Question

\[\large (x+y)dx + (x-y)dy = 0\] is the answer \[\large \frac{x^2}{2} + xy - \frac{y^2}{2} + c?\] im a little skeptic abt it :/

Answer 1

\[\frac{\partial}{\partial y}(x+y)=1\]\[\frac{\partial}{\partial x}(x-y)=1\]the equation is exact

Answer 2

yes it is exact

Answer 3

\[f(x,y)=\int(x-y)\text dy+g(x)\]\[\qquad =xy-\frac{y^2}{2}+g(x)\]
\[\frac{\partial f(x,y)}{\partial x}=y+g'(x)=(x+y)\]
\[g'(x)=x\]\[g(x)=\frac{x^2}2\]

\[f(x,y)=xy-\frac {y^2}2+\frac {x^2}2=c_1\]
\[f(x,y)=\frac {x^2}2 +xy-\frac {y^2}2+c=0\]

Answer 4

i hope my method is clear enough

Answer 5

it is...thanks...

Answer 6

yah, i only just learnt how to do these things myself

Answer 7

me too *brofist*

Answer 8

indeed