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solve: (x-y)y'=2x+y
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I think this is partial differential equation. I'm still taking the class so I'll see what I can do. I would first rewrite it as \[\frac{ dy }{ dx } (x-y) = 2x + y\] Then you can separate the variables and times both sides with dx to get: (x-y) dy = 2x + y (-y - y) dy = (2x - x) dx -2y dy = x dx Then you can integrate both sides and solve for y. Hope this helps.
I'm sorry the first line to my working should have been (x-y) dy = (2x+y) dx
Try substituting \(y=xv\), so that \(y'=v+xv'\): \[(x-y)y'=2x+y~~\iff~~(x-xv)(xv'+v)=2x+xv\] Simplifying a bit, you get something like \[v'=\frac{v^2+2}{x(1-v)}\] So a fairly simple separable equation.
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