Question

Does anyone want to take a stab using exact.
(y^3-x^2y)y'-xy^2=0

Answer 1

Click the attachment

Answer 2

Shows the original equation.

Answer 3

Well, you have that
\[ (y^3-x^2y)y'-xy^2=0 \]
i.e.,
\[ (y^3 - x^2y) dy - xy^2 dx = 0 \]
which is the differential of
\[ c = f(x,y) = \frac{y^4}{4} - \frac{x^2y^2}{2} \]
..which you can simplify a little further.

Answer 4

Make sense?

Answer 5

ya that's what I got when i integrated N(x,y)dy then Nx, simplified for h'(x) solved for h(x) to get f(x,y). but when i do it the other way M(x,y) i get a different answer. 1/3xy^3

Answer 6

how do you choose between M(x,y)dx or N(x,y)dy to start the problem.

Answer 7

What you should get from integrating M(x,y) WITH RESPECT TO x I might add, is
\[ -\frac{x^2y^2}{2} + f(y) \]
the function of y, f(y) is as far as differentiation w.r.t. x is concerned, just a constant. Hence we need to add it when we integrate.

Now, that is how we set the integral of the two sides equal. Here, f(y) = y^4/4

Answer 8

it doesn't matter which one you start with. Just make sure you set the integrals equal to each other, which you can always do if you've done the test to see if it was in fact an exact differential to begin with.

Answer 9

i got it i had a transpose error and that's why it didn't check doing it both ways.

Answer 10

Good, good.