### Author Topic: Question about how to deal with non exact equation  (Read 2178 times)

#### Boyu Zheng

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##### Question about how to deal with non exact equation
« on: September 19, 2018, 11:01:15 PM »
Hi professor Ivrii, You ask everybody, not just me. V.I.
When I do this question (3x+6/y) +(x^2/y+3y/x)dy/dx = 0 , I tried to get My and Nx and after take the partial derivative. I found My is not equal to Nx and its not exact.
I just wonder how to find the intergation factor in order to make the equation exact?

« Last Edit: September 19, 2018, 11:57:44 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Question about how to deal with non exact equation
« Reply #1 on: September 19, 2018, 11:59:12 PM »
Try follow three cases considered on the lecture. Basically what you wrote is useless since it could be interpreted in different ways

#### Boyu Zheng

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##### Re: Question about how to deal with non exact equation
« Reply #2 on: September 20, 2018, 12:29:07 AM »
I just reread your notes from the lecture but I still don't understand for try3 why do you use -(dM/dy - dN/dx)/Mx-Ny = f(xy) to get M+Ny'=0?

#### Victor Ivrii

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##### Re: Question about how to deal with non exact equation
« Reply #3 on: September 20, 2018, 04:10:19 AM »
I just reread your notes from the lecture but I still don't understand for try3 why do you use
$$-\frac{(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})}{(Mx-Ny)} = f(xy)$$
to get $M+Ny'=0$?
This is how to write math formulae: http://forum.math.toronto.edu/index.php?topic=610.0. On any quiz or test usage of "$d$" instead of "$\partial$" will be considered as an error.

This follows from our calculations, you need to look at them. It is also Problem 24 at https://q.utoronto.ca/courses/56504/files/1252344?module_item_id=318959

#### Pengyun Li

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##### Re: Question about how to deal with non exact equation
« Reply #4 on: October 01, 2018, 11:49:24 AM »
Let's see the equation first: (3x+6/y) + (x2/y +3y/x)dy/dx = 0.

Set M = 3x+6/y and N = x2/y +3y/x,
Then differentiate M w.r.t y and differentiate N w.r.t x, and then we derive that:
My= -6/(y2) and Nx = 2x/y - 3y/x2,
since My doesn't equal to Nx, then it is not exact.
Then we need to consider about the non-exact method.

However, note that:
(My-Nx)/N is not only a function of x, and (My-Nx)/M is not only a function of y.
Therefore we cannot use the normal method, so we have to use the formula:
(Nx - My)/(xM - yN) = R, where R = R(xy) depends on the quantity xy only (if you did not get it, then there must be something wrong, as if it is only function of x or only function of y, we would not need to use this method in the first place), then the differential equation had an integrating factor of the form μ(xy).

R = (Nx - My)/(xM - yN)
= (2x/y - 3y/x2 + 6/y2) / (3x2 + 6x/y - x2 - 3y2/x)
= 1/(xy)

Let t = xy, so R = 1/(xy) = (1/t)

μ(xy) = e(∫1/t dt) = e(lnt) = t = xy

Multiply μ(xy) on both sides of (3x+6/y) + (x2/y +3y/x)dy/dx = 0.

Then we derive that (3x2y+ 6x) + (x3 + 3y2) dy/dx = 0.

Set M = 3(x2)y+ 6x and N = x3 + 3y2,
Similarly, after differentiation, we can get that:
My = 3x2 = Nx, which is exact now!

So we can use the exact method.

There exists φ(x,y) s.t. φx = M, φy = N
Hence, φ = ∫M dx = ∫3x2y + 6x dx = yx3 + 3x2 + h(y)
(I write h(y) as it acts like a constant after we integrate w.r.t x only)
Therefore φy = x3 + h'(y) = x3 + 3y2
hence h'(y) = 3y2, h(y) = y3 + c

Therefore, φ = yx3 + 3x2 + y3 = c

Hope it helps