Problem 02 | Exact Equations

$M = 6x + y^2$

$N = y(2x - 3y) = 2xy - 3y^2$
 

Test for exactness
$\dfrac{\partial M}{\partial y} = 2y$

$\dfrac{\partial N}{\partial x} = 2y$

Exact!
 

Let
$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = 6x + y^2$

$\partial F = (6x + y^2) \, \partial x$
 

Integrate partially in x, holding y as constant
$\displaystyle \int \partial F = \int (6x + y^2) \, \partial x$

$F = 3x^2 + xy^2 + f(y)$       →   Equation (1)
 

Differentiate partially in y, holding x as constant
$\dfrac{\partial F}{\partial y} = 2xy + f'(y)$
 

Let
$\dfrac{\partial F}{\partial y} = N$

$2xy + f'(y) = 2xy - 3y^2$

$f'(y) = -3y^2$
 

Integrate partially in y, holding x as constant
$\displaystyle \int f'(y) = -3 \int y^2 \, \partial y$

$f(y) = -y^3$
 

Substitute f(y) to Equation (1)
$F = 3x^2 + xy^2 - y^3$
 

Equate F to c
$F = c$

$3x^2 + xy^2 - y^3 = c$           answer