# Initial value problem, end of problem Q

1. Jan 31, 2012

### csc2iffy

1. The problem statement, all variables and given/known data
Solve the I.V.P. x2(dy/dx) = (4x2-x-2)/((x+1)(y+1)) , y(1)=1

2. Relevant equations

3. The attempt at a solution
So far, I got to this:
y2/2 + y = log(x) + 2/x + 3log(x+1) + C
I used the initial conditions to solve for C and got:
C = -1/2 - 3log(2)
Substituting C back in:
y2/2 + y = log(x) + 2/x + 3log(x+1) + (-1/2 - 3log(2))
Now my question is, is it possible to solve for y or is this as far as I can go with the problem?

2. Jan 31, 2012

Well it is a quadratic, not the nicest one but you can solve it for y.

3. Jan 31, 2012

### csc2iffy

How do I set it up as a quadratic? I'm getting a little lost with all of the variables

4. Jan 31, 2012

$$y^2 + 2y -2[\frac{2}{x} + \ln(x(x+1)^3)] = 0$$

$$\text{Then quadratic formula } y = \frac{-2\pm\sqrt{4 - 4(1)(-2[\frac{2}{x} + \ln(x(x+1)^3)])}}{2(1)}$$

It's horrible, but it'll give you values of y.

Last edited: Jan 31, 2012
5. Jan 31, 2012

### csc2iffy

could i just complete the square to get:
y = +/- √[2(log(x) + (2/x) + 3log(x+1) + C) + 1] - 1
?

6. Jan 31, 2012