First integral

Patricia662

New member
Joined
Aug 30, 2020
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10
I dont know how to use this function m(x,y)=m(xy) in the equation. I just multiply with xy, so i get the identical partial equationsIMG_20200831_114321.jpg? Thank you
 

yoscar04

Full Member
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Jun 3, 2020
Messages
274
Hi, I am not sure what your question is. Is this part of a course? Which course? What have you tried?
 

Patricia662

New member
Joined
Aug 30, 2020
Messages
10
IMG_20200831_161249.jpgThats part of diferential equations. I dont know how to use this function µ.( (P(x)*µ)/dy)=( (Q(x)*µ)/dx). Because ( (P(x)/dy) is not equilavent to ( (Q(x)/dx) you need to.multiple with a function what we call integration multiplicator. And now my question is: is this function actually x*y or what? On the photo is something little i have tried. I understand concept but not with what i need to multiple that P(x) and Q(x) so it will be the same. I hope i wrote it understandable. Thank you
 

joshuaa

Junior Member
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Aug 21, 2020
Messages
57
excellent you found the integration factor

but i will mention an important thing

\(\displaystyle \int -x \tan(xy) \ d(xy) = x \ln|\cos(xy)|\)

here \(\displaystyle x\) is considered as a constant, so it is ok to ignore it

then, the integration factor is

\(\displaystyle \mu (xy) = e^{\ln|\cos(xy)|} = \cos(xy)\)

now you have

\(\displaystyle M = \cos(xy)(-xy\tan(xy) + y + 1)\)

\(\displaystyle N = \cos(xy)(x - x^2\tan(xy))\)

And you can define a new function \(\displaystyle g(y)\) as

\(\displaystyle g(y) = \int \left(N - \frac{\partial}{\partial y} \int M \ dx \right) \ dy\)

\(\displaystyle = \int \left(\cos(xy)(x - x^2\tan(xy)) - \frac{\partial}{\partial y} \int \cos(xy)(-xy\tan(xy) + y + 1) \ dx \right) \ dy\)

\(\displaystyle = \int \left(\cos(xy)(x - x^2\tan(xy)) - \frac{\partial}{\partial y} \ \sin(xy) + x \cos(xy) \right) \ dy\)

\(\displaystyle = \int \left(\cos(xy)(x - x^2\tan(xy)) - (x \cos(xy) - x^2 \sin(xy))\right) \ dy \)

\(\displaystyle = \int \left( x\cos(xy) - x^2\sin(xy) - x \cos(xy) + x^2 \sin(xy)\right) \ dy \)

\(\displaystyle = \int 0 \ dy = 0\)

the left side of the equation is

\(\displaystyle f(x, y) = \int M \ dx + g(y)\)

\(\displaystyle = \int \cos(xy)(-xy\tan(xy) + y + 1) \ dx + 0\)

\(\displaystyle = \int -xy\sin(xy) + y\cos(xy) + \cos(xy)) \ dx\)

\(\displaystyle = \sin(xy) + x\cos(xy)\)

the right side of the equation is just a constant

then the final solution is

\(\displaystyle \sin(xy) + x\cos(xy) = C\)
 
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