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

Options

A. ` ay^(2)=e^(x^(2)//y^(2))`
B. `ay= ex^(x//y)`
C. `y=e^(x^(2))+ e^(y^(2))+C`
D. `y=e^(x^(2)) +y^(2)+C`

Answer

b

Explanation

Given , `(dy)/(dx) = (xy)/(x^(2)+y^(2))`
Put `y = vx rArr (dy)/(dx) = v+x* (dv)/(dx)`
`:. v+ x (dv)/(dx) = ((x)(vx))/(x^(2)v^(2)+x^(2))`
` rArr v+x (dy)/(dx) = v/(1+v^(2))`
` rArr x (dv)/(dx) = (-v^(3))/(1+v^(2))`
` rArr x (dv)/(dx) = (-v^(3))/(1+v^(2))`
` rArr (1/v^(3)+1/v)dv= -(dx)/x`
On integrating both sides , we get
` -1/(2v^(2)) + log v = - log x - log C`
` rArr - (x^(2))/(2y^(2)) + log y = - log C`
` rArr log Cy = (x^(2))/(2y^(2)) rArr Cy = e^(x^(2)//2y^(2))`
` rArr C^(2) y^(2) = e^(x^(2)//y^(2))`
` rArr ay^(2) = e^(x^(2)//y^(2))` where , `C^(2) = a`