First integral

[Patricia662]

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 equations? Thank you

 

[yoscar04]

Hi, I am not sure what your question is. Is this part of a course? Which course? What have you tried?

 

[Patricia662]

Thats 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]

excellent you found the integration factor

but i will mention an important thing

[MATH]\int -x \tan(xy) \ d(xy) = x \ln|\cos(xy)|[/MATH]
here [MATH]x[/MATH] is considered as a constant, so it is ok to ignore it

then, the integration factor is

[MATH]\mu (xy) = e^{\ln|\cos(xy)|} = \cos(xy)[/MATH]
now you have

[MATH]M = \cos(xy)(-xy\tan(xy) + y + 1)[/MATH]
[MATH]N = \cos(xy)(x - x^2\tan(xy))[/MATH]
And you can define a new function [MATH]g(y)[/MATH] as

[MATH]g(y) = \int \left(N - \frac{\partial}{\partial y} \int M \ dx \right) \ dy[/MATH]
[MATH]= \int \left(\cos(xy)(x - x^2\tan(xy)) - \frac{\partial}{\partial y} \int \cos(xy)(-xy\tan(xy) + y + 1) \ dx \right) \ dy[/MATH]
[MATH]= \int \left(\cos(xy)(x - x^2\tan(xy)) - \frac{\partial}{\partial y} \ \sin(xy) + x \cos(xy) \right) \ dy[/MATH]
[MATH]= \int \left(\cos(xy)(x - x^2\tan(xy)) - (x \cos(xy) - x^2 \sin(xy))\right) \ dy [/MATH]
[MATH]= \int \left( x\cos(xy) - x^2\sin(xy) - x \cos(xy) + x^2 \sin(xy)\right) \ dy [/MATH]
[MATH]= \int 0 \ dy = 0[/MATH]
the left side of the equation is

[MATH]f(x, y) = \int M \ dx + g(y)[/MATH]
[MATH]= \int \cos(xy)(-xy\tan(xy) + y + 1) \ dx + 0[/MATH]
[MATH]= \int -xy\sin(xy) + y\cos(xy) + \cos(xy)) \ dx[/MATH]
[MATH]= \sin(xy) + x\cos(xy)[/MATH]
the right side of the equation is just a constant

then the final solution is

[MATH]\sin(xy) + x\cos(xy) = C[/MATH]