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Novark
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Finding the integrating factor (ODEs) [Solved]
Working on this problem, I can't figure out why we take the derivative of [tex]\mu[/tex] with respect to y, and what to do when our integrating factor is a function of both x and y. In the case below, it ended up being separable, but what can you do if it's not?
Find an integrating factor and solve the given equation.
[tex]y{\cdot}dx + (2xy - e^{-2y}){\cdot}dy = 0[/tex]
[tex]\frac{d\mu}{dy} = \frac{M_y - N_x}{N}{\cdot}\mu[/tex]
[tex]M(x,y) = y[/tex]
[tex]N(x,y) = (2xy - e^{-2y})[/tex]
So,
[tex]M_y = 1[/tex]
[tex]N_x = 2y[/tex]
[tex]\frac{d\mu}{dy} = \frac{1 - 2y}{(2xy - e^{-2y})}{\cdot}\mu \Rightarrow
\frac{d\mu}{\mu} = \frac{1 - 2y}{(2xy - e^{-2y})}{\cdot}dy[/tex]
[tex]ln(\mu) = \int{\frac{1 - 2y}{(2xy - e^{-2y})}}{\cdot}dy[/tex]
If someone could refresh me on the methods to integrate the above, that would also be much appreciated. What I really want to know, however, is why we choose [tex]\mu[/tex] with respect to y instead of x, and how to find our integrating factor when [tex]\mu[/tex] is not separable.
Thanks in advance :-)
Working on this problem, I can't figure out why we take the derivative of [tex]\mu[/tex] with respect to y, and what to do when our integrating factor is a function of both x and y. In the case below, it ended up being separable, but what can you do if it's not?
Homework Statement
Find an integrating factor and solve the given equation.
[tex]y{\cdot}dx + (2xy - e^{-2y}){\cdot}dy = 0[/tex]
Homework Equations
[tex]\frac{d\mu}{dy} = \frac{M_y - N_x}{N}{\cdot}\mu[/tex]
The Attempt at a Solution
[tex]M(x,y) = y[/tex]
[tex]N(x,y) = (2xy - e^{-2y})[/tex]
So,
[tex]M_y = 1[/tex]
[tex]N_x = 2y[/tex]
[tex]\frac{d\mu}{dy} = \frac{1 - 2y}{(2xy - e^{-2y})}{\cdot}\mu \Rightarrow
\frac{d\mu}{\mu} = \frac{1 - 2y}{(2xy - e^{-2y})}{\cdot}dy[/tex]
[tex]ln(\mu) = \int{\frac{1 - 2y}{(2xy - e^{-2y})}}{\cdot}dy[/tex]
If someone could refresh me on the methods to integrate the above, that would also be much appreciated. What I really want to know, however, is why we choose [tex]\mu[/tex] with respect to y instead of x, and how to find our integrating factor when [tex]\mu[/tex] is not separable.
Thanks in advance :-)
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