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

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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i 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

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just the same as what you did, i guess?

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no its not working im wondering where these equations come from...this one and prev are very curious...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large \int e^{\frac{y^3}{3}} dy\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but there is no closed form for this integral...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why? i'll try it in wolfram alpha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integrate++e%5E%28x%5E3%29

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0will i pass my midterm? :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure u will...dont think of these 2 prev problems

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aw. 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. :)
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