The solution of the differential equation `(x^(2)+y^(2))dx = 2xy dy ` is

Options

A. `x^(2)+y^(2)=Cy`
B. `C(x^(2)-y^(2))=x`
C. `x^(2)-y^(2)=Cy`
D. `x^(2)+y^(2)=Cx`

Answer

b

Explanation

Given , ` (dy)/(dx) = (x^(2) +y^(2))/(2xy) `
Put ` rArr v - log v = log x + log k`
` :. V + (dv)/(dx) = (x^(2)+v^(2)x^(2))/(2 xvx) `
`rArr x (dv)/(dx) = (1-v^(2))/(2v) `
` rArr (2v)/(1-v^(2))dv = (dx)/x`
On integrating both sides , we get
` rArr - log (1-v^(2)) = log x + log C`
`rArr log ( 1-v^(2))^(-1) = xC rArr (x^(2))/(x^(2)-y^(2))=xC`
` rArr x = C (x^(2) - y^(2))`