(x-y)y'=(x+y)

I've tried v=y/x and v=x-y, and both times it has just ended up as a mess.

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- February 23rd 2014, 02:04 PMohshiznit422Equation Help
(x-y)y'=(x+y)

I've tried v=y/x and v=x-y, and both times it has just ended up as a mess. - February 23rd 2014, 02:15 PMromsekRe: Equation Help
the solution Mathematica is pumping out seems to indicate that v=y/x was the right way to go.

you can certainly separate the equation that way

$$(x-y)y'=(x+y)$$

$$v=\frac{y}{x}$$

$$y=v x$$

$$y'=x v'+v$$

$$(x-v x)(x v'+v)=(x+v x)$$

$$v x - v^2 x+x^2 v' - v x^2 v' = x + v x$$

$$x^2(v' - v v')=x(1+v^2)$$

$$x v'(1-v)=(1+v^2)$$

$$\frac{1-v}{1+v^2}dv=\frac{dx}{x}$$