Integrating Factor for Exact Differential Equation

In summary, the student is having difficulty solving an equation involving integrating factors and is unsure if the equation can be solved using the original method or if he needs to add a term involving y.
  • #1
Wellesley
274
3
pr0me7heu2 said:

Homework Statement


,find an integrating factor and then solve the following:

[4(x3/y2)+(3/y)]dx + [3(x/y2)+4y]dy = 0


Homework Equations



u(y)=y2 is a valid integrating factor that yields a solution x4+3xy+y4=c

,no clue how though!


The Attempt at a Solution



I understand exact differential equations to really be a somewhat perverted form of a gradient of a multivariable surface defined by f=f(x,y)... and that solving an exact differential equations is comparable to finding the potential function using the fundamental theorem of line integrals.

,here with this equation I tried the usual methods

I first realized that the equation is NOT exact (My does not equal Nx), or in other words the two parts of the differential equation could not have possibly come from the same function f=f(x,y) in their current forms because fxy=fyx for any f(x,y).

So, I multiplied by an integrating factor u(x,y) such that the differential equation would be exact or such that (uM)y=(uN)x

,so far in my math education I've pretty much disregarded the idea of finding an integrating factor that is not solely a function of either y or x - it simply would be too hard.

,so in assuming u to be a function of either solely x or y I had hoped to find a linear and separable DE through the following:

du/dx=[u(My-Nx]/N or

du/dy=[u(Nx-My]/M

,unfortunately both of these yield seemingly impossibly, or at least algebraically disgusting results...

Any ideas on an easier or other method for solving for this/other integrating factors?

I can see why y2 is an integrating factor, but I don't know how this answer can be derived. Do you use:

[tex]
\frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)}
[/tex]
Where M (x,y) = 4(x3/y2)+(3/y)]dx, and N = 3(x/y2)+4y dy

When I try and solve this, I get a very complex integrating factor. Does anyone have any suggestions on solving these types of problems?
 
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  • #2
[tex]

\frac{M_{y}(x,y) - N_{x}(x,y)}{N(x,y)}

[/tex]

is equally valid
 
  • #3
djeitnstine said:
[tex]

\frac{M_{y}(x,y) - N_{x}(x,y)}{N(x,y)}

[/tex]

is equally valid

:confused: Isn't the integrating factor, a function of y (y2)? I know how to do other problems like this, but I just can't figure out this one. Would you be able to provide a better hint? Thanks.
 
  • #4
I still am having trouble with integrating factors...any ideas on the problem above?
 
  • #5
What did you get for [tex] \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)}[/tex]?

If your integrating factor say, [tex] \mu [/tex] is a function of x only, then

[tex] \frac{d \mu}{dx} = \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)} \mu [/tex]

From here you can solve for the exact equation after multiply the integrating factor by the original equation.
 
  • #6
konthelion said:
What did you get for [tex] \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)}[/tex]?

If your integrating factor say, [tex] \mu [/tex] is a function of x only, then

[tex] \frac{d \mu}{dx} = \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)} \mu [/tex]

From here you can solve for the exact equation after multiply the integrating factor by the original equation.

My=-8x3y-3-3y-2
Nx=3y-2

I think I see what I've been doing wrong now...I've always canceled out the two 3y-2, when I should have added. I can't believe I did that...

The reason I posted again, was partly due to another problem involving integrating factors.
-xsin(y) dy + (2+x)cos(y) dy ---- Find the integrating factor to make this exact.

Since I thought I didn't understand them well, I figured learning how to do the originally posted problem would help. Now, I'm not sure if this problem can be solved as written, since both derivates are with respect to y.
 

1. What is an integrating factor?

An integrating factor is a function used in the process of solving differential equations. It helps to transform a non-exact differential equation into an exact one, making it easier to solve.

2. How do I determine the integrating factor for a differential equation?

The integrating factor can be determined by multiplying both sides of the differential equation by an appropriate function, such as the reciprocal of the coefficient of the highest order derivative. This will make the equation exact and the integrating factor will be the function used.

3. Can an integrating factor always be found for a given differential equation?

No, there are some cases where an integrating factor cannot be found for a given differential equation. In these cases, other methods of solving the equation may need to be used.

4. How does using an integrating factor help in solving differential equations?

Using an integrating factor simplifies the process of solving a differential equation by making it exact. This allows for the use of standard methods for solving exact equations, making the solution more straightforward.

5. Are there any drawbacks to using an integrating factor?

One potential drawback of using an integrating factor is that it may not always lead to an exact solution. In some cases, the resulting equation may still be difficult to solve and alternative methods may need to be used.

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