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one more help please. some difficult DEs which I can't figure out. dx + (y^2x)dy = (x^2 y) dy

Mathematics
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it is : dx + (y^2 *x)dy = (x^2 *y) dy
i wonder: \[\frac{dx}{dy}+xy^2=x^2y\] \[x^{-2}x'+y^2x^{-1}=y\] \[z=x^{-1}:~z^{-1}=x;~z^{2}=x^{-2};~-z^{-2}z'=x';~\] \[z^2(-z^{-2}z')+y^2z=y\] \[-z'+y^2z=y\] \[z'-y^2z=-y\] justwondering, i got no idea if its right tho
\[e^{-\frac{1}{3}y^3}z'-y^2e^{-\frac{1}{3}y^3}z=-ye^{-\frac{1}{3}y^3}\] \[e^{-\frac{1}{3}y^3}z=-\int ye^{-\frac{1}{3}y^3}dy\] \[z=-~e^{\frac{1}{3}y^3}\int ye^{-\frac{1}{3}y^3}dy\]

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i doubt that was right; unless we were spose to go into the Gamma function :/
any ideas on how to start lilMiss ??
sorry. poor internet connection here. well, what i was thinking was to get the integrating factor of the equation.
just the same as what you did, i guess?
the twp dys have me iffy; i have never had to deal with them in that state
dx + (y^2x)dy = (x^2 y) dy dx = (x^2y - y^2x)dy dx/dy = x^2y - y^2x
http://www.wolframalpha.com/input/?i=dx%2Fdy+%3D+x%5E2y+-+y%5E2x i run into a gamma function at every turn :/
help! integrate this please. : int. e^(-1/3 *y^3)dy
i think there is no closed form for that integral ---------------------------------------------------------------------- \[M\text{d}x+N\text{d}y=0\]another method is multiplynig by \(x^a.y^b\) and searching for \(a\) and \(b\) for which\[\frac{\partial (x^a.y^b.M)}{\partial y}=\frac{\partial (x^a.y^b.N)}{\partial x}\]which turnes the equation to an exact diff equation let me try it
no its not working im wondering where these equations come from...this one and prev are very curious...
i already got this one, using bernoulli's equation with respect to x, are you familiar wit that? my only problem is how am i going to integrate e^(-1/3 *y^3)dy
\[\large \int e^{-\frac{y^3}{3}} dy\]?
but there is no closed form for this integral...
why? i'll try it in wolfram alpha
http://www.wolframalpha.com/input/?i=integrate++e%5E%28-x%5E3%29
:(
will i pass my midterm? :(
sure u will...dont think of these 2 prev problems
they were not normal
aw. thank you so much. really really thank you for everything that you've done just so i could have answers. one medal isn't enough, i guess. :)
welcome...:)

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