How can I determine if the equation (2x+3) + (2y-2)y' = 0 is exact or not?


by fehwqjkfdehqa
Tags: 2y2y, determine, equation, exact
fehwqjkfdehqa
fehwqjkfdehqa is offline
#1
Jul20-08, 08:16 PM
P: 1
How can I determine if the equation (2x+3) + (2y-2)y' = 0 is exact or not?

Now I know I need to take partial derivatives of certain terms of the equation, and call that M and N right?

1. How do I separate the terms?
2. How do I know which variable gets differentiated?

For example , if I separate it so that its:
partial x: (2x+3) = 2
partial y: (2y-2) = 2
2=2, so its exact.

BUT why can't I go:
partial x (2y-2) = 0
partial y (2x+3) = 0
0=0 so thats also exact.

So how do you determine which term is associated with what you are differentiating with repect to???

Also, why do textbooks use M and N??? Whats the point of using these when we can just say "partial differentiate with repect to x" and "partial differentiate with repect to y"

Also, does M ussually go with x and the N ussually go with y and why?

Please explain this to me my brain is about to explode.
Thanks
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jeffreydk
jeffreydk is offline
#2
Jul20-08, 10:24 PM
P: 136
It is common practice to write it as

[tex]M(x,y)dx+N(x,y)dy=0[/tex]

and then to differentiate N with respect to x and M with respect to y to check if the equation is exact. The whole method depends on the fact that there is some function where,

[tex] \frac{\partial \Phi}{\partial x}=M(x,y),\text{ }\frac{\partial \Phi}{\partial y}=N(x,y)[/tex]

You can't switch the respectful variables and have it still work so I think you found a special case. For example,

[tex](y^2-2x)dx+(2xy+1)dy=0[/tex]

[tex]M_y=2y, N_x=2y[/tex]

so it is exact, but the other way around you get,

[tex]M_x=-2, N_y=2x[/tex]

Hope that helps.


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