Exact Equations
Additional Examples
- We write the equation in the standard form,
M dx + N dy = 0 . $$ (-4xy + 5y + 4x) dx + (-6y^2 - 2x^2 - 8y + 5x + 9) dy = 0 $$ - We test for exactness. $$\frac{\partial}{\partial y}\left(-4xy + 5y + 4x\right) = -4x + 5 = \frac{\partial}{\partial x}\left(-6y^2 - 2x^2 - 8y + 5x + 9\right) $$ so the equation is exact.
- Write the partial differential equations. $$ \begin{align} \frac{\partial F}{\partial x} &= -4xy + 5y + 4x\\ \frac{\partial F}{\partial y} &= -6y^2 - 2x^2 - 8y + 5x + 9 \end{align}$$
- Integrate the first partial differential equation. $$ F(x,y) = \int (-4xy + 5y + 4x)\,\partial x = -2x^2y + 5xy + 2x^2 + C(y) $$
- Integrate the second partial differential equation. $$ F(x,y) = \int (-6y^2 - 2x^2 - 8y + 5x + 9)\,\partial y = -2y^3 - 2x^2y - 4y^2 + 5xy + 9y + \tilde{C}(x) $$
- Equate the expressions for F(x,y). Matching the expressions up, we find $C(y) = -2y^3 - 4y^2 + 9y$ and $ \tilde{C}(x) = 2x^2. $ So $$ F(x,y) = -2x^2y - 2y^3 + 5xy + 2x^2 - 4y^2 + 9y. $$
- The solution is $F(x,y) = K.$ $$ -2x^2y - 2y^3 + 5xy + 2x^2 - 4y^2 + 9y = K $$
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©2010, 2014 Andrew G. Bennett