Initial value problem, end of problem Q

[csc2iffy]

Homework Statement

Solve the I.V.P. x²(dy/dx) = (4x²-x-2)/((x+1)(y+1)) , y(1)=1

Homework Equations

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?

 

[Bread18]

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?

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

 

[csc2iffy]

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

 

[Bread18]

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

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

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

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

 

[csc2iffy]

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

 

[Bread18]

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

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

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

Oops forgot the + C, but you get the idea.

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

Completing the square would probably be a lot harder, but it should work.