Verifying an integrating factor -please check my work , thanks.

Click For Summary

Homework Help Overview

The discussion revolves around verifying that \( \frac{1}{y^4} \) is an integrating factor for the differential equation \( (3x^2 - y^2) \frac{dy}{dx} - 2xy = 0 \). Participants explore the implications of using this integrating factor and its effect on the separability of the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the transformation of the original equation after applying the integrating factor and question the disappearance of terms. There is an exploration of whether the resulting equation is separable and the implications of testing for exactness.

Discussion Status

Some participants have confirmed that the integrating factor makes the equation exact and are examining the conditions for exactness. There is ongoing clarification regarding the differentiation of the functions associated with the equation.

Contextual Notes

Participants note the importance of correctly identifying the functions \( M \) and \( N \) in the context of exact equations, as well as the continuity of partial derivatives in relation to the exactness condition.

darryw
Messages
124
Reaction score
0
Verifying an integrating factor --please check my work , thanks.

Homework Statement


Verify that 1/y^4 is an integrating factor for (3x^2-y^2)dy/dx - 2xy = 0, and then use it to solve the equation

(1/y^4)* (3x^2-y^2)dy/dx = 2xy* (1/y^4)

=3x^2 / y^4 dy = 2x/y^3 dx

=3x^2 / y dy = 2x dx

=integ (1/y)dy = (2/3)*integ (1/x)dx

=ln|y| = (2/3)ln|x| + c

=y = xe^(2/3 +c)


Homework Equations





The Attempt at a Solution


 
Physics news on Phys.org


darryw said:

Homework Statement


Verify that 1/y^4 is an integrating factor for (3x^2-y^2)dy/dx - 2xy = 0, and then use it to solve the equation

(1/y^4)* (3x^2-y^2)dy/dx = 2xy* (1/y^4)
What happened to the -y^2 term on the left side in the next line?

Also, don't connect equations with =. Expressions are equal to one another, but equations are not equal to equations.
darryw said:
=3x^2 / y^4 dy = 2x/y^3 dx

=3x^2 / y dy = 2x dx

=integ (1/y)dy = (2/3)*integ (1/x)dx

=ln|y| = (2/3)ln|x| + c

=y = xe^(2/3 +c)


Homework Equations





The Attempt at a Solution

 


(1/y^4)* (3x^2-y^2)dy/dx = 2xy* (1/y^4)

3x^2 / y^4 - 1/y^2 dy = 2x / y^3 dx

((3x^2/y) - y)dy = 2x dx

so, at this point i am finding out that it is not separable afterall, right? thanks
 


so i should test for exactness?

that is, M_y(x,y) ?=? (N_x(x,y)

?
thanks
 


darryw said:
so i should test for exactness?

that is, M_y(x,y) ?=? (N_x(x,y)
Yes. After you multiply by the integrating factor, test for exactness.
 


[3x^2 / y^4 - 1/y^2] dy - [2x / y^3] dx = 0
...M.........N...= 0

M_x = 6x/y^4 = N_y

ok so now i know it is exaxct and the integrating factor makes it exact, but i have question before i proceed.

If i were to partially differentiate M wrt y instead of x, (and then N wrt y) i get a totally different result.. The equation does not pass the exatness test.

How do i know which variable to differentiate wrt? thanks
 


You have your M and N switched. M is associated with dx and N is associated with dy. My = 6x/y^4 = Nx.
 


sorry i overlooked that..

but i still don't understand why M_x is not equal to N_y/?
 


There's no reason for Mx to be equal to Ny, but there is a good reason for My to be equal to Nx.

If the equation M dx + N dy = 0 is exact, then the left side represents the total differential of some unknown function F(x, y), so that we have dF(x, y) = 0 ==> F(x, y) = C.

In the differential equation M = Fx(x, y) and N = Fy(x, y). If F and its first and second partials are continuous, the mixed partials will be equal. When you check for exactness, you are checking to see whether My = Nx. IOW, you are verifying that
\frac{\partial ^2 F}{\partial x \partial y} = \frac{\partial ^2 F}{\partial y \partial x}
 
  • #10


thanks for the explanation. My final answer is now matching answer key, so I am moving on to my other prob..
thanks.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
4
Views
3K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K