Assume:
du= f1(u,v).dx+f2(u,v).dy (1)
dv= g1(u,v).dx+g2(u,v).dy (2)
f1,f2, and, g1,g2 are continuous and known functions.
We know further that implicitly h(u,v)=0 and hence we assume u=u(v) or v=v(u) is known and upon this fact, the equations (1), (2) are broken to:
du= f3(u).dx+f4(u).dy (3)
dv= g3(v).dx+g4(v).dy (4)
f3,f4, and, g3, g4 are continuous and known functions.
Moreover, we already know that dy/dx=v/u, hence needless to mention that dy/dx=v/u=m(u)=n(v), whilst m(.), n(.) are known continuous functions.
Holding these assumptions, please tell me how to solve (3), (4) to get u=u(x,y), and also v=v(x,y).
I don't understand your problem... if the functions f and g are known, being dx and dy the arbitrary differenzial, you have also that du and dv are known...
On the other hand, by definition of differential of u(x,y) and dv(x,y) you have also that
f1=du/dy, f2=du/dy, g3=dv/dx, g4=dv/dy
@Filippo Maria Denaro. You are missing the fact.
f1 is not du/dy, although it was typos, and you meant du/dx, but it is ∂u/∂x, and this could be critically game-changing.
f1=∂u/∂x, f2=∂u/∂y, g1=∂v/∂x, g2=∂v/∂y.
This is the delicate difference between partial differential
∂ and exact differential d.
Sure, I used d/dx and d/dy as brief symbol for partial derivatives, not for ordinary derivative. That does not change the fact that f and g are the derivatives in the expression for the total differential of u and v
du(x,y)=du/dx *dx + du/dy*dy
dv(x,y)=dv/dx *dx + dv/dy*dy
the RHS of the system is known
@Filippo Maria Denaro. Thanks to have your time on answering this question. So upon all the facts, what's your idea about getting the final solution.
Assume:
f1=∂u/∂x=u^3,
f2=∂u/∂y=u^2
then
du= u^3.dx+u^2.dy. (11)
If dy/dx=h(u) and h(u) is for example equal to sin(u), then
how do you solve the just recent equation (11), to get u=u(x,y)?
@Filippo Maria Denaro. Only give your own general solution methodology to the equation (11), and do not care much about h(u) =sin (u) or else.
You need to consider the differential calculus field, integration of (exact or not) differential.
A similar question is here
http://pleasemakeanote.blogspot.com/2010/07/integrating-totalexact-differentials.html
put X=(u,v) and
(f3(u) f4(u))
M(X)=matrix( )
( g3(v) g4(v))
The system equiv to X'=M(X) autonomous differential equation
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