Question

Solve the DE: (xy-1)dx+[(x^2)-xy]dy = 0

Answer 1

i've tried doing \[(\delta M/\delta y - \delta N/\delta x)/N\] where M = xy-1 and N = x^2 -xy

Answer 2

i am trying to solve the DE using Integrating Factors

Answer 3

Looks like itll work. \[M _{y}=x\]
\[N _{x}=2x-y\]
\[\frac{ M _{y}-N _{x} }{ N }=\frac{ x-(2x-y) }{ x^{2}-xy }\implies \frac{ -x+y }{ -x(-x+y) }= -\frac{ 1 }{ x } \]
So now using this result, our integrating factor is:
\[e ^{-\int\limits_{}^{}\frac{ 1 }{ x }}=\frac{ 1 }{ x } \]
With that as our integrating factor, could you finish from here?

Answer 4

oh ok, i guess my algebra was off when factoring, i got it from here thanks!

Answer 5

Awesome, np:3

Answer 6

I thought \[e ^{−∫1/x}=x\] considering \[−∫1/x = -\ln |x|\] and \[e^{-lnx} = -x\]

Answer 7

Nope. You can't just cancel out e and ln with the negative out there. You have to use log properties and bring it inside of the ln.

\[e^{-lnx}=e^{lnx^{-1}}=x^{-1} = \frac{ 1 }{ x } \]

Answer 8

oooo, gotcha. Sorry for that mistake.

Answer 9

No worries : 3