Question

can you solve (x+y)dx+(x-y)dy=0 please , exam in few hours

Answer 1

cant recall the details; but something about integrating one side and then deriving it down the other .....

Answer 2

M(x,y) + N(x,y) .... and some such

Answer 3

int(x+y)dx = (x^2)/2 + xy + g(y)

Answer 4

Well this is how I would approach it:
\[(x-y)dy=-(x+y)dx \rightarrow (x-y)\frac{dy}{dx}=-(x+y)\]
From here \[\frac{dy}{dx}=\frac{-(x+y)}{x-y}\]
Does that help?

Answer 5

Dy((x^2)/2 + xy + g(y)) = x + g'(y)

Answer 6

thank you very much

Answer 7

g'(y) = -y; so g(y) = -(y^2)/2 + C maybe?

to get us to:

(x^2)/2 + xy -(y^2)/2 + C ?

Answer 8

ok thx i'll try to approach it this way

Answer 9

thx you all

Answer 10

\(\color{red}{\text{youre welcome}}\) \(\color{magenta}{\text{:)}}\)