1. ## Inexact Differential Equation

I need help with this:

Given: [x - (sec y)(ln x)] dy + [tan y – (y/x)sec y] dx = 0
Required: General Solution
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Suppose that M = tan y – (y/x) sec y and N = x – (sec y) (ln x)
Then, partial derivative of N wrt x = 1 – ((sec y)/x)
And partial derivative of M wrt y = sec y (sec y – ((y tan y)/x) - 1/x)
f(y) = - 1/M (sec y (sec y – ((y tan y)/x) – 1/x))
from there, to get the integrating factor,
f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
and then, i’m kind of stuck.

2. Originally Posted by vividlyxinsane
I need help with this:

Given: [x - (sec y)(ln x)] dy [tan y – (y/x)sec y] dx = 0
Required: General Solution
----------------------------------------------------------------------

I can't seem to come up with the right answer.

Help? =]
are you sure that - is not + ?

3. Originally Posted by NonCommAlg
are you sure that - is not + ?

oh yeah, it's positive. sorry.

back to the problem,

Suppose that M = tan y –(y/x) sec y and N = x – (sec y) (ln x)
Then, partial derivative of N wrt x = 1 – ((sec y)/x)
And partial derivative of M wrt y = sec y (sec y – ((y tan y)/x) - 1/x)
f(y) = - 1/M (sec y (sec y – ((y tan y)/x) – 1/x))
from there, to get the integrating factor,
f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
and then, i’m kind of stuck.

4. Originally Posted by vividlyxinsane
oh yeah, it's positive. sorry.

back to the problem,

Suppose that M = tan y –(y/x) sec y and N = x – (sec y) (ln x)
Then, partial derivative of N wrt x = 1 – ((sec y)/x)
And partial derivative of M wrt y = sec y (sec y – ((y tan y)/x) - 1/x)
f(y) = - 1/M (sec y (sec y – ((y tan y)/x) – 1/x))
from there, to get the integrating factor,
f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
and then, i’m kind of stuck.
just multiply both sides of your original differential equation by $\cos y$ and you'll have an exact equation.

5. Originally Posted by NonCommAlg
just multiply both sides of your original differential equation by $\cos y$ and you'll have an exact equation.

cos y is the integrating factor?

before i get there, f(y) has to be purely in terms of y. i'm stuck at:

f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
i'm kind of clueless on what to do next.

6. Originally Posted by vividlyxinsane
cos y is the integrating factor?

before i get there, f(y) has to be purely in terms of y. i'm stuck at:

f(y) = (sec^2 y – ((y sec y tan y)/x) - 1) / (tan y – (y/x)sec y)
i'm kind of clueless on what to do next.
since $\sec^2y - 1 = \tan^2y,$ your function becomes: $\frac{x\tan^2 y - y \sec y \tan y}{x\tan y - y \sec y}=\tan y=f(y).$

7. Originally Posted by NonCommAlg
since $\sec^2y - 1 = \tan^2y,$ your function becomes: $\frac{x\tan^2 y - y \sec y \tan y}{x\tan y - y \sec y}=\tan y=f(y).$
so that was what i was missing. but shouldnt the IF = sec y?

since the integral of tan y is -ln(cos y) or ln(sec y)?

8. Originally Posted by vividlyxinsane
so that was what i was missing. but shouldnt the IF = sec y?

since the integral of tan y is -ln(cos y) or ln(sec y)?
the function that you called $f(y)$ is equal to $\frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}}{M}$ not $\frac{\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}}{M}.$ if $\mu$ is an integrating factor which is purely a function of $y,$ then: $\frac{d \mu}{\mu}= \frac{\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}}{M}dy= -f(y)dy=-\tan y \ dy.$ so $\mu=\cos y.$

9. Originally Posted by NonCommAlg
the function that you called $f(y)$ is equal to $\frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}}{M}$ not $\frac{\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}}{M}.$ if $\mu$ is an integrating factor which is purely a function of $y,$ then: $\frac{d \mu}{\mu}= \frac{\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y}}{M}dy= -f(y)dy=-\tan y \ dy.$ so $\mu=\cos y.$
oh, right.. f(y) IS NEGATIVE.
i got it.
THANKS A LOT! =]