differential equations?

Question

I need to find an equation in the form H(x,y) = c that satisfies given trajectories

For example:
dx/dt = 2y , dy/dt = 8x
For these trajectories I can do dy/dx = 8x/2y and use separation of variables to obtain: 4x^2 - y^2 = c

But for the trajectories
dx/dt = x^2 - x + y , dy/dt = y - 2xy
are not separable, so how can I find H(x,y)

Answer 1

You can form the DE
dy/dx = y(1-2x)/(x^2-x+y)
which, as you point out, is not separable but is EXACT.
That is, if you write the DE in the form
M dx + N dy = 0
we have partial M/partial y = partial N/partial x (partial derivatives)

Check: M = y(1-2x) and N = x-x^2-y so
partial M/partial y = 1-2x
and partial N/partial x = 1-2x.

For exact DEs we know the solution is of the form H(x,y) = c, because on differentiating with respect to x we get
(partial H/partial x) dx + (partial H/partial y) dy = 0
which is the DE if we identify
(partial H/partial x) = M
and (partial H/partial y) = N.

Here (partial H/partial x) = y(1-2x)
so H(x,y) = yx - yx^2 + f(y)
and (partial H/partial y) = x - x^2 +f '(y)
which means we need to take f '(y) = -y
or f(y) = -y^2/2 +c

Thus the solution is H(x,y) = yx(1-x) - y^2/2 = c.

Answer 2

It's not separable, but it's exact. You can write

dy/dx = (y - 2xy)/(x² - x + y) as

(x² - x + y)dy + (2xy - y)dx = 0

Note that

∂/∂x(x² - x + y) = 2x - 1
∂/∂y(2xy - y) = 2x - 1

Answer 3

well i know that the answer would be:

(x^2)*y + 1/2*(y^2) = x*y