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UnkleRhaukus
 4 years ago
\[(3xy+1)dx(6x2y3)dy=0\]
UnkleRhaukus
 4 years ago
\[(3xy+1)dx(6x2y3)dy=0\]

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UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[2{dy \over dx} ={6x2y+2 \over 6x2y3}\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \omega=6x2y+3\]\[{d \omega \over dx} =62{dy \over dx}\]\[2{dy \over dx} = 6  {d \omega \over dx}\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[6{d \omega \over dx} = {\omega +5 \over \omega}\]\[5{5 \over \omega} = {d\omega \over dx}\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[dx={1 \over 5 {5 \over \omega}}d \omega\] \[x=\int{1 \over 5 {5 \over \omega}}d \omega\] \[x={1 \over 5}(\omega+ln(\omega1))+c\] \[x={1 \over 5}(6x2y3)+{1 \over 5}ln(6x2y4)+c\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\omega=6x2y+3 \] it should be this? \[\omega=6x2y3\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0@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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the rest looks good..
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