Help with integrating factor

Wellesley

Quote by pr0me7heu2

1. The problem statement, all variables and given/known data
,find an integrating factor and then solve the following:

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

2. Relevant equations

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

,no clue how though!

3. 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 (M_y does not equal N_x), 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 f_xy=f_yx 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(M_y-N_x]/N or

du/dy=[u(N_x-M_y]/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?

djeitnstine

[tex]

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

[/tex]

is equally valid

Wellesley

Quote by djeitnstine

[tex]

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

[/tex]

is equally valid

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.

Wellesley

I still am having trouble with integrating factors....any ideas on the problem above?

konthelion

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.

Wellesley

Quote by konthelion

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.

M_y=-8x³y^-3-3y^-2
N_x=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.

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djeitnstine Recognitions: Gold Member	[tex] \frac{M_{y}(x,y) - N_{x}(x,y)}{N(x,y)} [/tex] is equally valid

Wellesley	Help with integrating factor I still am having trouble with integrating factors....any ideas on the problem above?

konthelion	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.