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

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  • #1
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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


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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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.
=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


Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #3
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(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
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so i should test for exactness?

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

?
thanks
 
  • #5
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6,208


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

but i still dont understand why M_x is not equal to N_y/?
 
  • #9
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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
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thanks for the explanation. My final answer is now matching answer key, so im moving on to my other prob..
thanks.
 

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