Solve for nth root first, or initial value first?

I know how to handle seperable first order ODEs, nth roots, and initial value problems. But I'm a touch confused about what to do when they all appear in the same problem.

For this:

$\displaystyle xy^3y' - 2x = 3$

where:

$\displaystyle x > 0$

$\displaystyle y(1) = -1$

We can get the general solution of

$\displaystyle y = (12ln(x) + 8x + C)^{\frac{1}{4}}$

But I'm a bit confused at to whether I should be getting the four roots of $\displaystyle y$ with $\displaystyle C$ first, or if I should be trying to solve for $\displaystyle C$ using the initial value given first, etc...

How is this best handled?

Re: Solve for nth root first, or initial value first?

Re: Solve for nth root first, or initial value first?

With, of course, the information you are **given**. I have no idea what "four roots of y" you are talking about!

Re: Solve for nth root first, or initial value first?

Quote:

Originally Posted by

**HallsofIvy** With, of course, the information you are **given**. I have no idea what "four roots of y" you are talking about!

Sorry, working with y^4 = ..., there are four roots of y.

Re: Solve for nth root first, or initial value first?

Quote:

Originally Posted by

**pickslides** Solve for C first.

I have this:

y(1) = -1 = (C + 8)^{1/4}

Do I solve for four values of C, do I just raise both sides by a power of 4 to get a single value of C, or something else?

Re: Solve for nth root first, or initial value first?

Quote:

Originally Posted by

**Lancet** I have this:

y(1) = -1 = (C + 8)^{1/4}

Do I solve for four values of C, do I just raise both sides by a power of 4 to get a single value of C, or something else?

The problem formulation that immediately conducts to the solution is: what is the number x one of the fourth roots of which is -1?...

Kind regards

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