Solve the differential equation (x^{2} + y^{2})dx- 2xydy = 0

#### Solution

(x^{2} + y^{2})dx- 2xydy = 0

(x^{2 }+ y^{2}) dx = 2xydy

`dy/dx = (x^2 + y^2)/(2xy)`.........(i)

The equation is a homogeneous equation

Let y= vx,

Differentiat ing w.r.t. x, we get,

`dy/dx=v+x(dv)/dx`

`dy/dx=(x^2+y^2)/(2xy) " from "(i)`

`v+x(dv)/dx=(x^2+(vx)^2)/(2x.(vx))`

`v+x(dv)/dx=(1+v^2)/(2v)`

`x(dv)/dx=(1+v^2)/(2v)-v`

`x(dv)/dx=(1+v^2-2v^2)/(2v)`

`x(dv)/dx=(1-v^2)/(2v)`

`(2v)/(1-v^2)dv=1/xdx`.......(ii)

Which is in variables separatable form

∴ Integrating both sides, we get

`int(2v)/(1-v^2)dv=int1/xdx + c_1`

`therefore -log|1-v^2|=log|x|+logc`

`therefore log|x(1-v^2)|=log|c|`

`therefore x(1-v^2)=c`

Resubstituting `v=y/x` we get

`x(1-y^2/x^2)=c`

`x((x^2-y^2)/x^2)=c `

`therefore x^2 - y^2 = cx`, where c is constant

which is the required general solution