Question

\[(3x-y+1)dx-(6x-2y-3)dy=0\]

Answer 1

\[2{dy \over dx} ={6x-2y+2 \over 6x-2y-3}\]

Answer 2

\[ \omega=6x-2y+3\]\[{d \omega \over dx} =6-2{dy \over dx}\]\[2{dy \over dx} = 6 - {d \omega \over dx}\]

Answer 3

\[6-{d \omega \over dx} = {\omega +5 \over \omega}\]\[5-{5 \over \omega} = {d\omega \over dx}\]

Answer 4

\[dx={1 \over 5 -{5 \over \omega}}d \omega\]
\[x=\int{1 \over 5 -{5 \over \omega}}d \omega\]
\[x={1 \over 5}(\omega+ln(\omega-1))+c\]
\[x={1 \over 5}(6x-2y-3)+{1 \over 5}ln(6x-2y-4)+c\]

Answer 5

when you differentiate the last equation you got in terms of x, and in terms of y each on their own, aren't you suppsed to get the equation you first strated with?

Answer 6

\[\omega=6x-2y+3 \]
it should be this?
\[\omega=6x-2y-3\]

Answer 7

@saljudieh07 i dont think the expression for y is simple
if it is (or if ive made an error) show me

@cinar you are absolutely correct, fortunately for me i haven't carried through that mistake

Answer 8

the rest looks good..