Factoring in terms of a variable

[GreatEscapist]

Homework Statement

((1/3)(y^3))+(1/y)=.5*ln((x^2)+1)

Solve in terms of y

Homework Equations

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?

 

[Bacle2]

There is a general formula for solving cubics. For a more formal/rigorous

approach, the implicit function theorem:

http://en.wikipedia.org/wiki/Implicit_function_theorem

Will tell you if it is possible and where so.

 

[Ray Vickson]

Homework Statement

((1/3)(y^3))+(1/y)=.5*ln((x^2)+1)

Solve in terms of y

Homework Equations

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?

If I understand correctly what you wrote, you want to solve the 4th degree equation
[tex] \frac{y^3}{3} + \frac{1}{y} = r,[/tex]
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.

 

[SammyS]

...

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?

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

 

[Mark44]

Homework Statement

((1/3)(y^3))+(1/y)=.5*ln((x^2)+1)
Solve in terms of y

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

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