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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
 one year ago
 one year 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
 one year ago
 one year 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
 one year 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
 one year ago

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

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

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

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

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

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

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

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

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.2
they were not normal
 one year 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. :)
 one year ago

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