Principle of Differentiation and Integration query

Hi Folks,

if we have an implicit function $\displaystyle z=x^2-4y^2+5xy-6=0$ eqn1

then we can find the standard derivative $\displaystyle \frac{dz}{dx}=2x-8y\frac{dy}{dx}+5x\frac{dy}{dx}+5y$ eqn 2 and

the partial derivative $\displaystyle \frac{\partial z}{\partial x}=2x+5y$ eqn3

However, I am trying to grasp the idea of having a standard/total derivation and partial derivation of functions as above yet we dont have standard or partial integration (ie the reverse). How is that?

If we integrate back eqn2 wrt to x we get eqn1 with a constant. This constant can be determined if BC's or IC's are known (ie 6) etc...but

if we integrate eqn3 wrt x we can get eqn1 with a constant also but we lose the $\displaystyle -4y^2$ term. How does one get this term back?

I dont think I understand this fully. Can anyone shed some light?

Thanks

Bugatti79

Principle of Differentiation and Integration query

Quote:

Originally Posted by

**Ackbeet** I'm saying that if I were given

$\displaystyle \frac{dz}{dx}=2x-8y\frac{dy}{dx}+5x\frac{dy}{dx}+5y$,

and told to integrate with respect to x, I would produce the following:

$\displaystyle z(x,y(x))=x^2 + 5\,x\,y(x) - 4\,y^{2}(x)+C.$

Just when I thought I understood!! (Headbang)

Integrating dz/dx, I get for each term

$\displaystyle \int2x dx = x^2$

$\displaystyle \int-8y\frac{dy}{dx} dx = \int -8y dy = -4y^2$ the dx's cancel

How does the third term go to 0 because I get

$\displaystyle \int 5x\frac{dy}{dx} dx= \int5x dy = 5xy$ the dx's cancel

$\displaystyle \int 5y dx= 5xy$

All of these term plus a constant etc. However, I now have two '5xy' terms. Should only have one. What have i done wrong?

I feel stupid now (Happy)

Thanks

Principle of Differentiation and Integration query

Quote:

Originally Posted by

**Ackbeet** The third term doesn't go to zero. I think you're going to see some interaction between terms there. You have to combine the last two terms, I think:

$\displaystyle \int (5x\frac{dy}{dx}+5y) dx=5\int(x y'(x)+y(x))\,dx

=5\int\frac{d}{dx}(x y(x))\,dx...$

which just becomes $\displaystyle 5xy(x)$...

think that makes sense. Thanks