Question

Solve the differential equation
(1+yx)x dy + (1-yx)y dx= 0

Answer 1

Let
`v(x)=xy(x)`, so that
`\frac yx=\frac v{x^2}`, and
`(\mathrm dv)/(\mathrm dx)=x(\mathrm dy)/(\mathrm dx)+y`, or
`(\mathrm dy)/(\mathrm dx)=\frac1x(\mathrm dv)/(\mathrm dx)-\frac v{x^2}`.

So the ODE is


`(1+yx)x\,\mathrm dy+(1-yx)y\,\mathrm dx=0\iff(\mathrm dy)/(\mathrm dx)=-((1-yx)y)/((1+yx)x)`

`\implies \frac1x(\mathrm dv)/(\mathrm dx)-\frac v{x^2}=-(1-v)/(1+v)\frac v{x^2}`

`\implies (\mathrm dv)/(\mathrm dx)=\frac vx\left(1-(1-v)/(1+v)\right)`

`\implies(\mathrm dv)/(\mathrm dx)=(2v)/(x(1+v))`

This ODE is separable, so you can write


`(1+v)/(2v)\,\mathrm dv=\frac{\mathrm dx}x`

and integrating both sides yields


`\displaystyle\int(1+v)/(2v)\,\mathrm dv=\int \frac{\mathrm dx}x`

`\frac12\left(v+\ln|v|\right)=\ln|x|+C`

`v+\ln|v|=2\ln|x|+C`

Replace
`v=xy` and you end up with


`xy-\ln|xy|=2\ln|x|+C`