use the exact method to solve the DE: 2xydx + x^2ydy=0

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use the exact method to solve the DE: 2xydx + x^2ydy=0

Mathematics
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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.

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Are you confused as to how it's done? Just wanting to check your answer or what?
i checked for exactness and it wasn't exact so I used the next method to check for exactness and it looks as thought I got something but unsure
It doesn't look exact to me because the term on the left has an extra y in it.

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Other answers:

using the test for exactness it doesnt come out exact but when you use integrating factors to solve then it works
its not an exact diff equation. divide throughout by x^2y and reduce it to separable form and solve it. its easy
if i used integrating factor method i was tuahgt to do if it doesnt come out exact then i get the integrating factor of e^(y-ln(y))
how did u check the exactness?
partial derivative of M with respect to y and partial derivative of N with respect to x and see if they are equal
P(x,y)=2xy Q(x,y)=x^2y for exactness partial derivative of p w.r.ty=partial derivative of Q w.r.t x
there is a method uzma ,differentiate the term with dx with respect to y and the term with dy with respect to x ,if both differentiated terms become equal then they r exact,otherwise non exact
yes right :)
com for chat uzma
so what do u guess about the exactness?
when doing that they come out non exact and then there is a method to use after that where you use integrating factors after applying an equation and then go from there
not exact initially
any help??
the eq is exact
non exact...m sorry :)
so the eq can made by multiplying with intgrating factor

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