# Integrating factors: How to turn 3xy^3 + (1+3x^2y^2)dy/dx=0 to dy/dx+P(x)=Q(x) form

## Homework Statement

Solve 3xy^3 + (1+3x^2y^2)dy/dx=0 using integrating factors

y' + p(x) = q(x)

## The Attempt at a Solution

I'm having trouble putting the equation to y' + p(x) = q(x)
I distributed dy/dx so it becomes 3xy^3dy/dx + 1dy/dx+3x^2y^2dy/dx=0
But I didn't know where to go from there.
So I multiplied both sides by dx and 3xy^3dx + (1+3x^2y^2)dy=0

Related Calculus and Beyond Homework Help News on Phys.org

For starters, it's y' + P(x)*y = q(x)

For this to be true, the DE has to be linear.

Do you think it is linear, separable or neither?

the definition i have for a linear DE is that it is a DE that can be written in the form y' + P(x)*y = q(x). I am trying to rewrite the DE in that form, but it looks like I can't. If I can't, then according to the definition I have, the equation is not linear, and therefore not separable. But there is an answer from the book's answer set, so it looks like it should be linear...

You won't need to rely upon integrating factors in this case.

we know dy/dx = -3xy^3/(1 + 3x^2y^2)

Thus: dx/dy = -1/3xy^3 - x/y

Making a simple substitution of u = xy

dx/dy = (y*du/dy - u)/y^2 when the substitution is made

The equation should become separable.

ohhh, i see it now. thank you!

HallsofIvy