# Problem 02 | Exact Equations

**Problem 02**

$(6x + y^2) \, dx + y(2x - 3y) \, dy = 0$

**Solution 02**

$M = 6x + y^2$

$N = y(2x - 3y) = 2xy - 3y^2$

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*

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