Consider the nominator and denominator as 2 lines.
Case 1: they intersect -> substitute so you move the intersection to the origine to lose the constant term, making it homogenous
Case 2: they're parallel -> there is a k that you can factor out (like in this case) to get the same coefficients in x and y, substitue for that expression.
then (1+y+2x)dy=-(4x+ 2y-1)dx
but then when you integrate both sides,
say on the left side, what do you do with "2x"? is it still possible to integrate that?
No, I did not intend that you try to integrate the two parts separately- not only is there a "2x" in the "dy" term but there is a "2y" in the "dx" term.
I said this was an exact equation. That means that the "differential form"
(1+ y+ 2x)dy+ (4x+ 2y-1)dx (and I want them on the same side of the equation!) IS an exact differential- that is, there exist some function F(x,y) so that
dF= (1+ y+ 2x)dy+ (4x+ 2y- 1)dx. That, in turn, means that
[tex]\frac{\partial F}{\partial x}= 4x+ 2y- 1[/tex] and
[tex]\frac{\partial F}{\partial y}= 1+ y+ 2x[/tex].
Since the "partial derivative of F with respect to x" treats y as if it were a constant, an "anti-partial-derivative" of that first would give F(x,y)=2x^{2}+ 2xy- x+ g(y). Notice that, since we are treating y as a constant here, the "constant of integration could be any function of y- that's the "g(y)".
Now differentiate that with respect to y:
[tex]2x+ g'(y)= \frac{\partial F}{\partial y}= 1+ y+ 2x[/tex].
Fortunately for us, the "2x" terms on each side cancel (that's because this was an "exact differential") leaving g'(y)= 1+ y so that g(y)= y+ (1/2)y^{2}+ C.
Putting that into the previous function, F(x,y)= 2x^{2}+ 2xy- x+ y+ (1/2)y^{2}+ C.
The original equation was "dF= 0" so F(x,y)= constant. Including the "C" above in that constant, 2x^{2}+ 2xy- x+ y+ (1/2)y^{2}= C is the general solution to the differential equation.
Your answer seems correct, it's just not in the same form I think.
Solved for y, both solutions give (apart from a factor 2 for the constant, but that's still a constant):