Here's the question you clicked on:
gerryliyana
solve y dx + (3xy - 1) dy = 0
\[y dx + (3xy - 1 ) dy = 0\] \[y dx + 3xy dy - dy = 0\]\[y \frac{ dx }{ dy }+3xy - 1 = 0\] \[\frac{ dx }{ dy }+ 3x = \frac{ 1 }{ y }\] I think, in order to satisfy the equation \(\frac{ dx }{ dy } + p(y)x = g(y) x\) \(p(y) = 3\) and \(g(y) = \frac{ 1 }{ y }\) \[\mu(y) = \exp(\int\limits\limits_{0}^{y} p(y) dy) = \exp (\int\limits\limits_{0}^{y} 3 dy) = e^{3y}\] \[x = \frac{ 1 }{ \mu(y) }\int\limits_{0}^{y} \mu(y) g(y) dy\] \[x = \frac{ 1 }{ e^{3y} } \int\limits_{0}^{y} e^{3y} \frac{ 1 }{ y } dy\]
I got \[xe^{3y} - \int\limits \frac{ e^{3y} }{ y } = C\] for the solution.
I'm not too sure, sorry.... But I think it should be ok...
Ok, np @Kira_Yamato i'm not sure for: \[\int\limits \frac{ e^{3y} }{ y } dy\] what results did you get?
This is what MatLab gave me
can i see ur script on matlab??
Sorry I mean Wolfram Alpha
ok.., thank u Kira :)