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

  • #1

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!
 

Answers and Replies

  • #2
24
0


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?
 
  • #3


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...
 
  • #4
24
0


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.
 
  • #5


ohhh, i see it now. thank you!
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
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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.
 

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