# Factoring in terms of a variable

1. Mar 13, 2013

### GreatEscapist

1. The problem statement, all variables and given/known data
((1/3)(y^3))+(1/y)=.5*ln((x^2)+1)

Solve in terms of y

2. Relevant equations

3. The attempt at a solution

I am in college differential equations, and i just can't solve in terms of y. y teacher wants that...i tried wolfram alpha, etc. How on earth do you factor this in terms of y?

Can you even?

2. Mar 13, 2013

### Bacle2

3. Mar 14, 2013

### Ray Vickson

If I understand correctly what you wrote, you want to solve the 4th degree equation
$$\frac{y^3}{3} + \frac{1}{y} = r,$$
where $r = (1/2) \ln( x^2 + 1).$
There are formulas for the solution of 4th degree polynomials, but they are not pretty. Maple obtains the following four solutions (assuming r > 0)

Solution 1:
y = 1/4*2^(1/2)*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+1/4*2^(1/2)*((-(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)-16*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+12*r*2^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3)/(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2))^(1/2)

Solution 2:
y = 1/4*2^(1/2)*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)-1/4*2^(1/2)*((-(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)-16*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+12*r*2^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3)/(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2))^(1/2)

Solution 3:
y = -1/4*2^(1/2)*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+1/4*(-(2*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+32*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+24*r*2^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3)/(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2))^(1/2)

Solution 4:
y = -1/4*2^(1/2)*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)-1/4*(-(2*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+32*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+24*r*2^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3)/(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2))^(1/2)

Note: Maple's notation is that a/b*c means (a/b)*c.

4. Mar 14, 2013

### SammyS

Staff Emeritus
If you mean solving for x in terms of y, that's not too difficult with this.

5. Mar 14, 2013

### Staff: Mentor

That's my take, too. "In terms of y" suggests solving for x as a function of y.

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