C. ` C (x^(2)-y^(2))=e^(tan^(-1)(y//x))`
D. None of the above

Answer

a

Explanation

Given , `(dy)/(dx) = (x+y)/(x-y) `
Put `y = vx rArr (dy)/(dx) = v+x (dv)/(dx)`
` :. v + x (dv)/(dx) = (1+v)/(1-v) rArr x (dv)/(dx) = (1+v^(2))/(1-v)`
` rArr 1/x dx = (1/(1+v^(2)) =v/(1+v^(2)))dv `
On integrating both sides , we get
` log _(e) x = tan ^(-1) v - 1/2log_(e) (1+v^(2)) - log _(e) C `
` rArr log_(e) x = tan ^(-1) v - 1/2 log_(e) (1+v^(2)) - log _(e)C`
` rArr log_(e) x = tan ^(-1) (y/x) =- 1/2 log_(e) [ 1+(y/x)^(2)]- log _(e) C`
` rArr log x + log sqrt((x^(2)+y^(2))/(x^(2)))+ log_(e) C = tan^(-1). y/x`
` rArr log C sqrt(x^(2)+y^(2)) = tan ^(-1). y/x`
` rArr C ( x^(2) + y^(2))^(1//2) = e^(tan^(-1)(y//x))`