Question

I'm confused as to how to start this problem.

Solve the differential equation

6xy dx + (4y + 9x^2)dy = 0

by finding an appropriate integrating factor to make it exact.

Answer 1

I'd first change it so that's its not implicit..

Answer 2

OK, so divide everything by dx, right?

Answer 3

Well, technically speaking, you're not "dividing", but yeah..

Answer 4

Actually there is probably an easier way.

Answer 5

hold on let me confirm

Answer 6

What do you think?

\(\dfrac{\partial}{\partial{y}}((6xy)\cdot r(x,y)) = \dfrac{\partial}{\partial{x}}((4y + 9x^{2})\cdot r(x,y))\)

So, doing the drudgery:

\([(6xy)\cdot r_{y}(x,y) + (6x)\cdot r(x,y)] = [(4y + 9x^{2})\cdot r_{x}(x,y) + (18x)\cdot r(x,y)]\)

Now some algebra... Let's see what you get.

Answer 7

Right, not really. OK, so if I do that, I get:\[6xy+(4y+9x^2)y'=0\]Then, I tested it for exactness by taking the partials of each term. I get \[6x\] and \[18x\] which are not equal. This fails the test for exactness, but I think they want me to use an integrating factor to change the terms into something that will be exact.

Answer 8

OK, stand by, I'll churn the algebra wheel.

Answer 9

OH! I see what you're saying! Give me a moment.

Answer 10

To solve it, we can integrate both groups, compare the answers, and construct the solution from that:

the integral of 6xy^3 dx is 3x^2y^3 + some function of y that would be obliterated by ordinary x differentiation.

the integral of (4y^3+9x^2y^2)dy is y^4 + 3x^2y^3 + some function of x that would be obliterated by ordinary y differentiation.

There's a term in common: the "hybrid" term 3x^2y^3 that has both x's and y's. The spare y^4 is part of the answer, too. I get

y^4 + 3x^2y^3 = C where C is some constant as an implicitly defined solution.

Answer 11

^ This was the easier way - found an explanation of it on yahoo

Answer 12

OK, thanks king. I'm going to follow tkhunny's method through to the end, because it looks like the technique we discussed in class.

Answer 13

Cool

Answer 14

OK, tkhunny, I'm kind of getting lost in the sea of letters here, but it looks like isolating the integrating factor r(x,y) is what I should do next. Doing this, I get \[12xr()=6xyr _{y}()-(4y+9x^2)r_x()\]
This looks like it could be a chain rule expansion, but I'm not sure. I could divide by the 12x, but until I know exactly where I'm going, I figured I'd leave it there with the r(). Am I on the right track?

Answer 15

I have to step out for class, but I'll come back to this in a little while. Thank you all.