# Need help solving an exact differential equation

1. Oct 15, 2009

### hachi_roku

1. The problem statement, all variables and given/known data
solve 2xy^3+(1+3x^2y^2)dy/dx=0

2. Relevant equations

3. 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 im not sure if im doing it right.

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

2. Oct 15, 2009

### hachi_roku

bump...anyone?

3. Oct 15, 2009

### Staff: Mentor

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

4. Oct 15, 2009

### hachi_roku

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

5. Oct 15, 2009

### Staff: Mentor

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: fx should be equal to M = 2xy3, and fy = 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?