What is purpose of an integrating factor and how do you get one?

[jenzao]

Homework Statement

Am I doing this correctly?
Is 2y^4dx +4xy^3dy exact or inexact?

Homework Equations

The Attempt at a Solution

So if its exact, then M(x,y)dx = N(x,y)dy right? (Euler)

But how can I take partial derivative of 2y^4 with respect to x, if there is no x to derive??

thats where I am stuck?

 

[rootX]

Am I doing this correctly?
Is 2y^4dx +4xy^3dy exact or inexact?

This is unclear.

So if its exact, then M(x,y)dx = N(x,y)dy right? (Euler)

Unclear

But how can I take partial derivative of 2y^4 with respect to x, if there is no x to derive??

2y^4 is constant does not change with respect to x

If the function f = 2y^4 does not change as x changes; it remains constant
d [f] / dx should be?

 

[dynamicsolo]

Homework Statement

Am I doing this correctly?
Is 2y^4dx +4xy^3dy exact or inexact?

Homework Equations

The Attempt at a Solution

So if its exact, then M(x,y)dx = N(x,y)dy right? (Euler)

You have omitted something here: this is supposed to be

M(x,y) dx + N(x,y) dy = 0

is an exact differential equation (in R) iff

M_y (x,y) = N_x (x,y) everywhere on R.

So you need to take partial derivatives to check this...

But how can I take partial derivative of 2y^4 with respect to x, if there is no x to derive??

... which you have reversed the order of in your work. You should be taking the partial derivative of 2y^4 with respect to y.

 

[HallsofIvy]

Homework Statement

Am I doing this correctly?
Is 2y^4dx +4xy^3dy exact or inexact?

Homework Equations

The Attempt at a Solution

So if its exact, then M(x,y)dx = N(x,y)dy right? (Euler)

No, you have it reversed. It is M(x,y)_y= N(x,y)_x- assuming you mean M(x,y)dx+ N(x,y)dy.

But how can I take partial derivative of 2y^4 with respect to x, if there is no x to derive??

Then it is a constant with respect to x and the derivative would be 0. However what you want is (2y⁴)_y and (4xy³)_x.

thats where I am stuck?

Remember WHY you are doing that. IF M(x,y)dx+ N(x,y)dy is an exact differential, then there exist some function f(x,y) such that df= M(x,y)dx+ N(x,y)dy. By the chain rule,
df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y} dy[/itex] <br /> so if M(x,y)dx + N(x,y)dy is exact, M(x,y)= f_x and N(x,y)= f_y. Taking the derivatives of M and N give the <b>mixed</b> derivatives: M(x,y)_y= f_xy and N(x,y)_x= f_yx. As long as the derivatives are all continuous, those mixed derivatives must be equal.<br /> <br /> That does NOT answer your title questin &quot;What is the purpose of an integrating factor and how do we find one&quot;. <br /> <br /> Given a function f(x,y), it &quot;differential&quot; is <br /> df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy<br /> Writing arbitrary functions for M(x,y) and N(x,y) in M(x,y)dx+ N(x,y)dy may LOOK like a differential but as I pointed out above, such an f exists if and only if the above criterion is satisfied- if and only if the differential is <b>exact</b>. It can be shown that, given anything of the form M(x,y)dx+ N(x,y)dy you can always multiply it by some function of x and y to <b>make</b> it exact- an &quot;integrating factor&quot;. There is, however, NO general method of finding an integrating factor.<br /> <br /> In this particular case, since both M(x,y) and N(x,y) involve only powers of x and y, I might <b>try</b> xⁿy^m as an integrating factor. Multiplying by that, I have 2xⁿy^m+4dx+ 4xⁿ⁺¹y^m+3dy. Now, (2xⁿy^m+4)_y= 2(m+4)xⁿy^m+3 and (4xⁿ⁺¹y^m+3)_x= 4(n+1)xⁿy^m+3. The powers of x and y are correct! In order that those be equal we must have 2(m+4)= 4(n+1) or 2m+ 8= 4n+ 4 so m= 2n- 2. It is easest to take n= 1 so m= 0. That means that x itself is an integrating factor: (2xy⁴)_{y[/sup]= 8xy³ and (4x²y³)_{x_{= 8xy³ also.<br /> <br /> That means there must exist some function f(x,y) so that df= f_xdx+ f_yydy= (2xy⁴)dx+ (4x²y³)dy. In particular, f_x= 2xy⁴ so, integrating with respect to x (while treating y as a constant since this is a partial derivative) we have f(x,y)= x²y⁴+ C(y). Notice that, because y is treated as a constant in <a href="https://www.physicsforums.com/insights/partial-differentiation-without-tears/" class="link link--internal">partial differentiation</a>, the &quot;constant of integration&quot; <b>might</b> be a function of y. However, differentiating that with respect to y we have (x²y⁴+ C(y))_y= 4x²y³+ C&#039;(y)= 4x²y³ so C&#039;(y)= 0- C really is a constant. f(x,y)= x² y⁴+ C.<br /> <br /> Now, whether multiplying by that &quot;integrating factor&quot; disrupts the <b>rest</b> of the problem depends upon what the problem is! dynamicsolo is incorrect to suggest that there <b>must</b> be an &quot;= 0&quot; on that. A &quot;differential&quot; is <b>not</b> a &quot;differential equation&quot; and an &quot;exact differential is not an &quot;exact differential equation&quot;. However, if the problem really were, as he suggested, to solve the (non-exact) differential equation 2y⁴dx+ 4xy³= 0, then multiply both sides by x gives 2xy⁴dx+ 4x²y³= d(x²y⁴)= 0 which has the general solution x²y⁴= C.}}}