Is Equation a Total Derivative?

[Lindsayyyy]

Homework Statement

Is the following equation a total derivative?

dz = 2ln(y)dx+{\frac x y}dy

Homework Equations

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The Attempt at a Solution

I would say no. I tried it with the symmetry of the second derivatives.

2ln(y) is {\frac {\partial z} {\partial x}}

when I derivate it now and say y is my variable I get:

{\frac {\partial z} {\partial y}}=\frac 2 y

same for the other expression

{\frac {\partial z} {\partial y}}=\frac x y

{\frac {\partial z} {\partial x}}=\frac 1 y


so they are different => no totale derivative. Am I right or where did I do mistakes?

Thanks for the help

 

[tiny-tim]

Hi Lindsayyyy!

That's completelyyyy right.

(just to check, I like to differentiate the obvious possibility, xlny, and that gives lnydx + xdy/y, which doesn't match)

 

[Lindsayyyy]

Thank you very much

 

[HallsofIvy]

Just to put my oar in, if that were a total derivative, there would exist a function, F(x,y) such that
\frac{dF}= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy
\frac{dF}= 2 ln(y)dx+ \frac{x}{y}dy
..
So we must have \partial F/\partial x= 2 ln(y) so F= 2x ln(y)+ \phi(y) where \phi(y) can be any differentiable function of y only.

From that \partial F/\partial y= 2x/y+ \phi'(y) and that must be equal to x/y: 2x/y+ \phi'(y)= x/y so \phi'(y)= -x/y which is impossible since \phi'(y) is not a function of y.

Note the "mixed second derivatives". If \partial F/\partial y= x/y+ \phi'(y)
then \partial^2 F/\partial x\partial y= 1/y and if \partial F/\partial x= 2ln(y) then \partial^2 F/\partial y\partial x= 2/y. Those mixed second derivatives should be the same and that was what you were checking.

 

[Lindsayyyy]

your solution is way more elegant than mine

I have another problem:

x^{\alpha}y^{\beta}(2ln(y)dx+{\frac x y}dy) is given. I shall find alpha and beta that the given expression is a total derivative.

My approach:

I found { \frac {\partial^{2}z} {\partial x \partial y}}=2\beta x^{\alpha}y^{\beta -1}ln(y)+{\frac 1 y}2x^{\alpha}y^{\beta}

and
{ \frac {\partial^{2}z} {\partial y \partial x}}=\alpha x^{\alpha -1}y^{\beta}{\frac x y}+{\frac {x^{\alpha}y^{\beta}} y}

I though I could equate them and solve the equation to get an expression for alpha and beta. But my problem is I can easily cancel some terms with potential rules like y^a-1=y^a/y etc.

And I get to the euqation:

2\beta lny+2 = \alpha+1

is my approach wrong, where are my mistakes? If I was right thus far, how should I go on?

Thanks for your help

 

[SammyS]

...
And I get to the euqation:

2\beta lny+2 = \alpha+1

is my approach wrong, where are my mistakes? If I was right thus far, how should I go on?

Thanks for your help

There is no y on the right hand side, so what must β be ?

 

[Lindsayyyy]

fail

\beta =0 \alpha =1

thanks, I'm blind, I know it's a bit too much and insolent to ask but you haven't calculated it by yourself in order to test if I was right until the last step, have you?

Thanks for your help

edit:

nevermind, I'm tired^^
I haven't thought about just to put it in and derive it haha. It works. thanks