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

1. ## 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?

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

Solve for C first.

3. ## 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!

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

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.

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

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?

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

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

$\chi$ $\sigma$