Solving 3xy^3 + (1+3x^2y^2)dy/dx=0: A Step-by-Step Guide

[jenettezone]

Homework Statement

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

Homework Equations

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
I don't know how to start this, please help!

 

[andylu224]

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?

 

[jenettezone]

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...

 

[andylu224]

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.

 

[jenettezone]

ohhh, i see it now. thank you!

 

[HallsofIvy]

Why did you say "the equation is not linear, and therefore not separable"? Most separable equations are not linear. An easy example is dy/dx= x/y.