Question

Does integrating a partial regenerate the original function? - I'm sure there's an addition function of the other variables that' got to be added.

Answer 1

I think we have to integrate w.r.t all variable and then take all terms. Like finding potential function from vector field

Answer 2

If its the first derivative that is being integrated then yes you would get the original function. If its the 2nd derivative then you get the first derivative when you integrate. Partial fractions is a technique used to integrate.

Answer 3

Let me me clarify :I think it should be \[\int\limits\limits\limits_{?}^{?}\frac{\partial P(x,y)}{\partial x} dx = P +f(y)\]

Answer 4

Furthermore i think the f(y) should be = -the parts of P consisting only of (depending purely on) y and constants + some constant

Answer 5

for example P = xy +y^2 :you cannot regenerate the pure y terms through an integragration across x.

Answer 6

for example P = xy +y^2 :you cannot regenerate the pure y terms through an integragration across x.

Answer 7

So partial of x y + y^2 is
wrt. x is\[y\]
wrt. y is
\[x+2 y \]

So we integral x partial with wr.t x
we get\[ yx+ C(y)\]

Now integrate y partial w.rt y we get
\[xy+y^2+C(x)\]

Now put it together, taking common term only once
\[xy+y^2\]

Answer 8

yes....i know there are methods of regeneration the original function i simply do not agree that \[ \int\limits\limits\limits\limits_{?}^{?}\frac{\partial P(x,y)}{\partial x} dx = P \]......
For those of you who have the book:I disagree with what is written on page 76-77 of dover's Ordinary differential equations by Tenenbaum and pollard