Question

\[(3x-y+1)\text d x-(6x-2y-3)\text d y=0\]

Answer 1

i have made a very small mistake, but i cannot see where

Answer 2

when i differentiate my solution i am not getting back the original equation

Answer 3

what's ur solution ?

Answer 4

\[0=x-2y+\ln(3x-y-2)+c\]
method is in the .pdf

Answer 5

ha ha the pdf answer is incorrect ! d/dx ( ln (3x-y-2 ) = (3-y')/(3x-y-2)

Answer 6

\[2y=x+\ln(3x-y-2)+c\]\[\frac{\text d }{\text dx}\left(2y\right)=\frac{\text d }{\text dx}\left(x+\ln(3x-y-2)+c\right)\]\[2\frac{\text d y }{\text d x}=1+\frac{3-\frac{\text d y }{\text d x}}{3x-y-2}\]\[2\frac{\text d y }{\text d x}=\frac{3x-y+1}{3x-y-2}-\frac{\frac{\text d y }{\text d x}}{3x-y-2}\]\[2\frac{\text d y }{\text d x}+\frac{{\frac{\text d y }{\text d x}}}{3x-y-2}=\frac{3x-y+1}{3x-y-2}\]\[\frac{\text d y }{\text d x}\left(2+\frac1{3x-y-2}\right)=\frac{3x-y+1}{3x-y-2}\]
\[\frac{\text d y }{\text d x}\left(\frac{2(3x-y-2)+1}{3x-y-2}\right)=\frac{3x-y+1}{3x-y-2}\]\[\frac{\text d y }{\text d x}\left(\frac{6x-2y-3}{3x-y-2}\right)=\frac{3x-y+1}{3x-y-2}\]\[\frac{\text d y }{\text d x}(6x-2y-3)=(3x-y+1)\]\[(3x-y+1)\text d x=(6x-2y-3)\text d y\]
thank you @Roya

Answer 7

u r welcome dear @UnkleRhaukus

Answer 8

it wasent really a small error was it?

Answer 9

of course it was . some times we solve all the problem but we didn't get the answer just because of a minus/ plus or something like that .

Answer 10

yeah but the error i was making was that y was independent of x, if this was true none of the question would make sense

Answer 11

first a should say apologize because i think the pdf is from a book , i didn't know it's ur solution . if i knew i won't say ha ha !
i'm so sri .
you are right . if y was independent of x , there won't be any function !

Answer 12

thats ok i can forgive you, you fixed my problem
i'd been stuck and confused with this problem for quite some time, and it great to be able to move on

Answer 13

u know i never make fun with others mistake . but if i find a mistake in a famous book i will laugh