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

In summary, verifying 1/y^4 as an integrating factor for (3x^2-y^2)dy/dx - 2xy = 0 and using it to solve the equation results in finding that the equation is exact. After checking for exactness, it is determined that My = Nx, and the final solution is y = xe^(2/3 + c).
  • #1
darryw
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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


 
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  • #2


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

 
  • #3


(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
 
  • #4


so i should test for exactness?

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

?
thanks
 
  • #5


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.
 
  • #6


[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
 
  • #7


You have your M and N switched. M is associated with dx and N is associated with dy. My = 6x/y^4 = Nx.
 
  • #8


sorry i overlooked that..

but i still don't understand why M_x is not equal to N_y/?
 
  • #9


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
[tex]\frac{\partial ^2 F}{\partial x \partial y} = \frac{\partial ^2 F}{\partial y \partial x}[/tex]
 
  • #10


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

1. What is an integrating factor?

An integrating factor is a function that is used to solve a differential equation by making it exact. It is multiplied by both sides of the equation to convert it into an exact differential equation, making it easier to solve.

2. How do you verify an integrating factor?

To verify an integrating factor, you need to multiply it by both sides of the differential equation and check if it becomes exact. If the equation becomes exact, then the integrating factor is correct. If not, you need to try a different integrating factor until the equation becomes exact.

3. Why is it important to verify an integrating factor?

Verifying an integrating factor is important because it ensures the accuracy of the solution to a differential equation. If the integrating factor is incorrect, the solution to the equation will also be incorrect.

4. What are some common mistakes when verifying an integrating factor?

Some common mistakes when verifying an integrating factor include forgetting to multiply it by both sides of the equation, using the wrong integrating factor, or making a mistake in the integration process.

5. Can an integrating factor be used for all types of differential equations?

No, an integrating factor can only be used for certain types of differential equations, specifically those that are not exact but can be made exact through multiplication by the integrating factor. It cannot be used for equations that are already exact or for equations that cannot be made exact.

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