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Calc 3... how do we find the integration factor of equation which is not exact.. (i want to know for in which both x and y is used as integration factor)

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do you have an example we can work through /
hmm no i don't have any example .. we were asked to look for it.. i just want to know any general method.. any link or inofrmation would appreciate it.. its basically part of diffrential equations.
\[\left(3x+\frac 6y\right)\text dx+\left(\frac {x^2}{y}+\frac{3y}{x}\right)\text dy =0\]

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Other answers:

\[\frac{\partial M}{\partial y}=\frac {-6}{y^{2}}\qquad\qquad \frac{\partial N}{\partial x}=\frac{2x}{y}-\frac{3y}{x^2}\]\[\qquad\qquad\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}\]
ok this is soemthing new.. is this the general form of the solution?
im using R for the integrating factor
\[\frac{\partial R(x,y)M}{\partial y}\quad= \frac{\partial R(x,y)N}{\partial x}\] using the product rule \[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]
ok i get this much
\[\small R_y(x,y)\left(3x+\frac 6y\right)+R(x,y)\left(\frac {-6}{y^{2}}\right)=R_x(x,y)\left(\frac {x^2}{y}+\frac{3y}{x}\right)+R(x,y)\left(\frac{2x}{y}-\frac{3y}{x^2}\right)\] \[\small nx^my^{n-1}\left(3x+\frac 6y\right)+x^my^n\left(\frac {-6}{y^{2}}-\frac{2x}{y}+\frac{3y}{x^2}\right)-mx^{m-1}y^n\left(\frac {x^2}{y}+\frac{3y}{x}\right)=0\]
.... somehow find n and m ,
how? any particular method?
something to do with the matching indices
hmm still dont get the last part..
me either
hmm ok its fine but thanks for the help bro.. :) will look on internet
if you find a really good sight can you tell me please
sure i will let u know..,articleId-19711.html @UnkleRhaukus this is a good one but it dont tell for the both x and y .. will post it if i find it

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