# Solve for nth root first, or initial value first?

• July 22nd 2011, 12:19 PM
Lancet
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:

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

where:

$x > 0$

$y(1) = -1$

We can get the general solution of

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

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

How is this best handled?
• July 22nd 2011, 01:25 PM
pickslides
Re: Solve for nth root first, or initial value first?
Solve for C first.
• July 22nd 2011, 01:37 PM
HallsofIvy
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!
• July 22nd 2011, 03:05 PM
Lancet
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.
• July 22nd 2011, 03:08 PM
Lancet
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?
• July 23rd 2011, 12:01 AM
chisigma
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

$\chi$ $\sigma$