Question

@amistre64 can you tell me how to use your method in exact DE for this?? \[(2x+y)dx + (x^2 - x)dy = 0\] it seems i end up in a dead end :/

Answer 1

linear DE!
i got integration factor=(x-1)/x

Answer 2

\[ e^{\int{\frac1{x(x-1)}}dx}\]

Answer 3

the ans is \[y(x-1/x)+logx=c\]
@experimentX is it correct?

Answer 4

(x-1)/x not x-1/x

Answer 5

lol not sure ...
\[ \frac{1}{x(x-1)} = \frac{-1}{x} + \frac{1}{x-1}\]

Answer 6

Looks like you are right!!

Answer 7

hmn!

Answer 8

does Fxy = Fyx?

Answer 9

2x NOTequal 2x-1; therefore its not "exact"

Answer 10

opps, well, same unequal outcome but i didnt do the left part right did i?

Answer 11

2x + y wrt y = 1

x^2-x wrt x = 2x -1

1 notequal 2x-1 so it aint exact

Answer 12

so it's not exact?? weird :/