Patricia662
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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
? Thank you
First integral
- Thread starter Patricia662
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Patricia662
New member
- Joined
- Aug 30, 2020
- Messages
- 10
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 youexcellent 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
\(\displaystyle \sin(xy) + x\cos(xy) = C\)
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
\(\displaystyle \sin(xy) + x\cos(xy) = C\)