Hi, I'm having trouble with a calculus question. It's for an assignment, and I've tried it myself for a while and have come to the conclusion I cannot see how it is done. However, I imagine for a better mathematician it should be fairly trivial.

The question is:

By introducing the new dependent variable z = y^2 solve the equation

(y^4 - 3x^2)(dy/dx) + xy = 0

Thanks in advance for your help!

Nevermind, I've solved it now

## Help with calculus

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- JamesCFraser
**Posts:**32**Joined:**Wed Nov 14, 2007 9:47 pm UTC

### Help with calculus

Last edited by JamesCFraser on Mon Oct 27, 2008 8:43 pm UTC, edited 1 time in total.

- Mathmagic
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### Re: Help with calculus

What are you solving for?

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- JamesCFraser
**Posts:**32**Joined:**Wed Nov 14, 2007 9:47 pm UTC

### Re: Help with calculus

The question just wants the implicit version of y(x) not as a differential equation.

- mochafairy
**Posts:**1098**Joined:**Tue Mar 25, 2008 11:27 pm UTC**Location:**Ohio

### Re: Help with calculus

I think it's asking you to sepa-grate. (move all x terms to one side, move all y terms to the other, then integrate both sides to get rid of the differential.)

I have no idea if that's what it's actually asking...especially with the whole 'z' term...

I have no idea if that's what it's actually asking...especially with the whole 'z' term...

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- JamesCFraser
**Posts:**32**Joined:**Wed Nov 14, 2007 9:47 pm UTC

### Re: Help with calculus

mochafairy wrote:I think it's asking you to sepa-grate. (move all x terms to one side, move all y terms to the other, then integrate both sides to get rid of the differential.)

I have no idea if that's what it's actually asking...especially with the whole 'z' term...

You had to let z = y^2, then find all derivatives of y in terms of z to create a new equation. The new equation could be divided through by x^2, then you could sub in v = z/x (z = xv) and do the same procedure. This left it as just partial fractions. Hot stuff!

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