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

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lilMissMindset Group TitleBest ResponseYou've already chosen the best response.0
it is : dx + (y^2 *x)dy = (x^2 *y) dy
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[e^{\frac{1}{3}y^3}z'y^2e^{\frac{1}{3}y^3}z=ye^{\frac{1}{3}y^3}\] \[e^{\frac{1}{3}y^3}z=\int ye^{\frac{1}{3}y^3}dy\] \[z=~e^{\frac{1}{3}y^3}\int ye^{\frac{1}{3}y^3}dy\]
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i doubt that was right; unless we were spose to go into the Gamma function :/
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
any ideas on how to start lilMiss ??
 2 years ago

lilMissMindset Group TitleBest ResponseYou've already chosen the best response.0
sorry. poor internet connection here. well, what i was thinking was to get the integrating factor of the equation.
 2 years ago

lilMissMindset Group TitleBest ResponseYou've already chosen the best response.0
just the same as what you did, i guess?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
the twp dys have me iffy; i have never had to deal with them in that state
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
dx + (y^2x)dy = (x^2 y) dy dx = (x^2y  y^2x)dy dx/dy = x^2y  y^2x
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=dx%2Fdy+%3D+x%5E2y++y%5E2x i run into a gamma function at every turn :/
 2 years ago

lilMissMindset Group TitleBest ResponseYou've already chosen the best response.0
help! integrate this please. : int. e^(1/3 *y^3)dy
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
no its not working im wondering where these equations come from...this one and prev are very curious...
 2 years ago

lilMissMindset Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
\[\large \int e^{\frac{y^3}{3}} dy\]?
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
but there is no closed form for this integral...
 2 years ago

lilMissMindset Group TitleBest ResponseYou've already chosen the best response.0
why? i'll try it in wolfram alpha
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
http://www.wolframalpha.com/input/?i=integrate++e%5E%28x%5E3%29
 2 years ago

lilMissMindset Group TitleBest ResponseYou've already chosen the best response.0
will i pass my midterm? :(
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
sure u will...dont think of these 2 prev problems
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
they were not normal
 2 years ago

lilMissMindset Group TitleBest ResponseYou've already chosen the best response.0
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. :)
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
welcome...:)
 2 years ago
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