Need help solving an exact differential equation

[hachi_roku]

Homework Statement

solve 2xy^3+(1+3x^2y^2)dy/dx=0

Homework Equations

The Attempt at a Solution

first i made sure this is exact by finding the partial on m with respect to y and the partial of n with respect to x. it is exact.

next i integrate M with respect to x...i get x^2y^3+g(y)

next i take the partial with respect to y and get 3x^2y^2+g'(y)

now this is were I am not sure if I am doing it right.

i set this equal to N(x,y) so i get g'(y) = 1 then integrate to get g(y) = y

my answer is x^2y^3+y but I am not sure. please help

 

[hachi_roku]

bump...anyone?

 

[Mark44]

Homework Statement

solve 2xy^3+(1+3x^2y^2)dy/dx=0

Homework Equations

The Attempt at a Solution

first i made sure this is exact by finding the partial on m with respect to y and the partial of n with respect to x. it is exact.

next i integrate M with respect to x...i get x^2y^3+g(y)

For the next line, don't you want to integrate N with respect to y?

next i take the partial with respect to y and get 3x^2y^2+g'(y)

now this is were I am not sure if I am doing it right.

i set this equal to N(x,y) so i get g'(y) = 1 then integrate to get g(y) = y

my answer is x^2y^3+y but I am not sure. please help

 

[hachi_roku]

i thought you only had to do one?
if i integrate N with respect to y, i get y+x^2y^3

 

[Mark44]

i thought you only had to do one?
if i integrate N with respect to y, i get y+x^2y^3

No, you have to do both.
When I integrate N with respect to y, I get something different, namely y + x²y³ + h(x), which is different from what you show. That h(x) is somewhat like the constant of integration that you're supposed to add on when you do an indefinite integral.

In your earlier work, you have f(x, y) = x²y³ + g(y), where g is a function of y alone.
In my work, I have f(x, y) = y + x²y³ + h(x).

Now we have to reconcile these two views of f(x, y). The g(y) you showed has to be equal to y. The h(x) that I show doesn't show up in the other view of f(x, y), so h(x) = 0.

This means that f(x, y) = x²y³ + y.

You can check this by taking both partials: f_x should be equal to M = 2xy³, and f_y = N = 1 + 3x²y.


Finally, your differential equation amounts to this:
df(x, y) = 0
I.e., the total derivative of f(x, y) = 0

If the derivative of some function = 0, what can you say about that function?