Question

show that if \[\frac{N_x-M_y}{xM-yN}=R\]where R depends on the quantity xy only. Then M+Ny'=0 has an integrating factor of the form\[\mu(xy)\]. Find a general formula for this integrating factor.
Please, help.

Answer 1

From definition for equation to have intgrating factor:
\((M\mu)_y =( N\mu )_x\)
differentiating
\(M_y\mu+M\mu_y=N_x\mu+N\mu_x\)
so
\( \large \mu=\frac{M\mu_y-N\mu_x}{N_x-M_y}\)
now take \(\huge \frac{1}{\mu}\)
so
if this quanitiy we call R and it depends only on xy, we see that 1/R=\(\mu\) which will also depend on xy

Answer 2

so, we can conclude that the form of that integrating factor is \[R= \frac {1}{\mu}\], right?

Answer 3

yes

Answer 4

just there is one point:
R(xy) can be anything. For example: \(3(xy)^2+e^{xy}\)
So \(R(xy)_{x}=R_xy\) and \(R(xy)_y=R_yx\)
and you have to take this into the account for \(\mu\)

Answer 5

you would check befor starting to look for integrating factor, :)

Answer 6

this problem is showing one. It means it is general case. I have another problem. They asked me apply this concept. I will carefully do it, If I get stuck, I tag you to get help.
Thanks a lot.

Answer 7

My question is when Myko take \[R= \frac{1}{\mu}\] it doesn't form the form question asked for, I need x*M - y*N at denominator,

Answer 8

@miko @druminjosh . I have class now and have to go. Whenever you guys came online, please explain me steps. Thanks in advance.

Answer 9

continuing with my 1º comment:
\(\huge \frac{1}{\mu}=\frac{N_x-M_y}{M\mu_y-N\mu_x}\)
this is already almost what your question states. Just need one more step.
Now \(\mu(xy)\) is any function. So taking it's derivative respect to x and later also y will look something like this:
\(\large\mu_y=\mu_{xy}x\) .This is just chain rule, if you wonder, :).
Same for the other: \(\large\mu_x=\mu_{xy}y\).
Now put this into the expretion for \(\huge \frac{1}{\mu}\) and rearange it a bit.
\(\huge \frac{\mu_{xy}}{\mu}=\frac{N_x-M_y}{xM-yN}\)
This is now exactly what you question states. So the relation between R and \(\mu\) is:
\(\huge\frac{\mu_{xy}}{\mu}=R\)
Hope it is ok now, :).

Answer 10

@Loser66

Answer 11

what @myko said. don't have anything to add, he beat me to it :) good job @myko (and thanks! save me a lot of work hehe)