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Wellesley
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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?