Question

solve y dx + (3xy - 1) dy = 0

Answer 1

solve for y or dy/dx?

Answer 2

solve for y :)

Answer 3

\[y dx + (3xy - 1 ) dy = 0\]
\[y dx + 3xy dy - dy = 0\]\[y \frac{ dx }{ dy }+3xy - 1 = 0\]
\[\frac{ dx }{ dy }+ 3x = \frac{ 1 }{ y }\]
I think, in order to satisfy the equation \(\frac{ dx }{ dy } + p(y)x = g(y) x\)
\(p(y) = 3\) and \(g(y) = \frac{ 1 }{ y }\)
\[\mu(y) = \exp(\int\limits\limits_{0}^{y} p(y) dy) = \exp (\int\limits\limits_{0}^{y} 3 dy) = e^{3y}\]
\[x = \frac{ 1 }{ \mu(y) }\int\limits_{0}^{y} \mu(y) g(y) dy\]
\[x = \frac{ 1 }{ e^{3y} } \int\limits_{0}^{y} e^{3y} \frac{ 1 }{ y } dy\]

Answer 4

is it correct??

Answer 5

I got \[xe^{3y} - \int\limits \frac{ e^{3y} }{ y } = C\] for the solution.

Answer 6

I'm not too sure, sorry.... But I think it should be ok...

Answer 7

Ok, np @Kira_Yamato
i'm not sure for:
\[\int\limits \frac{ e^{3y} }{ y } dy\]
what results did you get?

Answer 8

This is what MatLab gave me

Answer 9

can i see ur script on matlab??

Answer 10

Sorry I mean Wolfram Alpha

Answer 11

ok.., thank u Kira :)