# Help solving a first order DE

1. Jan 19, 2009

### Outlaw747

1. The problem statement, all variables and given/known data
Solve the following DE:
(6x$$^{4}+x^{3}$$+10x$$^{2}$$+2x-4)yy' = (y$$^{2}$$+y+2)(12x$$^{4}$$+2x$$^{3}$$+20x$$^{2}$$+6x-7)

2. Relevant equations
All we really know technique wise is seperation of variables then integrate. Basically this is the first week of DE class.

3. The attempt at a solution
I divided both sides by the left parantheses part and did division. Put the y's on the left and I am stuck at integration.

$$\int\frac{y}{y^2+y+2}$$dy =2$$\int\frac{2x+1}{6x^4+x^3+10x^2+2x-4}$$dx

There is where I left off. I am having difficulty with both integrals, if they are even possible. Did I miss something in an early step? Not sure where to proceed or if I am heading in the right direction. Help would be great.

2. Jan 19, 2009

### Dick

That's a pretty masochistic problem. One place you did go wrong is that the ratio of x polynomials is 2+(2x+1)/(6*x^4+x^3+10*x^2+2*x-4). The 2 doesn't come out as a multiplicative factor. Now you just have a couple of moderately nasty (but doable) integrals. For the y integral you need to start by completing the square in the denominator, do a ordinary substitution and then a trig substitution. The x integral is a partial fractions project. You started it out right by doing the division.

3. Jan 19, 2009

### Staff: Mentor

The right side should be
$$\int (2 + \frac{2x + 1}{6x^4 + x^3 + 10x^2 + 2x - 4}) dx$$

not 2 time the integral that you show.

For the integral on the left, you could rewrite the numerator as y + 1/2 - 1/2, and then split it into two integrals, with y + 1/2 in the numerator of the first, and -1/2 in the numerator of the second. After integration, these would be K*ln(y^2 + y + 2) and arctan(something).

For the integral on the right, you'll get 2x + something, where the "something" is going to take some work. You should start by factoring the 6x^4 + ... expression in the denominator, if possible. If you can get this expression factored, you can use partial fractions to break up (2x + 1)/(6x^4 + ... - 4) into the sum of easier to work with rational functions.

4. Jan 19, 2009

### Outlaw747

When you the complete the square do you add 1/(1/4) for completing the square in the bottom as (y+1/2)^2 so add 4 to the right?

5. Jan 19, 2009

### Dick

Wouldn't that mean you subtract 1/4 from the 2? y^2+y+2=(y+1/2)^2+7/4, isn't that it?