Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

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
\[e^{-\frac{1}{3}y^3}z'-y^2e^{-\frac{1}{3}y^3}z=-ye^{-\frac{1}{3}y^3}\] \[e^{-\frac{1}{3}y^3}z=-\int ye^{-\frac{1}{3}y^3}dy\] \[z=-~e^{\frac{1}{3}y^3}\int ye^{-\frac{1}{3}y^3}dy\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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 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
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. :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question