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

1. May 31, 2010

darryw

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

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 31, 2010

Staff: Mentor

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

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.

3. May 31, 2010

darryw

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

(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. May 31, 2010

darryw

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

so i should test for exactness?

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

?
thanks

5. May 31, 2010

Staff: Mentor

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

Yes. After you multiply by the integrating factor, test for exactness.

6. May 31, 2010

darryw

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

[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. May 31, 2010

Staff: Mentor

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

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

8. May 31, 2010

darryw

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

sorry i overlooked that..

but i still dont understand why M_x is not equal to N_y/?

9. May 31, 2010

Staff: Mentor

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

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. May 31, 2010

darryw

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

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