Question

\[(y^{2}-x^{2}-z)\frac{ dx }{ dt }+(z-y^{2}-xy)\frac{dy}{dt}+x\frac{dz}{dt}=0\]

how to integrate the eqn? earlier this day someone pointed out that this is a total differential of some fn, but a solution in the book says that the original eqn is \[z=x^{2}+y^{2}-xy\]. Even if i try to take a total differential of the original eqn i can't even remotely get to the eqn to be solved. any suggestions?

Answer 1

it would be really helpful if you posted what it is you are actually trying to do (eg a copy of the actual question) but here is a way of reconciling your "answer" to what you originally posted.

you want:
\[(y^{2}-x^{2}-z) \ dx +(z-y^{2}-xy) \ dy+x \ dz=0\]
you have:
\[z=x^2+y^2−xy\]
so, using differentials:
\[dz=2x \ dx+2 y \ dy−x \ dy - y \ dx = (2x - y)dx + (2y - x) \ dy \]
times by x: \[x \ dz= (2x^2 - xy) \ dx + (2xy - x^2) \ dy \]
we know:
\[x^2 = z - y^2+xy\]
\[x \ dz= (x^2 + z - y^2 +xy - xy) \ dx + (2xy - z + y^2 - xy) \ dy \]
\[x \ dz= (x^2 + z - y^2 ) \ dx + (xy - z + y^2 ) \ dy \]
\[(y^{2}-x^{2}-z) \ dx +(z-y^{2}-xy) \ dy+x \ dz=0\]

Answer 2

irishboy thanks, as an insight it was helpful. however, my question was "how to integrate the equation?" that is what i am trying to accomplish. So it's the other way around: i have \[(y^{2}−x^{2}−z)\frac{dx}{dt}+(z−y^{2}−xy)\frac{dy}{dt}+x\frac{dz}{dt}=0\] and by integration i want\[z=x^{2}+y^{2}-xy\]

Answer 3

you previously explained that you could not even connect them:

""Even if i try to take a total differential of the original eqn i can't even remotely get to the eqn to be solved. any suggestions?.""

now you can :p

and, as stated above:

**it would be really helpful if you posted what it is you are actually trying to do (eg a copy of the actual question)**


do you actually understand what you are being asked to do?

Answer 4

the assignment says: "Decide if the the coupling is integrable and find it's final form"

Answer 5

thanks @DeLaHruza, that's exactly what i needed to know.

now i'll have a look and see if i can be of any further help.

Answer 6

Hi @DeLaHruza

i have really wrestled with this yet feel i have got nowhere :(

i thought it might be a scalar line integral, but there is no specified path allowing parameterisation or such like to give a solution.

i thought it might be a line integral of a vector field, which it isn't because the field itself is non conservative and so there is no generalised solution..you can get very close to one but i can't figure out what to do to z to force a solution, and that is part of my dilemma, z is not independent in this equation.

what puzzles me, most, is that as soon as you feed the solution into the D.E. as presented, it suddenly looks like a conservative field. ie, the D.E. becomes
\[(y - 2x) \ dx + (x - 2y) \ dy + dz = 0 \ and \\ \left|\begin{matrix}\hat i & \hat j & \hat k\\ ∂/∂x & ∂/∂y & ∂/∂z \\ (y - 2x) & (x - 2y) & 1\end{matrix}\right| = 0 \ !!!!\]
maybe the answer is an integrating factor -- i don't know. the solution is a surface whereas the DE looks to me like a 3-D scalar or vector field. clearly out of my depth.

i hope you find a solution; and, if so, please post it here if you have the time.

good luck.

i can also tag some people here if you want.

Answer 7

thanks for your time, i will keep searching for the solution

Answer 8

man i think i've found the solution, although i don't completely understand why it's the way it is. i will try to outline it. as you said, the integrating factor was the way. i've found the solution in a book "A treatise on infinitesimal calculus" by Bartholomew Price (btw, it's available free on Google Books) on pg. 380

Answer 9

1) first i've put:
\[M(x,y,z)=y ^{2}-x ^{2}-z\]\[N(x,y,z)=z-y ^{2}-xy\]\[O(x,y,z)=x\]\[u=u(x,y,z)\]
2) then exactness conditions:
\[\frac{ d }{ dy}(uO)=\frac{ d }{ dz }(uN)\]\[\frac{ d }{ dz }(uM)=\frac{ d }{ dx }(uO)\]\[\frac{ d }{ dx }(uN)=\frac{ d }{ dy }(uM)\]
3) now:
\[O(\frac{ du }{ dy })-N(\frac{ du }{ dz })=u(\frac{ dN }{ dz }-\frac{ dO }{ dy })\]\[M(\frac{ du }{ dz })-O(\frac{ du }{ dx })=u(\frac{ dO }{ dx }-\frac{ dM }{ dz })\]\[N(\frac{ du }{ dx })-M(\frac{ du }{ dy })=u(\frac{ dM }{ dy }-\frac{ dN }{ dx })\]
4) now in 2) multiply first eqn by M, second eqn by N and third by O and add those three together yielding:
\[M(\frac{ dN }{ dz }-\frac{ dO }{ dy })+N(\frac{ dO }{ dx }-\frac{ dM }{ dz })+O(\frac{ dM }{ dy }-\frac{ dN }{ dx })=0\]
5) substitute for M,N,O and differentiate resulting in:
\[-y ^{2}-x ^{2}+z+xy=0 \rightarrow z=x ^{2}+y ^{2}-xy\]
which is the solution according to my book.

Answer 10

there are few things i don't understand:
a) in 3) i don't yet understand why the signs on both sides of the equations are switched
b) when i plug the result 5) into the original DE, it is suddenly exact. what i don't understand is why it is considered a result...why in the hell i don't have to solve the now exact DE?

Answer 11

Hi @DeLaHruza

it's really very kind of you to get back on this, and i am glad that you have made some real progress

i will pound away at it too. please keep me informed. :p