# Exact equations

1. Sep 17, 2010

### renathy

I have the following exact equations, however, teacher said it is incorrect. Cannot find a mistake. Could you please help me?

(3x^2 - 2x - y) dx + (2y - x + 3y^2)dy = 0

This is exact equation, because:
P = (3x^2 - 2x - y)
Q = (2y - x + 3y^2)
P'y = Q'x = -1

Then integrate Intx P = x^3 - x^2 - yx + fi(y).

Find derivative: (x^3 - x^2 - yx + fi(y)'y = -x+ fi'(y) = Q = (2y - x + 3y^2).
Find fi'(y) = 2y + 3y^2. Iegūst fi(y) = y^3 + y^2.

Result: x^3 - x^2 - yx + y^3 + y^2 + C.

Where is a mistake? Mabybe the whole idea is incorrect? Please, help me.

Last edited: Sep 17, 2010
2. Sep 17, 2010

### Eynstone

I don't see any mistakes.

3. Sep 17, 2010

### HallsofIvy

A better notation of "the derivative with respect to y" than ( )'y is ()_y. The former is too likely to be confused with "the derivative times y".

The mistake is that you have an expression but no function! You need to write
$x^2- x^2- yx+ y^3+ y^2= C$
or
$x^2- x^2- yx+ y^3+ y^2+ C= 0$
(or any other constant on the right side.)
Those are now "implicit functions" that could, theoretically, be solve for x or y. Just the expression $x^3- x^2- yx+ y^3+ y^2+ C$ is not. Your teacher is being very strict but you need to learn to be precise in mathematics.