Going backwards: finding a function from differential

[Yankel]

Hello,

I have this expression:

\[4xy^{3}\cdot dx+6x^{2}y^{2}\cdot dy\]

and I am asked to say if this expression can be a total differential of a function with two variables. If so, I need to find the function.

How can I do it ? It must involve integration somehow...

Thank you !

 

[chisigma]

Hello,

I have this expression:

\[4xy^{3}\cdot dx+6x^{2}y^{2}\cdot dy\]

and I am asked to say if this expression can be a total differential of a function with two variables. If so, I need to find the function.

How can I do it ? It must involve integration somehow...

Thank you !

You have two differential eqution...

$\displaystyle \frac{\partial {f}}{\partial{x}} = 4\ x\ y^{3}\ (1)$

$\displaystyle \frac{\partial {f}}{\partial{y}} = 6\ x^{2}\ y^{2}\ (2)$

... and the solution both (1) an (2) is $\displaystyle f(x,y) = 2\ x^{2}\ y^{3} + c $ [c is an arbitrary constant...], that is the requested function...Kind regards$\chi$ $\sigma$

 

[MarkFL]

If you are given the equation:

\(\displaystyle 4xy^3\,dx+6x^2y^2\,dy=0\)

Then you may test for exactness by seeing if:

\(\displaystyle \frac{\partial}{\partial y}\left(4xy^3\right)=\frac{\partial}{\partial x}\left(6x^2y^2\right)\)

Is this condition true?

 

[Yankel]

Thank you both. Can you think of an expression of the form Adx+Bdy which is not a total differential of a function, so I know how an answer of 'no' to the original question looks like ?

 

[chisigma]

Thank you both. Can you think of an expression of the form Adx+Bdy which is not a total differential of a function, so I know how an answer of 'no' to the original question looks like ?

An expression like...

$\displaystyle \alpha(x,y)\ dx + \beta(x,y)\ d y \ (1)$

.. . is called 'exact differential' if it exists a function f(x,y) so that is...

$\displaystyle \frac{\partial {f}}{\partial{x}}= \alpha (x,y)$

... and...

$\displaystyle \frac{\partial{f}}{\partial{y}}= \beta(x,y)$

If the conditions of the Schwartz theorem are satisfied, the a sufficient condition for the existence of f(x, y) is...

$\displaystyle \frac{\partial{\alpha}}{\partial{y}} = \frac{\partial{\beta}}{\partial{x}}\ (2)$

Kind regards$\chi$ $\sigma$

 

[Fantini]

If you don't mind me here consider $$\color{black}4xy^3 \, dx + (6x^2y^2 \color{red}+x^2 \color{black}) \, dy.$$ Then $$\frac{\partial}{\partial y} 4xy^3 = 12xy^2 \text{ and } \frac{\partial}{\partial x} 6x^2y^2 +x^2 = 12xy^2 + 2x,$$ therefore this is not an exact differential.